Why is $frac{987654321}{123456789} = 8.0000000729?!$
$begingroup$
Many years ago,
I noticed that $987654321/123456789 = 8.0000000729ldots$.
I sent it in to Martin Gardner at Scientific American
and he published it in his column!!!
My life has gone downhill since then:)
My questions are:
Why is this so?
What happens beyond the "$729$"?
What happens in bases other than $10$?
arithmetic rational-numbers
edited Mar 23 '17 at 10:56
Leila
3,49553156
asked May 19 '13 at 7:06
marty cohenmarty cohen
74.1k549128
$endgroup$
|
show 13 more comments
$begingroup$
Many years ago,
I noticed that $987654321/123456789 = 8.0000000729ldots$.
I sent it in to Martin Gardner at Scientific American
and he published it in his column!!!
My life has gone downhill since then:)
My questions are:
Why is this so?
What happens beyond the "$729$"?
What happens in bases other than $10$?
arithmetic rational-numbers
edited Mar 23 '17 at 10:56
Leila
3,49553156
asked May 19 '13 at 7:06
marty cohenmarty cohen
74.1k549128
$endgroup$
13
$begingroup$
8.000000072900000663390006036849054935326399911470239194379176...
$endgroup$
– copper.hat
May 19 '13 at 7:10
116
$begingroup$
I don't see it, why is this number interesting?
$endgroup$
– Matsemann
May 19 '13 at 9:43
20
$begingroup$
8.0000000729000006633900060368490549353263999114702391943791766688505076865396199475105415223459278533479434654662855357431983752631052148942574555377428453934598930804850270324137459949650885541823058430589831718367468637143964598010077841891708361214546087052369392176561468806709366141055231883602610140783752281132145758302526400552990245032211229793122191117411939168448646432882682539232411107014941073835963771907270324356159951641055559933605595395810918101879354727102128016629364951327221057077711619407175736605299203108222748284827009391925785466524647745374294482906079794445326129452467
$endgroup$
– Double AA
May 19 '13 at 10:29
15
$begingroup$
Unrelated, but ever notice that Sqrt(9.87654321) approximates pi? The number system is full of this kind of stuff.
$endgroup$
– cobaltduck
May 19 '13 at 11:44
23
$begingroup$
@cobaltduck: It is a terrible approximation given how many digits you're putting in. It has more than $10$ times the error of $sqrt{9.87}$
$endgroup$
– Jonas Meyer
May 19 '13 at 18:28
|
show 13 more comments
$begingroup$
Many years ago,
I noticed that $987654321/123456789 = 8.0000000729ldots$.
I sent it in to Martin Gardner at Scientific American
and he published it in his column!!!
My life has gone downhill since then:)
My questions are:
Why is this so?
What happens beyond the "$729$"?
What happens in bases other than $10$?
arithmetic rational-numbers
edited Mar 23 '17 at 10:56
Leila
3,49553156
asked May 19 '13 at 7:06
marty cohenmarty cohen
74.1k549128
$endgroup$
Many years ago,
I noticed that $987654321/123456789 = 8.0000000729ldots$.
I sent it in to Martin Gardner at Scientific American
and he published it in his column!!!
My life has gone downhill since then:)
My questions are:
Why is this so?
What happens beyond the "$729$"?
What happens in bases other than $10$?
arithmetic rational-numbers
arithmetic rational-numbers
edited Mar 23 '17 at 10:56
Leila
3,49553156
asked May 19 '13 at 7:06
marty cohenmarty cohen
74.1k549128
edited Mar 23 '17 at 10:56
Leila
3,49553156
asked May 19 '13 at 7:06
marty cohenmarty cohen
74.1k549128
edited Mar 23 '17 at 10:56
Leila
3,49553156
edited Mar 23 '17 at 10:56
Leila
3,49553156
edited Mar 23 '17 at 10:56
Leila
3,49553156
3,49553156
asked May 19 '13 at 7:06
marty cohenmarty cohen
74.1k549128
asked May 19 '13 at 7:06
marty cohenmarty cohen
74.1k549128
asked May 19 '13 at 7:06
marty cohenmarty cohen
74.1k549128
74.1k549128
13
$begingroup$
8.000000072900000663390006036849054935326399911470239194379176...
$endgroup$
– copper.hat
May 19 '13 at 7:10
116
$begingroup$
I don't see it, why is this number interesting?
$endgroup$
– Matsemann
May 19 '13 at 9:43
20
$begingroup$
8.0000000729000006633900060368490549353263999114702391943791766688505076865396199475105415223459278533479434654662855357431983752631052148942574555377428453934598930804850270324137459949650885541823058430589831718367468637143964598010077841891708361214546087052369392176561468806709366141055231883602610140783752281132145758302526400552990245032211229793122191117411939168448646432882682539232411107014941073835963771907270324356159951641055559933605595395810918101879354727102128016629364951327221057077711619407175736605299203108222748284827009391925785466524647745374294482906079794445326129452467
$endgroup$
– Double AA
May 19 '13 at 10:29
15
$begingroup$
Unrelated, but ever notice that Sqrt(9.87654321) approximates pi? The number system is full of this kind of stuff.
$endgroup$
– cobaltduck
May 19 '13 at 11:44
23
$begingroup$
@cobaltduck: It is a terrible approximation given how many digits you're putting in. It has more than $10$ times the error of $sqrt{9.87}$
$endgroup$
– Jonas Meyer
May 19 '13 at 18:28
|
show 13 more comments
13
$begingroup$
8.000000072900000663390006036849054935326399911470239194379176...
$endgroup$
– copper.hat
May 19 '13 at 7:10
116
$begingroup$
I don't see it, why is this number interesting?
$endgroup$
– Matsemann
May 19 '13 at 9:43
20
$begingroup$
8.0000000729000006633900060368490549353263999114702391943791766688505076865396199475105415223459278533479434654662855357431983752631052148942574555377428453934598930804850270324137459949650885541823058430589831718367468637143964598010077841891708361214546087052369392176561468806709366141055231883602610140783752281132145758302526400552990245032211229793122191117411939168448646432882682539232411107014941073835963771907270324356159951641055559933605595395810918101879354727102128016629364951327221057077711619407175736605299203108222748284827009391925785466524647745374294482906079794445326129452467
$endgroup$
– Double AA
May 19 '13 at 10:29
15
$begingroup$
Unrelated, but ever notice that Sqrt(9.87654321) approximates pi? The number system is full of this kind of stuff.
$endgroup$
– cobaltduck
May 19 '13 at 11:44
23
$begingroup$
@cobaltduck: It is a terrible approximation given how many digits you're putting in. It has more than $10$ times the error of $sqrt{9.87}$
$endgroup$
– Jonas Meyer
May 19 '13 at 18:28
13
13
$begingroup$
8.000000072900000663390006036849054935326399911470239194379176...
$endgroup$
– copper.hat
May 19 '13 at 7:10
$begingroup$
8.000000072900000663390006036849054935326399911470239194379176...
$endgroup$
– copper.hat
May 19 '13 at 7:10
116
116
$begingroup$
I don't see it, why is this number interesting?
$endgroup$
– Matsemann
May 19 '13 at 9:43
$begingroup$
I don't see it, why is this number interesting?
$endgroup$
– Matsemann
May 19 '13 at 9:43
20
20
$begingroup$
8.0000000729000006633900060368490549353263999114702391943791766688505076865396199475105415223459278533479434654662855357431983752631052148942574555377428453934598930804850270324137459949650885541823058430589831718367468637143964598010077841891708361214546087052369392176561468806709366141055231883602610140783752281132145758302526400552990245032211229793122191117411939168448646432882682539232411107014941073835963771907270324356159951641055559933605595395810918101879354727102128016629364951327221057077711619407175736605299203108222748284827009391925785466524647745374294482906079794445326129452467
$endgroup$
– Double AA
May 19 '13 at 10:29
$begingroup$
8.0000000729000006633900060368490549353263999114702391943791766688505076865396199475105415223459278533479434654662855357431983752631052148942574555377428453934598930804850270324137459949650885541823058430589831718367468637143964598010077841891708361214546087052369392176561468806709366141055231883602610140783752281132145758302526400552990245032211229793122191117411939168448646432882682539232411107014941073835963771907270324356159951641055559933605595395810918101879354727102128016629364951327221057077711619407175736605299203108222748284827009391925785466524647745374294482906079794445326129452467
$endgroup$
– Double AA
May 19 '13 at 10:29
15
15
$begingroup$
Unrelated, but ever notice that Sqrt(9.87654321) approximates pi? The number system is full of this kind of stuff.
