$1 < a$ and $bne0$ imply $1<a^b$
$1 < a$ and $bne0$ imply $1<a^b$ when $a,b$ are arbitrary nonnegative integers.
I've tried to prove it by induction.
I've assumed that $b < a$ (Is valid my assumption?)
I'm using this definition for $a<b$
begin{align*}
a<b Leftrightarrow (exists kin mathbb N)a + (k + 1) = b\
end{align*}
Base P(1)
$0<1le 1 < a implies1<a^1$
$1<a^1 equiv (1+(k+1))^1=a^1$
Hyphotesis P(n)
$0<1le n < a implies1<a^n$
$1<a^n equiv (1+(k+1))^n=a^n$
Induction step
$0<1le n + 1 < a implies1<a^{n+1}$
we have $(1+(k+1))^n*(1+(k+1))=a^n*a$ (hyphotesis)
then $(1+(k+1))^n = a^n$, because $(1+(k+1)) = a$ (base)
(here I used the property $ac= bc => a = b$ with $c ne 0$)
proof-verification proof-writing order-theory natural-numbers
asked Dec 7 at 21:48
adriana634
416
add a comment |
$1 < a$ and $bne0$ imply $1<a^b$ when $a,b$ are arbitrary nonnegative integers.
I've tried to prove it by induction.
I've assumed that $b < a$ (Is valid my assumption?)
I'm using this definition for $a<b$
begin{align*}
a<b Leftrightarrow (exists kin mathbb N)a + (k + 1) = b\
end{align*}
Base P(1)
$0<1le 1 < a implies1<a^1$
$1<a^1 equiv (1+(k+1))^1=a^1$
Hyphotesis P(n)
$0<1le n < a implies1<a^n$
$1<a^n equiv (1+(k+1))^n=a^n$
Induction step
$0<1le n + 1 < a implies1<a^{n+1}$
we have $(1+(k+1))^n*(1+(k+1))=a^n*a$ (hyphotesis)
then $(1+(k+1))^n = a^n$, because $(1+(k+1)) = a$ (base)
(here I used the property $ac= bc => a = b$ with $c ne 0$)
proof-verification proof-writing order-theory natural-numbers
asked Dec 7 at 21:48
adriana634
416
@ZacharySelk b is positive since I am working with natural numbers and $b≠0$.
– adriana634
Dec 7 at 22:06
@Bernard I've missed this "a,b,c are arbitrary nonnegative integers".
– adriana634
Dec 7 at 22:10
1
"I've assumed that $b<a$" This is an invalid assumption. You will need to use double-induction here and prove it for all $a,b$ pairs, not just those where $b$ is smaller. Have you proven yet the property that given $xgeq 1$ and $ygeq 0$ that $xygeq y$? Using this you have $a^{n+1}=a^n cdot a geq a^n > 1$ where the first inequality is from the mentioned property and the final inequality is from the induction hypothesis when inducting over the exponent. And then inducting over $a$ is straightforward since if $a>1$ it should be clear that $a+1$ is also greater than $1$.
– JMoravitz
Dec 7 at 22:14
May be you could start with the opposite and prove its impossible. $ 1> a^{b}$ implies $log(1)> b.log(a)$. However, log a is not negative for $a >=1$...You can complete the proof. Note the case of $b <0$.
– NoChance
Dec 7 at 22:40
add a comment |
$1 < a$ and $bne0$ imply $1<a^b$ when $a,b$ are arbitrary nonnegative integers.
I've tried to prove it by induction.
I've assumed that $b < a$ (Is valid my assumption?)
