Why is $5$ the remainder of $5 over 13$?
I don't understand why the remainder of $5 over 13$ is $5$. I know that the DA tells us that $5 = 0(13) + r$ so the remainder has to be $5$ based on this, but I'm a little unsure of why/how it works
elementary-number-theory divisibility
edited Dec 9 at 8:25
Batominovski
33.7k33292
asked Dec 9 at 0:55
ming
3165
add a comment |
I don't understand why the remainder of $5 over 13$ is $5$. I know that the DA tells us that $5 = 0(13) + r$ so the remainder has to be $5$ based on this, but I'm a little unsure of why/how it works
elementary-number-theory divisibility
edited Dec 9 at 8:25
Batominovski
33.7k33292
asked Dec 9 at 0:55
ming
3165
By definition the remainder of $,5div 13,$ is the least nonnegative element of $, 5+13,Bbb Z$ $ $
– Bill Dubuque
Dec 9 at 1:06
add a comment |
I don't understand why the remainder of $5 over 13$ is $5$. I know that the DA tells us that $5 = 0(13) + r$ so the remainder has to be $5$ based on this, but I'm a little unsure of why/how it works
elementary-number-theory divisibility
edited Dec 9 at 8:25
Batominovski
33.7k33292
asked Dec 9 at 0:55
ming
3165
I don't understand why the remainder of $5 over 13$ is $5$. I know that the DA tells us that $5 = 0(13) + r$ so the remainder has to be $5$ based on this, but I'm a little unsure of why/how it works
elementary-number-theory divisibility
elementary-number-theory divisibility
edited Dec 9 at 8:25
Batominovski
33.7k33292
asked Dec 9 at 0:55
ming
3165
edited Dec 9 at 8:25
Batominovski
33.7k33292
asked Dec 9 at 0:55
ming
3165
edited Dec 9 at 8:25
Batominovski
33.7k33292
edited Dec 9 at 8:25
Batominovski
33.7k33292
edited Dec 9 at 8:25
Batominovski
33.7k33292
33.7k33292
asked Dec 9 at 0:55
ming
3165
asked Dec 9 at 0:55
ming
3165
asked Dec 9 at 0:55
ming
3165
3165
By definition the remainder of $,5div 13,$ is the least nonnegative element of $, 5+13,Bbb Z$ $ $
– Bill Dubuque
Dec 9 at 1:06
add a comment |
By definition the remainder of $,5div 13,$ is the least nonnegative element of $, 5+13,Bbb Z$ $ $
– Bill Dubuque
Dec 9 at 1:06
By definition the remainder of $,5div 13,$ is the least nonnegative element of $, 5+13,Bbb Z$ $ $
– Bill Dubuque
Dec 9 at 1:06
By definition the remainder of $,5div 13,$ is the least nonnegative element of $, 5+13,Bbb Z$ $ $
– Bill Dubuque
Dec 9 at 1:06
add a comment |
3 Answers
3
active
oldest
votes
For positive integers, "divide by $13$" means "subtract the largest multiple of $13$ you can while still remaining nonnegative". The remainder is what's left. If you think in terms of objects, you remove them from the set you start with, in batches of $13$.
If you start with $5$ you can't subtract any multiples of $13$ so all $5$ are left.
You don't need a formal "division algorithm" for this. It works fine if all you know about division is repeated subtraction (which is what the division algorithm formalizes).
answered Dec 9 at 1:00
Ethan Bolker
41.2k547108
add a comment |
For an intuitive approach: If you have five candies to give out equally to 13 children, each child gets 0 candies, and there remain 5 undistributed candies.
answered Dec 9 at 1:04
paw88789
29k12349
add a comment |
Well $r$ has to be in a set ${0,1,2...,12}$. Which of this fullfiles equation you wrote?
And this is always true if you divide $a$ by $b$ and $a<b$. You will get $r=a$:
$$ a = 0cdot b +rimplies r=a$$
answered Dec 9 at 0:57
greedoid
37.7k114794
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
For positive integers, "divide by $13$" means "subtract the largest multiple of $13$ you can while still remaining nonnegative". The remainder is what's left. If you think in terms of objects, you remove them from the set you start with, in batches of $13$.
If you start with $5$ you can't subtract any multiples of $13$ so all $5$ are left.
You don't need a formal "division algorithm" for this. It works fine if all you know about division is repeated subtraction (which is what the division algorithm formalizes).
answered Dec 9 at 1:00
Ethan Bolker
41.2k547108
add a comment |
For positive integers, "divide by $13$" means "subtract the largest multiple of $13$ you can while still remaining nonnegative". The remainder is what's left. If you think in terms of objects, you remove them from the set you start with, in batches of $13$.