$endgroup$
– cobaltduck
May 19 '13 at 11:44
$begingroup$
Unrelated, but ever notice that Sqrt(9.87654321) approximates pi? The number system is full of this kind of stuff.
$endgroup$
– cobaltduck
May 19 '13 at 11:44
23
23
$begingroup$
@cobaltduck: It is a terrible approximation given how many digits you're putting in. It has more than $10$ times the error of $sqrt{9.87}$
$endgroup$
– Jonas Meyer
May 19 '13 at 18:28
$begingroup$
@cobaltduck: It is a terrible approximation given how many digits you're putting in. It has more than $10$ times the error of $sqrt{9.87}$
$endgroup$
– Jonas Meyer
May 19 '13 at 18:28
|
show 13 more comments
7 Answers
7
active
oldest
votes
$begingroup$
In base $n$ the numerator is $$p = n^{n-1} - frac{n^{n-1}-1}{(n-1)^2}$$ and the denominator is $$q = frac{n(n^{n-1}-1)}{(n-1)^2}-1.$$
Note that $p = (n-2)q + n-1$ and for the quotient we get
$$
frac{p}{q} = n-2 + frac{(n-1)^3}{n^n} frac{1}{1 - frac{n^2-n+1}{n^n}} = n-2 + frac{(n-1)^3}{n^n} sum_{k=0}^{infty} left(frac{n^2-n+1}{n^n}right)^k.
$$
Indeed for $n=10$ this is
$$frac{987654321}{123456789} = 8 + frac{729}{10^{10}}sum_{k=0}^{infty}left(frac{91}{10^{10}}right)^k
$$
answered May 19 '13 at 8:35
WimCWimC
24.2k23063
$endgroup$
37
$begingroup$
You and Frederica would make some scary babies...
$endgroup$
– Sheljohn
May 19 '13 at 8:44
42
$begingroup$
@Sh3ljohn So you know how I look... :-(
$endgroup$
– WimC
May 19 '13 at 8:45
2
$begingroup$
so methodical. Neat.
$endgroup$
– oldrinb
May 19 '13 at 13:52
2
$begingroup$
@Sh3ljohn It's "Federica", there is no $R$ after $F$.
$endgroup$
– Billy Rubina
Sep 23 '14 at 23:34
add a comment |
$begingroup$
$$729=9^3$$
$$66339=9^3cdot 91$$
$$6036849=9^3cdot 91^2$$
$$...$$
$$987654321/123456789=8+9^3cdot 10^{-10}cdotdisplaystylesum_{n=0}^{infty}(91cdot 10^{-10})^n$$
answered May 19 '13 at 7:24
Federica MaggioniFederica Maggioni
5,97111336
$endgroup$
8
$begingroup$
@FedericaMaggioni Please tell me you didn't just "find" that...!
$endgroup$
– Sheljohn
May 19 '13 at 7:40
36
$begingroup$
Could you add a bit more detail?
$endgroup$
– nbubis
May 19 '13 at 8:08
8
$begingroup$
Is it just me who is too stupid to understand what the pattern is?
$endgroup$
– Parth Kohli
May 20 '13 at 10:11
1
$begingroup$
@ΠάρτηΚοχλί look at Double AA's expansion in the comments above..
$endgroup$
– Federica Maggioni
May 20 '13 at 10:26
1
$begingroup$
Ah, gotcha. Nice answer.
$endgroup$
– Parth Kohli
May 20 '13 at 11:27
|
show 2 more comments
$begingroup$
Let
$$S_n(a)=1 +2a+ldots +na^{n-1}=frac{na^{n+1}-(n+1)a^n+1}{(a-1)^2},$$
$$T_n(a)=a^{n-1}+2a^{n-2}+ldots +n=a^{n-1}S_n(a^{-1}).$$
Then
$$
frac{S_n(a)}{T_n(a)}=frac{na^{n+1}-(n+1)a^n+1}{a^{n+1}-(n+1)a+n}.$$
For $a=10,n=9$ we have
$$
frac{S_n(a)}{T_n(a)}approxfrac{8cdot 10^{10}+1}{10^{10}}.
$$
answered May 19 '13 at 7:46
Boris NovikovBoris Novikov
16.1k11529
$endgroup$
add a comment |
$begingroup$
Just to add to the excellent answers above, some examples:
${987654321,/,123456789}approx 8.00000007290000066339$
${{87654321}_9,/,{12345678}_9}approx {7.000000628000056238}_9$
${{7654321}_8,/,{1234567}_8}approx {6.0000052700046137}_8$
${{654321}_7,/,{123456}_7}approx {5.00004260036036}_7$
${{mathrm{fedcba987654321}}_{16},/,{mathrm{123456789abcdef}}_{16}}approx {mathrm{e.0000000000000d2f00000000000c693f}}_{16}$
answered May 19 '13 at 17:14
Mark AdlerMark Adler
919713
$endgroup$
$begingroup$
And what do these numbers show??
$endgroup$
– Matsemann
May 21 '13 at 19:04
8
$begingroup$
They show empirically that the behavior seen in base 10 is present for all bases.
$endgroup$
– Mark Adler
May 21 '13 at 20:13
2
$begingroup$
And what behavior is that? I asked in a comment to the question what's so interesting about this, no one has answered.
$endgroup$
– Matsemann
Jun 13 '13 at 8:59
9
$begingroup$
The behavior is that the result is extremely close to the base minus two. Like a joke, if it has to be explained to you, it won't be funny. If you weren't intrigued by this when you first saw it, then you probably never will be.
$endgroup$
– Mark Adler
Jun 13 '13 at 15:00
$begingroup$
No, they show empirically that the behavior seen in base 10 is present for bases 9, 8, 7 and 16.
$endgroup$
– jwg
Jun 28 '17 at 10:09
add a comment |
$begingroup$
$98765432 / 12345679 = 8$, exactly. You can see how the pattern works by multiplying out $12345679 * 8$ starting at either end.
This explains why your fraction is close to an integer. If you think the $729$ is interesting (I don't), it can be explained by some of the other answers here.
Edit:
What can we say about the fact that $12345679 * 8 = 98765432$? I have been aware of this 'factlet' for about 20 years, and remember it being used to 'demonstrate' calculators (which often had 8 digit displays back in the day).
I just recently realised that:
$$
frac{1}{81} = left(frac{1}{9}right)^2 = left(sum_{k=1}^{infty}frac{1}{10^k}right)^2 = sum_{k=1}^{infty} sum_{m=1}^{k-1} frac{1}{10^m} frac{1}{10^{k-m}} = sum_{k=1}^{infty} frac{k-1}{10^k}
$$
In other words, while $frac{1}{9} = 0.1111111ldots$
$$
frac{1}{81} = 0.01 + 0.002 + 0.0003 + 0.0004 + 0.00005 ldots
$$
It is pretty easy to see that this infinite sum is going to converge to something starting $0.012345ldots$. If you keep on adding, or work out $frac{1}{81}$ by division, you get
$$
0.012345679012345679012345679ldots
$$
When you get to the point where you add $frac{10}{10^{11}}$, the first carry happens, which leads to the 9 where you might expect an 8. After that every addition carries and the decimal expansion repeats every 9 digits (not every ten - because the amount we carry keeps on getting bigger and bigger).
Now, $frac{8}{81} = frac{9}{81} - frac{1}{81}$, or
$$
frac{8}{81} = 0.11111111ldots - 0.012345679012345ldots
$$
Think of each '1' digit in $0.111ldots$ as being a '10' in the next column. This means that we can work out $frac{8}{81}$ as the "10's complement" of $frac{1}{81}$, since we are subtracting a digit between $1$ and $9$ from $10$, to get another single digit which appears in the same place. So $frac{8}{81}$ starts $0.098765ldots$. The only break in the pattern is when you get to the digit '0' - subtracting 0 from 10 leaves you with 10, or a '1' in the next digit on the left, changing the 1 to a 2.