I'm using this definition for $a<b$
begin{align*}
a<b Leftrightarrow (exists kin mathbb N)a + (k + 1) = b\
end{align*}
Base P(1)
$0<1le 1 < a implies1<a^1$
$1<a^1 equiv (1+(k+1))^1=a^1$
Hyphotesis P(n)
$0<1le n < a implies1<a^n$
$1<a^n equiv (1+(k+1))^n=a^n$
Induction step
$0<1le n + 1 < a implies1<a^{n+1}$
we have $(1+(k+1))^n*(1+(k+1))=a^n*a$ (hyphotesis)
then $(1+(k+1))^n = a^n$, because $(1+(k+1)) = a$ (base)
(here I used the property $ac= bc => a = b$ with $c ne 0$)
proof-verification proof-writing order-theory natural-numbers
asked Dec 7 at 21:48
adriana634
416
$1 < a$ and $bne0$ imply $1<a^b$ when $a,b$ are arbitrary nonnegative integers.
I've tried to prove it by induction.
I've assumed that $b < a$ (Is valid my assumption?)
I'm using this definition for $a<b$
begin{align*}
a<b Leftrightarrow (exists kin mathbb N)a + (k + 1) = b\
end{align*}
Base P(1)
$0<1le 1 < a implies1<a^1$
$1<a^1 equiv (1+(k+1))^1=a^1$
Hyphotesis P(n)
$0<1le n < a implies1<a^n$
$1<a^n equiv (1+(k+1))^n=a^n$
Induction step
$0<1le n + 1 < a implies1<a^{n+1}$
we have $(1+(k+1))^n*(1+(k+1))=a^n*a$ (hyphotesis)
then $(1+(k+1))^n = a^n$, because $(1+(k+1)) = a$ (base)
(here I used the property $ac= bc => a = b$ with $c ne 0$)
proof-verification proof-writing order-theory natural-numbers
proof-verification proof-writing order-theory natural-numbers
asked Dec 7 at 21:48
adriana634
416
asked Dec 7 at 21:48
adriana634
416
edited Dec 8 at 0:06
asked Dec 7 at 21:48
adriana634
416
asked Dec 7 at 21:48
adriana634
416
asked Dec 7 at 21:48
adriana634
416
416
@ZacharySelk b is positive since I am working with natural numbers and $b≠0$.
– adriana634
Dec 7 at 22:06
@Bernard I've missed this "a,b,c are arbitrary nonnegative integers".
– adriana634
Dec 7 at 22:10
1
"I've assumed that $b<a$" This is an invalid assumption. You will need to use double-induction here and prove it for all $a,b$ pairs, not just those where $b$ is smaller. Have you proven yet the property that given $xgeq 1$ and $ygeq 0$ that $xygeq y$? Using this you have $a^{n+1}=a^n cdot a geq a^n > 1$ where the first inequality is from the mentioned property and the final inequality is from the induction hypothesis when inducting over the exponent. And then inducting over $a$ is straightforward since if $a>1$ it should be clear that $a+1$ is also greater than $1$.
– JMoravitz
Dec 7 at 22:14
May be you could start with the opposite and prove its impossible. $ 1> a^{b}$ implies $log(1)> b.log(a)$. However, log a is not negative for $a >=1$...You can complete the proof. Note the case of $b <0$.
– NoChance
Dec 7 at 22:40
add a comment |
@ZacharySelk b is positive since I am working with natural numbers and $b≠0$.
– adriana634
Dec 7 at 22:06
@Bernard I've missed this "a,b,c are arbitrary nonnegative integers".
– adriana634
Dec 7 at 22:10
1
"I've assumed that $b<a$" This is an invalid assumption. You will need to use double-induction here and prove it for all $a,b$ pairs, not just those where $b$ is smaller. Have you proven yet the property that given $xgeq 1$ and $ygeq 0$ that $xygeq y$? Using this you have $a^{n+1}=a^n cdot a geq a^n > 1$ where the first inequality is from the mentioned property and the final inequality is from the induction hypothesis when inducting over the exponent. And then inducting over $a$ is straightforward since if $a>1$ it should be clear that $a+1$ is also greater than $1$.
– JMoravitz
Dec 7 at 22:14
May be you could start with the opposite and prove its impossible. $ 1> a^{b}$ implies $log(1)> b.log(a)$. However, log a is not negative for $a >=1$...You can complete the proof. Note the case of $b <0$.