If you start with $5$ you can't subtract any multiples of $13$ so all $5$ are left.
You don't need a formal "division algorithm" for this. It works fine if all you know about division is repeated subtraction (which is what the division algorithm formalizes).
answered Dec 9 at 1:00
Ethan Bolker
41.2k547108
add a comment |
For positive integers, "divide by $13$" means "subtract the largest multiple of $13$ you can while still remaining nonnegative". The remainder is what's left. If you think in terms of objects, you remove them from the set you start with, in batches of $13$.
If you start with $5$ you can't subtract any multiples of $13$ so all $5$ are left.
You don't need a formal "division algorithm" for this. It works fine if all you know about division is repeated subtraction (which is what the division algorithm formalizes).
answered Dec 9 at 1:00
Ethan Bolker
41.2k547108
For positive integers, "divide by $13$" means "subtract the largest multiple of $13$ you can while still remaining nonnegative". The remainder is what's left. If you think in terms of objects, you remove them from the set you start with, in batches of $13$.
If you start with $5$ you can't subtract any multiples of $13$ so all $5$ are left.
You don't need a formal "division algorithm" for this. It works fine if all you know about division is repeated subtraction (which is what the division algorithm formalizes).
answered Dec 9 at 1:00
Ethan Bolker
41.2k547108
answered Dec 9 at 1:00
Ethan Bolker
41.2k547108
answered Dec 9 at 1:00
Ethan Bolker
41.2k547108
answered Dec 9 at 1:00
Ethan Bolker
41.2k547108
41.2k547108
add a comment |
add a comment |
For an intuitive approach: If you have five candies to give out equally to 13 children, each child gets 0 candies, and there remain 5 undistributed candies.
answered Dec 9 at 1:04
paw88789
29k12349
add a comment |
For an intuitive approach: If you have five candies to give out equally to 13 children, each child gets 0 candies, and there remain 5 undistributed candies.
answered Dec 9 at 1:04
paw88789
29k12349
add a comment |
For an intuitive approach: If you have five candies to give out equally to 13 children, each child gets 0 candies, and there remain 5 undistributed candies.
answered Dec 9 at 1:04
paw88789
29k12349
For an intuitive approach: If you have five candies to give out equally to 13 children, each child gets 0 candies, and there remain 5 undistributed candies.
answered Dec 9 at 1:04
paw88789
29k12349
answered Dec 9 at 1:04
paw88789
29k12349
answered Dec 9 at 1:04
paw88789
29k12349
answered Dec 9 at 1:04
paw88789
29k12349
29k12349
add a comment |
add a comment |
Well $r$ has to be in a set ${0,1,2...,12}$. Which of this fullfiles equation you wrote?
And this is always true if you divide $a$ by $b$ and $a<b$. You will get $r=a$:
$$ a = 0cdot b +rimplies r=a$$
answered Dec 9 at 0:57
greedoid
37.7k114794
add a comment |
Well $r$ has to be in a set ${0,1,2...,12}$. Which of this fullfiles equation you wrote?
And this is always true if you divide $a$ by $b$ and $a<b$. You will get $r=a$:
$$ a = 0cdot b +rimplies r=a$$
answered Dec 9 at 0:57
greedoid
37.7k114794
add a comment |
Well $r$ has to be in a set ${0,1,2...,12}$. Which of this fullfiles equation you wrote?
And this is always true if you divide $a$ by $b$ and $a<b$. You will get $r=a$:
$$ a = 0cdot b +rimplies r=a$$
answered Dec 9 at 0:57
greedoid
37.7k114794
Well $r$ has to be in a set ${0,1,2...,12}$. Which of this fullfiles equation you wrote?
And this is always true if you divide $a$ by $b$ and $a<b$. You will get $r=a$:
$$ a = 0cdot b +rimplies r=a$$
answered Dec 9 at 0:57
greedoid
37.7k114794
answered Dec 9 at 0:57
greedoid
37.7k114794
answered Dec 9 at 0:57
greedoid
37.7k114794
answered Dec 9 at 0:57
greedoid
37.7k114794
37.7k114794
add a comment |
add a comment |
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By definition the remainder of $,5div 13,$ is the least nonnegative element of $, 5+13,Bbb Z$ $ $
– Bill Dubuque
Dec 9 at 1:06