So
$$
frac{8}{81} = 0.098765432098765432098765ldots
$$
and therefore
$$
0.0123456790123456790ldots * 8 = 0.0987654320987654320ldots
$$
and clearly this gets you that
$$
12345679 * 8 = 98765432
$$
answered May 20 '13 at 10:01
jwgjwg
2,3471627
$endgroup$
$begingroup$
A bit late but notetimes($times$) instead of *.
$endgroup$
– TheSimpliFire
Apr 4 '18 at 18:01
add a comment |
$begingroup$
I think @BorisNovikov's answer is the best here, but I had to do some work to understand it; this is my attempt at clarifying his answer, hopefully this can help others too:
In the following, we adopt the notation $123456789 = [ 1, 2, ..., B-1 ]_B$ in base $B=10$.
Boris introduced the following sum:
$$
S_d(B) = [d, d-1, ..., 1]_B = 1 + 2B + ... + dB^{d-1} = sum_{k=1}^d kB^{k-1} = frac{dB^{d+1} - (d+1)B^d + 1}{(B-1)^2}
$$
where $d$ is the number of digits, $B$ the base, and the final formula is obtained by derivation of the usual geometric sum.
From there, it is easy to see that:
$$
[1, 2, ..., d]_B = d + (d-1)B + ... + 1B^{d-1} = B^{d-1} S_d(1/B)
$$
and putting both results together, we have:
$$
frac{ [B-1, B-2, ..., 1]_B }{ [1, 2, ..., B-1]_B } = frac{ S_{B-1}(B) }{ B^{B-2} S_{B-1}(1/B) } = frac{1 + (B-2)B^B}{B^B - B(B-1) - 1}
$$
Clearly both the numerator and denominator are dominated by the terms in $B^B$, so this fraction is equivalent to $B-2$ as $B$ tends to infinity, and we can see that the exact value is very close already for $B=10$ (see also @Mark Adler's post for other examples). Beautiful, isn't it? :)
answered May 31 '18 at 15:08
SheljohnSheljohn
1,276723
$endgroup$
add a comment |
$begingroup$
I think one should also think about how this is related to 9 being before 10. Notice the two following striking facts:
$12345679*8=98765432$ and $12345679*9=111111111$
answered Sep 8 '18 at 0:26
Sergio LuceroSergio Lucero
1115
$endgroup$
$begingroup$
See @jwg's answer.
$endgroup$
– Sheljohn
Jan 1 at 14:07
add a comment |
protected by Community♦ May 22 '13 at 13:20
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?
7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In base $n$ the numerator is $$p = n^{n-1} - frac{n^{n-1}-1}{(n-1)^2}$$ and the denominator is $$q = frac{n(n^{n-1}-1)}{(n-1)^2}-1.$$
Note that $p = (n-2)q + n-1$ and for the quotient we get
$$
frac{p}{q} = n-2 + frac{(n-1)^3}{n^n} frac{1}{1 - frac{n^2-n+1}{n^n}} = n-2 + frac{(n-1)^3}{n^n} sum_{k=0}^{infty} left(frac{n^2-n+1}{n^n}right)^k.
$$
Indeed for $n=10$ this is
$$frac{987654321}{123456789} = 8 + frac{729}{10^{10}}sum_{k=0}^{infty}left(frac{91}{10^{10}}right)^k
$$
answered May 19 '13 at 8:35
WimCWimC
24.2k23063
$endgroup$
37
$begingroup$
You and Frederica would make some scary babies...
$endgroup$
– Sheljohn
May 19 '13 at 8:44
42
$begingroup$
@Sh3ljohn So you know how I look... :-(
$endgroup$
– WimC
May 19 '13 at 8:45
2
$begingroup$
so methodical. Neat.
$endgroup$
– oldrinb
May 19 '13 at 13:52
2
$begingroup$
@Sh3ljohn It's "Federica", there is no $R$ after $F$.
$endgroup$
– Billy Rubina
Sep 23 '14 at 23:34
add a comment |
$begingroup$
In base $n$ the numerator is $$p = n^{n-1} - frac{n^{n-1}-1}{(n-1)^2}$$ and the denominator is $$q = frac{n(n^{n-1}-1)}{(n-1)^2}-1.$$
Note that $p = (n-2)q + n-1$ and for the quotient we get
$$
frac{p}{q} = n-2 + frac{(n-1)^3}{n^n} frac{1}{1 - frac{n^2-n+1}{n^n}} = n-2 + frac{(n-1)^3}{n^n} sum_{k=0}^{infty} left(frac{n^2-n+1}{n^n}right)^k.
$$
Indeed for $n=10$ this is
$$frac{987654321}{123456789} = 8 + frac{729}{10^{10}}sum_{k=0}^{infty}left(frac{91}{10^{10}}right)^k
$$
answered May 19 '13 at 8:35
WimCWimC
24.2k23063
$endgroup$
37
$begingroup$
You and Frederica would make some scary babies...
$endgroup$
– Sheljohn
May 19 '13 at 8:44
42
$begingroup$
@Sh3ljohn So you know how I look... :-(
$endgroup$
– WimC
May 19 '13 at 8:45
2
$begingroup$
so methodical. Neat.
$endgroup$
– oldrinb
May 19 '13 at 13:52
2
$begingroup$
@Sh3ljohn It's "Federica", there is no $R$ after $F$.
$endgroup$
– Billy Rubina
Sep 23 '14 at 23:34
add a comment |
$begingroup$
In base $n$ the numerator is $$p = n^{n-1} - frac{n^{n-1}-1}{(n-1)^2}$$ and the denominator is $$q = frac{n(n^{n-1}-1)}{(n-1)^2}-1.$$
Note that $p = (n-2)q + n-1$ and for the quotient we get
$$
frac{p}{q} = n-2 + frac{(n-1)^3}{n^n} frac{1}{1 - frac{n^2-n+1}{n^n}} = n-2 + frac{(n-1)^3}{n^n} sum_{k=0}^{infty} left(frac{n^2-n+1}{n^n}right)^k.
$$
Indeed for $n=10$ this is
$$frac{987654321}{123456789} = 8 + frac{729}{10^{10}}sum_{k=0}^{infty}left(frac{91}{10^{10}}right)^k
$$
answered May 19 '13 at 8:35
WimCWimC
24.2k23063
$endgroup$
In base $n$ the numerator is $$p = n^{n-1} - frac{n^{n-1}-1}{(n-1)^2}$$ and the denominator is $$q = frac{n(n^{n-1}-1)}{(n-1)^2}-1.$$
Note that $p = (n-2)q + n-1$ and for the quotient we get
$$
frac{p}{q} = n-2 + frac{(n-1)^3}{n^n} frac{1}{1 - frac{n^2-n+1}{n^n}} = n-2 + frac{(n-1)^3}{n^n} sum_{k=0}^{infty} left(frac{n^2-n+1}{n^n}right)^k.
$$
Indeed for $n=10$ this is
$$frac{987654321}{123456789} = 8 + frac{729}{10^{10}}sum_{k=0}^{infty}left(frac{91}{10^{10}}right)^k
$$
answered May 19 '13 at 8:35
WimCWimC
24.2k23063
answered May 19 '13 at 8:35
WimCWimC
24.2k23063
answered May 19 '13 at 8:35
WimCWimC
24.2k23063
answered May 19 '13 at 8:35
WimCWimC
24.2k23063
24.2k23063
37
$begingroup$
You and Frederica would make some scary babies...
$endgroup$
– Sheljohn
May 19 '13 at 8:44
42
$begingroup$
@Sh3ljohn So you know how I look... :-(
$endgroup$
– WimC
May 19 '13 at 8:45
2
$begingroup$
so methodical. Neat.
$endgroup$
– oldrinb
May 19 '13 at 13:52
2
$begingroup$
@Sh3ljohn It's "Federica", there is no $R$ after $F$.