– NoChance
Dec 7 at 22:40
@ZacharySelk b is positive since I am working with natural numbers and $b≠0$.
– adriana634
Dec 7 at 22:06
@ZacharySelk b is positive since I am working with natural numbers and $b≠0$.
– adriana634
Dec 7 at 22:06
@Bernard I've missed this "a,b,c are arbitrary nonnegative integers".
– adriana634
Dec 7 at 22:10
@Bernard I've missed this "a,b,c are arbitrary nonnegative integers".
– adriana634
Dec 7 at 22:10
1
1
"I've assumed that $b<a$" This is an invalid assumption. You will need to use double-induction here and prove it for all $a,b$ pairs, not just those where $b$ is smaller. Have you proven yet the property that given $xgeq 1$ and $ygeq 0$ that $xygeq y$? Using this you have $a^{n+1}=a^n cdot a geq a^n > 1$ where the first inequality is from the mentioned property and the final inequality is from the induction hypothesis when inducting over the exponent. And then inducting over $a$ is straightforward since if $a>1$ it should be clear that $a+1$ is also greater than $1$.
– JMoravitz
Dec 7 at 22:14
"I've assumed that $b<a$" This is an invalid assumption. You will need to use double-induction here and prove it for all $a,b$ pairs, not just those where $b$ is smaller. Have you proven yet the property that given $xgeq 1$ and $ygeq 0$ that $xygeq y$? Using this you have $a^{n+1}=a^n cdot a geq a^n > 1$ where the first inequality is from the mentioned property and the final inequality is from the induction hypothesis when inducting over the exponent. And then inducting over $a$ is straightforward since if $a>1$ it should be clear that $a+1$ is also greater than $1$.
– JMoravitz
Dec 7 at 22:14
May be you could start with the opposite and prove its impossible. $ 1> a^{b}$ implies $log(1)> b.log(a)$. However, log a is not negative for $a >=1$...You can complete the proof. Note the case of $b <0$.
– NoChance
Dec 7 at 22:40
May be you could start with the opposite and prove its impossible. $ 1> a^{b}$ implies $log(1)> b.log(a)$. However, log a is not negative for $a >=1$...You can complete the proof. Note the case of $b <0$.
– NoChance
Dec 7 at 22:40
add a comment |
2 Answers
2
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It wasn't valid to assume $b < a$ and you never used it
I can't follow your induction step.
If we assume $1 < a^n$ then $a^{n+1}=a^n*a$ and we want to prove $a^n*a > 1$ or that there is a $kin mathbb N$ so that $1 + (k+1)= a^n*a$
We know that $1 < a$ so there is a $j$ so that $1 +(j+1)=a$. And we know that $1 < a^n$ and that there is an $m$ so that $1+(m+1) =a^n$ so $a^{n+1} = (1 + (j+1))(1 + (m+1)) = 1 + (j+m+2) + (j+1)(m+1)$. So let $k = (j+m+2) + (j+1)(m+1)in mathbb N$ and we are done.
answered Dec 7 at 23:23
fleablood
68k22684
I understand what you did here but there's a detail I don't know why is here: $1 + (k+1)= a^n*an$ why is that last n here? I expected it to be $1 + (k+1)= a^n*a$, or am I wrong?
– adriana634
Dec 8 at 15:32
add a comment |
I'm not sure that induction is required for this proof. Proof by two cases:
Case 1: If $b=1$ then it is obvious that $a^b=a>1$.
Case 2: If $b>1$ then $a^b=acdot a^{b-1}geq a>1$
The first inequality, that $a^{b-1}geq 1$, is justified by the fact that any power of a nonzero natural number is itself a nonzero natural number (since $mathbb{N}$ equipped with multiplication has no non-zero zero divisors).
answered Dec 7 at 22:46
RandomMathGuy
1872
add a comment |
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2 Answers
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It wasn't valid to assume $b < a$ and you never used it
I can't follow your induction step.