$endgroup$
– Billy Rubina
Sep 23 '14 at 23:34
add a comment |
37
$begingroup$
You and Frederica would make some scary babies...
$endgroup$
– Sheljohn
May 19 '13 at 8:44
42
$begingroup$
@Sh3ljohn So you know how I look... :-(
$endgroup$
– WimC
May 19 '13 at 8:45
2
$begingroup$
so methodical. Neat.
$endgroup$
– oldrinb
May 19 '13 at 13:52
2
$begingroup$
@Sh3ljohn It's "Federica", there is no $R$ after $F$.
$endgroup$
– Billy Rubina
Sep 23 '14 at 23:34
37
37
$begingroup$
You and Frederica would make some scary babies...
$endgroup$
– Sheljohn
May 19 '13 at 8:44
$begingroup$
You and Frederica would make some scary babies...
$endgroup$
– Sheljohn
May 19 '13 at 8:44
42
42
$begingroup$
@Sh3ljohn So you know how I look... :-(
$endgroup$
– WimC
May 19 '13 at 8:45
$begingroup$
@Sh3ljohn So you know how I look... :-(
$endgroup$
– WimC
May 19 '13 at 8:45
2
2
$begingroup$
so methodical. Neat.
$endgroup$
– oldrinb
May 19 '13 at 13:52
$begingroup$
so methodical. Neat.
$endgroup$
– oldrinb
May 19 '13 at 13:52
2
2
$begingroup$
@Sh3ljohn It's "Federica", there is no $R$ after $F$.
$endgroup$
– Billy Rubina
Sep 23 '14 at 23:34
$begingroup$
@Sh3ljohn It's "Federica", there is no $R$ after $F$.
$endgroup$
– Billy Rubina
Sep 23 '14 at 23:34
add a comment |
$begingroup$
$$729=9^3$$
$$66339=9^3cdot 91$$
$$6036849=9^3cdot 91^2$$
$$...$$
$$987654321/123456789=8+9^3cdot 10^{-10}cdotdisplaystylesum_{n=0}^{infty}(91cdot 10^{-10})^n$$
answered May 19 '13 at 7:24
Federica MaggioniFederica Maggioni
5,97111336
$endgroup$
8
$begingroup$
@FedericaMaggioni Please tell me you didn't just "find" that...!
$endgroup$
– Sheljohn
May 19 '13 at 7:40
36
$begingroup$
Could you add a bit more detail?
$endgroup$
– nbubis
May 19 '13 at 8:08
8
$begingroup$
Is it just me who is too stupid to understand what the pattern is?
$endgroup$
– Parth Kohli
May 20 '13 at 10:11
1
$begingroup$
@ΠάρτηΚοχλί look at Double AA's expansion in the comments above..
$endgroup$
– Federica Maggioni
May 20 '13 at 10:26
1
$begingroup$
Ah, gotcha. Nice answer.
$endgroup$
– Parth Kohli
May 20 '13 at 11:27
|
show 2 more comments
$begingroup$
$$729=9^3$$
$$66339=9^3cdot 91$$
$$6036849=9^3cdot 91^2$$
$$...$$
$$987654321/123456789=8+9^3cdot 10^{-10}cdotdisplaystylesum_{n=0}^{infty}(91cdot 10^{-10})^n$$
answered May 19 '13 at 7:24
Federica MaggioniFederica Maggioni
5,97111336
$endgroup$
8
$begingroup$
@FedericaMaggioni Please tell me you didn't just "find" that...!
$endgroup$
– Sheljohn
May 19 '13 at 7:40
36
$begingroup$
Could you add a bit more detail?
$endgroup$
– nbubis
May 19 '13 at 8:08
8
$begingroup$
Is it just me who is too stupid to understand what the pattern is?
$endgroup$
– Parth Kohli
May 20 '13 at 10:11
1
$begingroup$
@ΠάρτηΚοχλί look at Double AA's expansion in the comments above..
$endgroup$
– Federica Maggioni
May 20 '13 at 10:26
1
$begingroup$
Ah, gotcha. Nice answer.
$endgroup$
– Parth Kohli
May 20 '13 at 11:27
|
show 2 more comments
$begingroup$
$$729=9^3$$
$$66339=9^3cdot 91$$
$$6036849=9^3cdot 91^2$$
$$...$$
$$987654321/123456789=8+9^3cdot 10^{-10}cdotdisplaystylesum_{n=0}^{infty}(91cdot 10^{-10})^n$$
answered May 19 '13 at 7:24
Federica MaggioniFederica Maggioni
5,97111336
$endgroup$
$$729=9^3$$
$$66339=9^3cdot 91$$
$$6036849=9^3cdot 91^2$$
$$...$$
$$987654321/123456789=8+9^3cdot 10^{-10}cdotdisplaystylesum_{n=0}^{infty}(91cdot 10^{-10})^n$$
answered May 19 '13 at 7:24
Federica MaggioniFederica Maggioni
5,97111336
answered May 19 '13 at 7:24
Federica MaggioniFederica Maggioni
5,97111336
answered May 19 '13 at 7:24
Federica MaggioniFederica Maggioni
5,97111336
answered May 19 '13 at 7:24
Federica MaggioniFederica Maggioni
5,97111336
5,97111336
8
$begingroup$
@FedericaMaggioni Please tell me you didn't just "find" that...!
$endgroup$
– Sheljohn
May 19 '13 at 7:40
36
$begingroup$
Could you add a bit more detail?
$endgroup$
– nbubis
May 19 '13 at 8:08
8
$begingroup$
Is it just me who is too stupid to understand what the pattern is?
$endgroup$
– Parth Kohli
May 20 '13 at 10:11
1
$begingroup$
@ΠάρτηΚοχλί look at Double AA's expansion in the comments above..
$endgroup$
– Federica Maggioni
May 20 '13 at 10:26
1
$begingroup$
Ah, gotcha. Nice answer.
$endgroup$
– Parth Kohli
May 20 '13 at 11:27
|
show 2 more comments
8
$begingroup$
@FedericaMaggioni Please tell me you didn't just "find" that...!
$endgroup$
– Sheljohn
May 19 '13 at 7:40
36
$begingroup$
Could you add a bit more detail?
$endgroup$
– nbubis
May 19 '13 at 8:08
8
$begingroup$
Is it just me who is too stupid to understand what the pattern is?
$endgroup$
– Parth Kohli
May 20 '13 at 10:11
1
$begingroup$
@ΠάρτηΚοχλί look at Double AA's expansion in the comments above..
$endgroup$
– Federica Maggioni
May 20 '13 at 10:26
1
$begingroup$
Ah, gotcha. Nice answer.
$endgroup$
– Parth Kohli
May 20 '13 at 11:27
8
8
$begingroup$
@FedericaMaggioni Please tell me you didn't just "find" that...!
$endgroup$
– Sheljohn
May 19 '13 at 7:40
$begingroup$
@FedericaMaggioni Please tell me you didn't just "find" that...!
$endgroup$
– Sheljohn
May 19 '13 at 7:40
36
36
$begingroup$
Could you add a bit more detail?
$endgroup$
– nbubis
May 19 '13 at 8:08
$begingroup$
Could you add a bit more detail?
$endgroup$
– nbubis
May 19 '13 at 8:08
8
8
$begingroup$
Is it just me who is too stupid to understand what the pattern is?
$endgroup$
– Parth Kohli
May 20 '13 at 10:11
$begingroup$
Is it just me who is too stupid to understand what the pattern is?
$endgroup$
– Parth Kohli
May 20 '13 at 10:11
1
1
$begingroup$
@ΠάρτηΚοχλί look at Double AA's expansion in the comments above..
$endgroup$
– Federica Maggioni
May 20 '13 at 10:26
$begingroup$
@ΠάρτηΚοχλί look at Double AA's expansion in the comments above..
$endgroup$
– Federica Maggioni
May 20 '13 at 10:26
1
1
$begingroup$
Ah, gotcha. Nice answer.
$endgroup$
– Parth Kohli
May 20 '13 at 11:27
$begingroup$
Ah, gotcha. Nice answer.
$endgroup$
– Parth Kohli
May 20 '13 at 11:27
|
show 2 more comments
$begingroup$
Let
$$S_n(a)=1 +2a+ldots +na^{n-1}=frac{na^{n+1}-(n+1)a^n+1}{(a-1)^2},$$
$$T_n(a)=a^{n-1}+2a^{n-2}+ldots +n=a^{n-1}S_n(a^{-1}).$$
Then
$$
frac{S_n(a)}{T_n(a)}=frac{na^{n+1}-(n+1)a^n+1}{a^{n+1}-(n+1)a+n}.$$
For $a=10,n=9$ we have
$$
frac{S_n(a)}{T_n(a)}approxfrac{8cdot 10^{10}+1}{10^{10}}.
$$
answered May 19 '13 at 7:46
Boris NovikovBoris Novikov
16.1k11529
$endgroup$
add a comment |
$begingroup$
Let
$$S_n(a)=1 +2a+ldots +na^{n-1}=frac{na^{n+1}-(n+1)a^n+1}{(a-1)^2},$$
$$T_n(a)=a^{n-1}+2a^{n-2}+ldots +n=a^{n-1}S_n(a^{-1}).$$
Then
$$
frac{S_n(a)}{T_n(a)}=frac{na^{n+1}-(n+1)a^n+1}{a^{n+1}-(n+1)a+n}.$$
For $a=10,n=9$ we have
$$
frac{S_n(a)}{T_n(a)}approxfrac{8cdot 10^{10}+1}{10^{10}}.
$$
answered May 19 '13 at 7:46
Boris NovikovBoris Novikov
16.1k11529
$endgroup$
add a comment |
$begingroup$
Let
$$S_n(a)=1 +2a+ldots +na^{n-1}=frac{na^{n+1}-(n+1)a^n+1}{(a-1)^2},$$
$$T_n(a)=a^{n-1}+2a^{n-2}+ldots +n=a^{n-1}S_n(a^{-1}).$$
Then
$$
frac{S_n(a)}{T_n(a)}=frac{na^{n+1}-(n+1)a^n+1}{a^{n+1}-(n+1)a+n}.$$
For $a=10,n=9$ we have
$$
frac{S_n(a)}{T_n(a)}approxfrac{8cdot 10^{10}+1}{10^{10}}.
$$
answered May 19 '13 at 7:46
Boris NovikovBoris Novikov
16.1k11529
$endgroup$
Let
$$S_n(a)=1 +2a+ldots +na^{n-1}=frac{na^{n+1}-(n+1)a^n+1}{(a-1)^2},$$
$$T_n(a)=a^{n-1}+2a^{n-2}+ldots +n=a^{n-1}S_n(a^{-1}).$$
Then
$$
frac{S_n(a)}{T_n(a)}=frac{na^{n+1}-(n+1)a^n+1}{a^{n+1}-(n+1)a+n}.$$
For $a=10,n=9$ we have
$$
frac{S_n(a)}{T_n(a)}approxfrac{8cdot 10^{10}+1}{10^{10}}.
$$
answered May 19 '13 at 7:46
Boris NovikovBoris Novikov
16.1k11529
answered May 19 '13 at 7:46
Boris NovikovBoris Novikov
16.1k11529
answered May 19 '13 at 7:46
Boris NovikovBoris Novikov
16.1k11529
answered May 19 '13 at 7:46
Boris NovikovBoris Novikov
16.1k11529
16.1k11529
add a comment |
add a comment |
$begingroup$
Just to add to the excellent answers above, some examples:
${987654321,/,123456789}approx 8.00000007290000066339$
${{87654321}_9,/,{12345678}_9}approx {7.000000628000056238}_9$
${{7654321}_8,/,{1234567}_8}approx {6.0000052700046137}_8$
${{654321}_7,/,{123456}_7}approx {5.00004260036036}_7$
${{mathrm{fedcba987654321}}_{16},/,{mathrm{123456789abcdef}}_{16}}approx {mathrm{e.0000000000000d2f00000000000c693f}}_{16}$
answered May 19 '13 at 17:14
Mark AdlerMark Adler
919713
$endgroup$
$begingroup$
And what do these numbers show??
$endgroup$
– Matsemann
May 21 '13 at 19:04
8
$begingroup$
They show empirically that the behavior seen in base 10 is present for all bases.
$endgroup$
– Mark Adler
May 21 '13 at 20:13
2
$begingroup$
And what behavior is that? I asked in a comment to the question what's so interesting about this, no one has answered.
$endgroup$
– Matsemann
Jun 13 '13 at 8:59
9
$begingroup$
The behavior is that the result is extremely close to the base minus two. Like a joke, if it has to be explained to you, it won't be funny. If you weren't intrigued by this when you first saw it, then you probably never will be.
$endgroup$
– Mark Adler
Jun 13 '13 at 15:00
$begingroup$
No, they show empirically that the behavior seen in base 10 is present for bases 9, 8, 7 and 16.
$endgroup$
– jwg
Jun 28 '17 at 10:09
add a comment |
$begingroup$
Just to add to the excellent answers above, some examples:
${987654321,/,123456789}approx 8.00000007290000066339$
${{87654321}_9,/,{12345678}_9}approx {7.000000628000056238}_9$
${{7654321}_8,/,{1234567}_8}approx {6.0000052700046137}_8$
${{654321}_7,/,{123456}_7}approx {5.00004260036036}_7$
${{mathrm{fedcba987654321}}_{16},/,{mathrm{123456789abcdef}}_{16}}approx {mathrm{e.0000000000000d2f00000000000c693f}}_{16}$
answered May 19 '13 at 17:14
Mark AdlerMark Adler
919713
$endgroup$
$begingroup$
And what do these numbers show??
$endgroup$
– Matsemann
May 21 '13 at 19:04
8
$begingroup$
They show empirically that the behavior seen in base 10 is present for all bases.
$endgroup$
– Mark Adler
May 21 '13 at 20:13
2
$begingroup$
And what behavior is that? I asked in a comment to the question what's so interesting about this, no one has answered.
$endgroup$
– Matsemann
Jun 13 '13 at 8:59
9
$begingroup$
The behavior is that the result is extremely close to the base minus two. Like a joke, if it has to be explained to you, it won't be funny. If you weren't intrigued by this when you first saw it, then you probably never will be.
$endgroup$
– Mark Adler
Jun 13 '13 at 15:00
$begingroup$
No, they show empirically that the behavior seen in base 10 is present for bases 9, 8, 7 and 16.
$endgroup$
– jwg
Jun 28 '17 at 10:09
add a comment |
$begingroup$
Just to add to the excellent answers above, some examples:
${987654321,/,123456789}approx 8.00000007290000066339$
${{87654321}_9,/,{12345678}_9}approx {7.000000628000056238}_9$
${{7654321}_8,/,{1234567}_8}approx {6.0000052700046137}_8$
${{654321}_7,/,{123456}_7}approx {5.00004260036036}_7$
${{mathrm{fedcba987654321}}_{16},/,{mathrm{123456789abcdef}}_{16}}approx {mathrm{e.0000000000000d2f00000000000c693f}}_{16}$
answered May 19 '13 at 17:14
Mark AdlerMark Adler
919713
$endgroup$
Just to add to the excellent answers above, some examples:
${987654321,/,123456789}approx 8.00000007290000066339$
${{87654321}_9,/,{12345678}_9}approx {7.000000628000056238}_9$
${{7654321}_8,/,{1234567}_8}approx {6.0000052700046137}_8$
${{654321}_7,/,{123456}_7}approx {5.00004260036036}_7$
${{mathrm{fedcba987654321}}_{16},/,{mathrm{123456789abcdef}}_{16}}approx {mathrm{e.0000000000000d2f00000000000c693f}}_{16}$
answered May 19 '13 at 17:14
Mark AdlerMark Adler
919713
edited May 19 '13 at 17:21
answered May 19 '13 at 17:14
Mark AdlerMark Adler
919713
answered May 19 '13 at 17:14
Mark AdlerMark Adler
919713
answered May 19 '13 at 17:14
Mark AdlerMark Adler
919713
919713
$begingroup$
And what do these numbers show??