If we assume $1 < a^n$ then $a^{n+1}=a^n*a$ and we want to prove $a^n*a > 1$ or that there is a $kin mathbb N$ so that $1 + (k+1)= a^n*a$
We know that $1 < a$ so there is a $j$ so that $1 +(j+1)=a$. And we know that $1 < a^n$ and that there is an $m$ so that $1+(m+1) =a^n$ so $a^{n+1} = (1 + (j+1))(1 + (m+1)) = 1 + (j+m+2) + (j+1)(m+1)$. So let $k = (j+m+2) + (j+1)(m+1)in mathbb N$ and we are done.
answered Dec 7 at 23:23
fleablood
68k22684
I understand what you did here but there's a detail I don't know why is here: $1 + (k+1)= a^n*an$ why is that last n here? I expected it to be $1 + (k+1)= a^n*a$, or am I wrong?
– adriana634
Dec 8 at 15:32
add a comment |
It wasn't valid to assume $b < a$ and you never used it
I can't follow your induction step.
If we assume $1 < a^n$ then $a^{n+1}=a^n*a$ and we want to prove $a^n*a > 1$ or that there is a $kin mathbb N$ so that $1 + (k+1)= a^n*a$
We know that $1 < a$ so there is a $j$ so that $1 +(j+1)=a$. And we know that $1 < a^n$ and that there is an $m$ so that $1+(m+1) =a^n$ so $a^{n+1} = (1 + (j+1))(1 + (m+1)) = 1 + (j+m+2) + (j+1)(m+1)$. So let $k = (j+m+2) + (j+1)(m+1)in mathbb N$ and we are done.
answered Dec 7 at 23:23
fleablood
68k22684
I understand what you did here but there's a detail I don't know why is here: $1 + (k+1)= a^n*an$ why is that last n here? I expected it to be $1 + (k+1)= a^n*a$, or am I wrong?
– adriana634
Dec 8 at 15:32
add a comment |
It wasn't valid to assume $b < a$ and you never used it
I can't follow your induction step.
If we assume $1 < a^n$ then $a^{n+1}=a^n*a$ and we want to prove $a^n*a > 1$ or that there is a $kin mathbb N$ so that $1 + (k+1)= a^n*a$
We know that $1 < a$ so there is a $j$ so that $1 +(j+1)=a$. And we know that $1 < a^n$ and that there is an $m$ so that $1+(m+1) =a^n$ so $a^{n+1} = (1 + (j+1))(1 + (m+1)) = 1 + (j+m+2) + (j+1)(m+1)$. So let $k = (j+m+2) + (j+1)(m+1)in mathbb N$ and we are done.
answered Dec 7 at 23:23
fleablood
68k22684
It wasn't valid to assume $b < a$ and you never used it
I can't follow your induction step.
If we assume $1 < a^n$ then $a^{n+1}=a^n*a$ and we want to prove $a^n*a > 1$ or that there is a $kin mathbb N$ so that $1 + (k+1)= a^n*a$
We know that $1 < a$ so there is a $j$ so that $1 +(j+1)=a$. And we know that $1 < a^n$ and that there is an $m$ so that $1+(m+1) =a^n$ so $a^{n+1} = (1 + (j+1))(1 + (m+1)) = 1 + (j+m+2) + (j+1)(m+1)$. So let $k = (j+m+2) + (j+1)(m+1)in mathbb N$ and we are done.
answered Dec 7 at 23:23
fleablood
68k22684
edited Dec 8 at 16:55
answered Dec 7 at 23:23
fleablood
68k22684
answered Dec 7 at 23:23
fleablood
68k22684
answered Dec 7 at 23:23
fleablood
68k22684
68k22684
I understand what you did here but there's a detail I don't know why is here: $1 + (k+1)= a^n*an$ why is that last n here? I expected it to be $1 + (k+1)= a^n*a$, or am I wrong?
– adriana634
Dec 8 at 15:32
add a comment |
I understand what you did here but there's a detail I don't know why is here: $1 + (k+1)= a^n*an$ why is that last n here? I expected it to be $1 + (k+1)= a^n*a$, or am I wrong?