$endgroup$
– Matsemann
May 21 '13 at 19:04
8
$begingroup$
They show empirically that the behavior seen in base 10 is present for all bases.
$endgroup$
– Mark Adler
May 21 '13 at 20:13
2
$begingroup$
And what behavior is that? I asked in a comment to the question what's so interesting about this, no one has answered.
$endgroup$
– Matsemann
Jun 13 '13 at 8:59
9
$begingroup$
The behavior is that the result is extremely close to the base minus two. Like a joke, if it has to be explained to you, it won't be funny. If you weren't intrigued by this when you first saw it, then you probably never will be.
$endgroup$
– Mark Adler
Jun 13 '13 at 15:00
$begingroup$
No, they show empirically that the behavior seen in base 10 is present for bases 9, 8, 7 and 16.
$endgroup$
– jwg
Jun 28 '17 at 10:09
add a comment |
$begingroup$
And what do these numbers show??
$endgroup$
– Matsemann
May 21 '13 at 19:04
8
$begingroup$
They show empirically that the behavior seen in base 10 is present for all bases.
$endgroup$
– Mark Adler
May 21 '13 at 20:13
2
$begingroup$
And what behavior is that? I asked in a comment to the question what's so interesting about this, no one has answered.
$endgroup$
– Matsemann
Jun 13 '13 at 8:59
9
$begingroup$
The behavior is that the result is extremely close to the base minus two. Like a joke, if it has to be explained to you, it won't be funny. If you weren't intrigued by this when you first saw it, then you probably never will be.
$endgroup$
– Mark Adler
Jun 13 '13 at 15:00
$begingroup$
No, they show empirically that the behavior seen in base 10 is present for bases 9, 8, 7 and 16.
$endgroup$
– jwg
Jun 28 '17 at 10:09
$begingroup$
And what do these numbers show??
$endgroup$
– Matsemann
May 21 '13 at 19:04
$begingroup$
And what do these numbers show??
$endgroup$
– Matsemann
May 21 '13 at 19:04
8
8
$begingroup$
They show empirically that the behavior seen in base 10 is present for all bases.
$endgroup$
– Mark Adler
May 21 '13 at 20:13
$begingroup$
They show empirically that the behavior seen in base 10 is present for all bases.
$endgroup$
– Mark Adler
May 21 '13 at 20:13
2
2
$begingroup$
And what behavior is that? I asked in a comment to the question what's so interesting about this, no one has answered.
$endgroup$
– Matsemann
Jun 13 '13 at 8:59
$begingroup$
And what behavior is that? I asked in a comment to the question what's so interesting about this, no one has answered.
$endgroup$
– Matsemann
Jun 13 '13 at 8:59
9
9
$begingroup$
The behavior is that the result is extremely close to the base minus two. Like a joke, if it has to be explained to you, it won't be funny. If you weren't intrigued by this when you first saw it, then you probably never will be.
$endgroup$
– Mark Adler
Jun 13 '13 at 15:00
$begingroup$
The behavior is that the result is extremely close to the base minus two. Like a joke, if it has to be explained to you, it won't be funny. If you weren't intrigued by this when you first saw it, then you probably never will be.
$endgroup$
– Mark Adler
Jun 13 '13 at 15:00
$begingroup$
No, they show empirically that the behavior seen in base 10 is present for bases 9, 8, 7 and 16.
$endgroup$
– jwg
Jun 28 '17 at 10:09
$begingroup$
No, they show empirically that the behavior seen in base 10 is present for bases 9, 8, 7 and 16.
$endgroup$
– jwg
Jun 28 '17 at 10:09
add a comment |
$begingroup$
$98765432 / 12345679 = 8$, exactly. You can see how the pattern works by multiplying out $12345679 * 8$ starting at either end.
This explains why your fraction is close to an integer. If you think the $729$ is interesting (I don't), it can be explained by some of the other answers here.
Edit:
What can we say about the fact that $12345679 * 8 = 98765432$? I have been aware of this 'factlet' for about 20 years, and remember it being used to 'demonstrate' calculators (which often had 8 digit displays back in the day).
I just recently realised that:
$$
frac{1}{81} = left(frac{1}{9}right)^2 = left(sum_{k=1}^{infty}frac{1}{10^k}right)^2 = sum_{k=1}^{infty} sum_{m=1}^{k-1} frac{1}{10^m} frac{1}{10^{k-m}} = sum_{k=1}^{infty} frac{k-1}{10^k}
$$
In other words, while $frac{1}{9} = 0.1111111ldots$
$$
frac{1}{81} = 0.01 + 0.002 + 0.0003 + 0.0004 + 0.00005 ldots
$$
It is pretty easy to see that this infinite sum is going to converge to something starting $0.012345ldots$. If you keep on adding, or work out $frac{1}{81}$ by division, you get
$$
0.012345679012345679012345679ldots
$$
When you get to the point where you add $frac{10}{10^{11}}$, the first carry happens, which leads to the 9 where you might expect an 8. After that every addition carries and the decimal expansion repeats every 9 digits (not every ten - because the amount we carry keeps on getting bigger and bigger).
Now, $frac{8}{81} = frac{9}{81} - frac{1}{81}$, or
$$
frac{8}{81} = 0.11111111ldots - 0.012345679012345ldots
$$
Think of each '1' digit in $0.111ldots$ as being a '10' in the next column. This means that we can work out $frac{8}{81}$ as the "10's complement" of $frac{1}{81}$, since we are subtracting a digit between $1$ and $9$ from $10$, to get another single digit which appears in the same place. So $frac{8}{81}$ starts $0.098765ldots$. The only break in the pattern is when you get to the digit '0' - subtracting 0 from 10 leaves you with 10, or a '1' in the next digit on the left, changing the 1 to a 2.
So
$$
frac{8}{81} = 0.098765432098765432098765ldots
$$
and therefore
$$
0.0123456790123456790ldots * 8 = 0.0987654320987654320ldots
$$
and clearly this gets you that
$$
12345679 * 8 = 98765432
$$
answered May 20 '13 at 10:01
jwgjwg
2,3471627
$endgroup$
$begingroup$
A bit late but notetimes($times$) instead of *.
$endgroup$
– TheSimpliFire
Apr 4 '18 at 18:01
add a comment |
$begingroup$
$98765432 / 12345679 = 8$, exactly. You can see how the pattern works by multiplying out $12345679 * 8$ starting at either end.
This explains why your fraction is close to an integer. If you think the $729$ is interesting (I don't), it can be explained by some of the other answers here.
Edit:
What can we say about the fact that $12345679 * 8 = 98765432$? I have been aware of this 'factlet' for about 20 years, and remember it being used to 'demonstrate' calculators (which often had 8 digit displays back in the day).
I just recently realised that:
$$
frac{1}{81} = left(frac{1}{9}right)^2 = left(sum_{k=1}^{infty}frac{1}{10^k}right)^2 = sum_{k=1}^{infty} sum_{m=1}^{k-1} frac{1}{10^m} frac{1}{10^{k-m}} = sum_{k=1}^{infty} frac{k-1}{10^k}
$$
In other words, while $frac{1}{9} = 0.1111111ldots$
$$
frac{1}{81} = 0.01 + 0.002 + 0.0003 + 0.0004 + 0.00005 ldots
$$
It is pretty easy to see that this infinite sum is going to converge to something starting $0.012345ldots$. If you keep on adding, or work out $frac{1}{81}$ by division, you get
$$
0.012345679012345679012345679ldots
$$
When you get to the point where you add $frac{10}{10^{11}}$, the first carry happens, which leads to the 9 where you might expect an 8. After that every addition carries and the decimal expansion repeats every 9 digits (not every ten - because the amount we carry keeps on getting bigger and bigger).
Now, $frac{8}{81} = frac{9}{81} - frac{1}{81}$, or
$$
frac{8}{81} = 0.11111111ldots - 0.012345679012345ldots
$$
Think of each '1' digit in $0.111ldots$ as being a '10' in the next column. This means that we can work out $frac{8}{81}$ as the "10's complement" of $frac{1}{81}$, since we are subtracting a digit between $1$ and $9$ from $10$, to get another single digit which appears in the same place. So $frac{8}{81}$ starts $0.098765ldots$. The only break in the pattern is when you get to the digit '0' - subtracting 0 from 10 leaves you with 10, or a '1' in the next digit on the left, changing the 1 to a 2.
So
$$
frac{8}{81} = 0.098765432098765432098765ldots
$$
and therefore
$$
0.0123456790123456790ldots * 8 = 0.0987654320987654320ldots
$$
and clearly this gets you that
$$
12345679 * 8 = 98765432
$$
answered May 20 '13 at 10:01
jwgjwg
2,3471627
$endgroup$
$begingroup$
A bit late but notetimes($times$) instead of *.
$endgroup$
– TheSimpliFire
Apr 4 '18 at 18:01
add a comment |
$begingroup$
$98765432 / 12345679 = 8$, exactly. You can see how the pattern works by multiplying out $12345679 * 8$ starting at either end.
This explains why your fraction is close to an integer. If you think the $729$ is interesting (I don't), it can be explained by some of the other answers here.
Edit:
What can we say about the fact that $12345679 * 8 = 98765432$? I have been aware of this 'factlet' for about 20 years, and remember it being used to 'demonstrate' calculators (which often had 8 digit displays back in the day).
I just recently realised that:
$$
frac{1}{81} = left(frac{1}{9}right)^2 = left(sum_{k=1}^{infty}frac{1}{10^k}right)^2 = sum_{k=1}^{infty} sum_{m=1}^{k-1} frac{1}{10^m} frac{1}{10^{k-m}} = sum_{k=1}^{infty} frac{k-1}{10^k}
$$
In other words, while $frac{1}{9} = 0.1111111ldots$
$$
frac{1}{81} = 0.01 + 0.002 + 0.0003 + 0.0004 + 0.00005 ldots
$$
It is pretty easy to see that this infinite sum is going to converge to something starting $0.012345ldots$. If you keep on adding, or work out $frac{1}{81}$ by division, you get
$$
0.012345679012345679012345679ldots
$$
When you get to the point where you add $frac{10}{10^{11}}$, the first carry happens, which leads to the 9 where you might expect an 8. After that every addition carries and the decimal expansion repeats every 9 digits (not every ten - because the amount we carry keeps on getting bigger and bigger).
Now, $frac{8}{81} = frac{9}{81} - frac{1}{81}$, or
$$
frac{8}{81} = 0.11111111ldots - 0.012345679012345ldots
$$
Think of each '1' digit in $0.111ldots$ as being a '10' in the next column. This means that we can work out $frac{8}{81}$ as the "10's complement" of $frac{1}{81}$, since we are subtracting a digit between $1$ and $9$ from $10$, to get another single digit which appears in the same place. So $frac{8}{81}$ starts $0.098765ldots$. The only break in the pattern is when you get to the digit '0' - subtracting 0 from 10 leaves you with 10, or a '1' in the next digit on the left, changing the 1 to a 2.
So
$$
frac{8}{81} = 0.098765432098765432098765ldots
$$
and therefore
$$
0.0123456790123456790ldots * 8 = 0.0987654320987654320ldots
$$
and clearly this gets you that
$$
12345679 * 8 = 98765432
$$
answered May 20 '13 at 10:01
jwgjwg
2,3471627
$endgroup$
$98765432 / 12345679 = 8$, exactly. You can see how the pattern works by multiplying out $12345679 * 8$ starting at either end.
This explains why your fraction is close to an integer. If you think the $729$ is interesting (I don't), it can be explained by some of the other answers here.
Edit:
What can we say about the fact that $12345679 * 8 = 98765432$? I have been aware of this 'factlet' for about 20 years, and remember it being used to 'demonstrate' calculators (which often had 8 digit displays back in the day).
I just recently realised that:
$$
frac{1}{81} = left(frac{1}{9}right)^2 = left(sum_{k=1}^{infty}frac{1}{10^k}right)^2 = sum_{k=1}^{infty} sum_{m=1}^{k-1} frac{1}{10^m} frac{1}{10^{k-m}} = sum_{k=1}^{infty} frac{k-1}{10^k}
$$
In other words, while $frac{1}{9} = 0.1111111ldots$
$$
frac{1}{81} = 0.01 + 0.002 + 0.0003 + 0.0004 + 0.00005 ldots
$$
It is pretty easy to see that this infinite sum is going to converge to something starting $0.012345ldots$. If you keep on adding, or work out $frac{1}{81}$ by division, you get
$$
0.012345679012345679012345679ldots
$$
When you get to the point where you add $frac{10}{10^{11}}$, the first carry happens, which leads to the 9 where you might expect an 8. After that every addition carries and the decimal expansion repeats every 9 digits (not every ten - because the amount we carry keeps on getting bigger and bigger).
Now, $frac{8}{81} = frac{9}{81} - frac{1}{81}$, or
$$
frac{8}{81} = 0.11111111ldots - 0.012345679012345ldots
$$
Think of each '1' digit in $0.111ldots$ as being a '10' in the next column. This means that we can work out $frac{8}{81}$ as the "10's complement" of $frac{1}{81}$, since we are subtracting a digit between $1$ and $9$ from $10$, to get another single digit which appears in the same place. So $frac{8}{81}$ starts $0.098765ldots$. The only break in the pattern is when you get to the digit '0' - subtracting 0 from 10 leaves you with 10, or a '1' in the next digit on the left, changing the 1 to a 2.
So
$$
frac{8}{81} = 0.098765432098765432098765ldots
$$
and therefore
$$
0.0123456790123456790ldots * 8 = 0.0987654320987654320ldots
$$
and clearly this gets you that
$$
12345679 * 8 = 98765432
$$
answered May 20 '13 at 10:01
jwgjwg
2,3471627
edited Apr 15 '15 at 21:08
answered May 20 '13 at 10:01
jwgjwg
2,3471627
answered May 20 '13 at 10:01
jwgjwg
2,3471627
answered May 20 '13 at 10:01
jwgjwg
2,3471627
2,3471627
$begingroup$
A bit late but notetimes($times$) instead of *.
$endgroup$
– TheSimpliFire
Apr 4 '18 at 18:01
add a comment |
$begingroup$
A bit late but notetimes($times$) instead of *.
$endgroup$
– TheSimpliFire
Apr 4 '18 at 18:01
$begingroup$
A bit late but note
times ($times$) instead of *.$endgroup$
– TheSimpliFire
Apr 4 '18 at 18:01
$begingroup$
A bit late but note
times ($times$) instead of *.$endgroup$
– TheSimpliFire
Apr 4 '18 at 18:01
add a comment |
$begingroup$
I think @BorisNovikov's answer is the best here, but I had to do some work to understand it; this is my attempt at clarifying his answer, hopefully this can help others too:
In the following, we adopt the notation $123456789 = [ 1, 2, ..., B-1 ]_B$ in base $B=10$.
Boris introduced the following sum:
$$
S_d(B) = [d, d-1, ..., 1]_B = 1 + 2B + ... + dB^{d-1} = sum_{k=1}^d kB^{k-1} = frac{dB^{d+1} - (d+1)B^d + 1}{(B-1)^2}
$$
where $d$ is the number of digits, $B$ the base, and the final formula is obtained by derivation of the usual geometric sum.
From there, it is easy to see that:
$$
[1, 2, ..., d]_B = d + (d-1)B + ... + 1B^{d-1} = B^{d-1} S_d(1/B)
$$
and putting both results together, we have:
$$
frac{ [B-1, B-2, ..., 1]_B }{ [1, 2, ..., B-1]_B } = frac{ S_{B-1}(B) }{ B^{B-2} S_{B-1}(1/B) } = frac{1 + (B-2)B^B}{B^B - B(B-1) - 1}
$$
Clearly both the numerator and denominator are dominated by the terms in $B^B$, so this fraction is equivalent to $B-2$ as $B$ tends to infinity, and we can see that the exact value is very close already for $B=10$ (see also @Mark Adler's post for other examples). Beautiful, isn't it? :)
answered May 31 '18 at 15:08
SheljohnSheljohn
1,276723
$endgroup$
add a comment |
$begingroup$
I think @BorisNovikov's answer is the best here, but I had to do some work to understand it; this is my attempt at clarifying his answer, hopefully this can help others too:
In the following, we adopt the notation $123456789 = [ 1, 2, ..., B-1 ]_B$ in base $B=10$.