– adriana634
Dec 8 at 15:32
I understand what you did here but there's a detail I don't know why is here: $1 + (k+1)= a^n*an$ why is that last n here? I expected it to be $1 + (k+1)= a^n*a$, or am I wrong?
– adriana634
Dec 8 at 15:32
I understand what you did here but there's a detail I don't know why is here: $1 + (k+1)= a^n*an$ why is that last n here? I expected it to be $1 + (k+1)= a^n*a$, or am I wrong?
– adriana634
Dec 8 at 15:32
add a comment |
I'm not sure that induction is required for this proof. Proof by two cases:
Case 1: If $b=1$ then it is obvious that $a^b=a>1$.
Case 2: If $b>1$ then $a^b=acdot a^{b-1}geq a>1$
The first inequality, that $a^{b-1}geq 1$, is justified by the fact that any power of a nonzero natural number is itself a nonzero natural number (since $mathbb{N}$ equipped with multiplication has no non-zero zero divisors).
answered Dec 7 at 22:46
RandomMathGuy
1872
add a comment |
I'm not sure that induction is required for this proof. Proof by two cases:
Case 1: If $b=1$ then it is obvious that $a^b=a>1$.
Case 2: If $b>1$ then $a^b=acdot a^{b-1}geq a>1$
The first inequality, that $a^{b-1}geq 1$, is justified by the fact that any power of a nonzero natural number is itself a nonzero natural number (since $mathbb{N}$ equipped with multiplication has no non-zero zero divisors).
answered Dec 7 at 22:46
RandomMathGuy
1872
add a comment |
I'm not sure that induction is required for this proof. Proof by two cases:
Case 1: If $b=1$ then it is obvious that $a^b=a>1$.
Case 2: If $b>1$ then $a^b=acdot a^{b-1}geq a>1$
The first inequality, that $a^{b-1}geq 1$, is justified by the fact that any power of a nonzero natural number is itself a nonzero natural number (since $mathbb{N}$ equipped with multiplication has no non-zero zero divisors).
answered Dec 7 at 22:46
RandomMathGuy
1872
I'm not sure that induction is required for this proof. Proof by two cases:
Case 1: If $b=1$ then it is obvious that $a^b=a>1$.
Case 2: If $b>1$ then $a^b=acdot a^{b-1}geq a>1$
The first inequality, that $a^{b-1}geq 1$, is justified by the fact that any power of a nonzero natural number is itself a nonzero natural number (since $mathbb{N}$ equipped with multiplication has no non-zero zero divisors).
answered Dec 7 at 22:46
RandomMathGuy
1872
edited Dec 7 at 23:00
answered Dec 7 at 22:46
RandomMathGuy
1872
answered Dec 7 at 22:46
RandomMathGuy
1872
answered Dec 7 at 22:46
RandomMathGuy
1872
1872
add a comment |
add a comment |
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@ZacharySelk b is positive since I am working with natural numbers and $b≠0$.
– adriana634
Dec 7 at 22:06
@Bernard I've missed this "a,b,c are arbitrary nonnegative integers".
– adriana634
Dec 7 at 22:10
1
"I've assumed that $b<a$" This is an invalid assumption. You will need to use double-induction here and prove it for all $a,b$ pairs, not just those where $b$ is smaller. Have you proven yet the property that given $xgeq 1$ and $ygeq 0$ that $xygeq y$? Using this you have $a^{n+1}=a^n cdot a geq a^n > 1$ where the first inequality is from the mentioned property and the final inequality is from the induction hypothesis when inducting over the exponent. And then inducting over $a$ is straightforward since if $a>1$ it should be clear that $a+1$ is also greater than $1$.
– JMoravitz
Dec 7 at 22:14
May be you could start with the opposite and prove its impossible. $ 1> a^{b}$ implies $log(1)> b.log(a)$. However, log a is not negative for $a >=1$...You can complete the proof. Note the case of $b <0$.
– NoChance
Dec 7 at 22:40