Boris introduced the following sum:
$$
S_d(B) = [d, d-1, ..., 1]_B = 1 + 2B + ... + dB^{d-1} = sum_{k=1}^d kB^{k-1} = frac{dB^{d+1} - (d+1)B^d + 1}{(B-1)^2}
$$
where $d$ is the number of digits, $B$ the base, and the final formula is obtained by derivation of the usual geometric sum.
From there, it is easy to see that:
$$
[1, 2, ..., d]_B = d + (d-1)B + ... + 1B^{d-1} = B^{d-1} S_d(1/B)
$$
and putting both results together, we have:
$$
frac{ [B-1, B-2, ..., 1]_B }{ [1, 2, ..., B-1]_B } = frac{ S_{B-1}(B) }{ B^{B-2} S_{B-1}(1/B) } = frac{1 + (B-2)B^B}{B^B - B(B-1) - 1}
$$
Clearly both the numerator and denominator are dominated by the terms in $B^B$, so this fraction is equivalent to $B-2$ as $B$ tends to infinity, and we can see that the exact value is very close already for $B=10$ (see also @Mark Adler's post for other examples). Beautiful, isn't it? :)
answered May 31 '18 at 15:08
SheljohnSheljohn
1,276723
$endgroup$
add a comment |
$begingroup$
I think @BorisNovikov's answer is the best here, but I had to do some work to understand it; this is my attempt at clarifying his answer, hopefully this can help others too:
In the following, we adopt the notation $123456789 = [ 1, 2, ..., B-1 ]_B$ in base $B=10$.
Boris introduced the following sum:
$$
S_d(B) = [d, d-1, ..., 1]_B = 1 + 2B + ... + dB^{d-1} = sum_{k=1}^d kB^{k-1} = frac{dB^{d+1} - (d+1)B^d + 1}{(B-1)^2}
$$
where $d$ is the number of digits, $B$ the base, and the final formula is obtained by derivation of the usual geometric sum.
From there, it is easy to see that:
$$
[1, 2, ..., d]_B = d + (d-1)B + ... + 1B^{d-1} = B^{d-1} S_d(1/B)
$$
and putting both results together, we have:
$$
frac{ [B-1, B-2, ..., 1]_B }{ [1, 2, ..., B-1]_B } = frac{ S_{B-1}(B) }{ B^{B-2} S_{B-1}(1/B) } = frac{1 + (B-2)B^B}{B^B - B(B-1) - 1}
$$
Clearly both the numerator and denominator are dominated by the terms in $B^B$, so this fraction is equivalent to $B-2$ as $B$ tends to infinity, and we can see that the exact value is very close already for $B=10$ (see also @Mark Adler's post for other examples). Beautiful, isn't it? :)
answered May 31 '18 at 15:08
SheljohnSheljohn
1,276723
$endgroup$
I think @BorisNovikov's answer is the best here, but I had to do some work to understand it; this is my attempt at clarifying his answer, hopefully this can help others too:
In the following, we adopt the notation $123456789 = [ 1, 2, ..., B-1 ]_B$ in base $B=10$.
Boris introduced the following sum:
$$
S_d(B) = [d, d-1, ..., 1]_B = 1 + 2B + ... + dB^{d-1} = sum_{k=1}^d kB^{k-1} = frac{dB^{d+1} - (d+1)B^d + 1}{(B-1)^2}
$$
where $d$ is the number of digits, $B$ the base, and the final formula is obtained by derivation of the usual geometric sum.
From there, it is easy to see that:
$$
[1, 2, ..., d]_B = d + (d-1)B + ... + 1B^{d-1} = B^{d-1} S_d(1/B)
$$
and putting both results together, we have:
$$
frac{ [B-1, B-2, ..., 1]_B }{ [1, 2, ..., B-1]_B } = frac{ S_{B-1}(B) }{ B^{B-2} S_{B-1}(1/B) } = frac{1 + (B-2)B^B}{B^B - B(B-1) - 1}
$$
Clearly both the numerator and denominator are dominated by the terms in $B^B$, so this fraction is equivalent to $B-2$ as $B$ tends to infinity, and we can see that the exact value is very close already for $B=10$ (see also @Mark Adler's post for other examples). Beautiful, isn't it? :)
answered May 31 '18 at 15:08
SheljohnSheljohn
1,276723
edited Jan 1 at 13:54
answered May 31 '18 at 15:08
SheljohnSheljohn
1,276723
answered May 31 '18 at 15:08
SheljohnSheljohn
1,276723
answered May 31 '18 at 15:08
SheljohnSheljohn
1,276723
1,276723
add a comment |
add a comment |
$begingroup$
I think one should also think about how this is related to 9 being before 10. Notice the two following striking facts:
$12345679*8=98765432$ and $12345679*9=111111111$
answered Sep 8 '18 at 0:26
Sergio LuceroSergio Lucero
1115
$endgroup$
$begingroup$
See @jwg's answer.
$endgroup$
– Sheljohn
Jan 1 at 14:07
add a comment |
$begingroup$
I think one should also think about how this is related to 9 being before 10. Notice the two following striking facts:
$12345679*8=98765432$ and $12345679*9=111111111$
answered Sep 8 '18 at 0:26
Sergio LuceroSergio Lucero
1115
$endgroup$
$begingroup$
See @jwg's answer.
$endgroup$
– Sheljohn
Jan 1 at 14:07
add a comment |
$begingroup$
I think one should also think about how this is related to 9 being before 10. Notice the two following striking facts:
$12345679*8=98765432$ and $12345679*9=111111111$
answered Sep 8 '18 at 0:26
Sergio LuceroSergio Lucero
1115
$endgroup$
I think one should also think about how this is related to 9 being before 10. Notice the two following striking facts:
$12345679*8=98765432$ and $12345679*9=111111111$
answered Sep 8 '18 at 0:26
Sergio LuceroSergio Lucero
1115
answered Sep 8 '18 at 0:26
Sergio LuceroSergio Lucero
1115
answered Sep 8 '18 at 0:26
Sergio LuceroSergio Lucero
1115
answered Sep 8 '18 at 0:26
Sergio LuceroSergio Lucero
1115
1115
$begingroup$
See @jwg's answer.
$endgroup$
– Sheljohn
Jan 1 at 14:07
add a comment |
$begingroup$
See @jwg's answer.
$endgroup$
– Sheljohn
Jan 1 at 14:07
$begingroup$
See @jwg's answer.
$endgroup$
– Sheljohn
Jan 1 at 14:07
$begingroup$
See @jwg's answer.
$endgroup$
– Sheljohn
Jan 1 at 14:07
add a comment |
protected by Community♦ May 22 '13 at 13:20
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?
13
$begingroup$
8.000000072900000663390006036849054935326399911470239194379176...
$endgroup$
– copper.hat
May 19 '13 at 7:10
116
$begingroup$
I don't see it, why is this number interesting?
$endgroup$
– Matsemann
May 19 '13 at 9:43
20
$begingroup$
8.0000000729000006633900060368490549353263999114702391943791766688505076865396199475105415223459278533479434654662855357431983752631052148942574555377428453934598930804850270324137459949650885541823058430589831718367468637143964598010077841891708361214546087052369392176561468806709366141055231883602610140783752281132145758302526400552990245032211229793122191117411939168448646432882682539232411107014941073835963771907270324356159951641055559933605595395810918101879354727102128016629364951327221057077711619407175736605299203108222748284827009391925785466524647745374294482906079794445326129452467
$endgroup$
– Double AA
May 19 '13 at 10:29
15
$begingroup$
Unrelated, but ever notice that Sqrt(9.87654321) approximates pi? The number system is full of this kind of stuff.
$endgroup$
– cobaltduck
May 19 '13 at 11:44
23
$begingroup$
@cobaltduck: It is a terrible approximation given how many digits you're putting in. It has more than $10$ times the error of $sqrt{9.87}$
$endgroup$
– Jonas Meyer
May 19 '13 at 18:28