Mixing
$forall{x}inmathbb{Z}$, if $forall{x}inmathbb{N}$, $a^xmid{x}$, then $a=1$ or $a=-1$.
$begingroup$
Prove or disprove:
$forall{x}inmathbb{Z}$, if $forall{x}inmathbb{N}$, $a^xmid{x}$, then $a=1$ or $a=-1$.
I tried proving the positive but I'm not sure if it is the right approach.
proof-writing divisibility
edited Jan 21 at 5:31
David G. Stork
11k41432
asked Jan 21 at 5:26
macymacy
305
$endgroup$
add a comment |
$begingroup$
Prove or disprove:
$forall{x}inmathbb{Z}$, if $forall{x}inmathbb{N}$, $a^xmid{x}$, then $a=1$ or $a=-1$.
I tried proving the positive but I'm not sure if it is the right approach.
proof-writing divisibility
edited Jan 21 at 5:31
David G. Stork
11k41432
asked Jan 21 at 5:26
macymacy
305
$endgroup$
3
$begingroup$
Why do you have both $forall x in mathbb{Z}$ and $forall x in mathbb{N}$? I don't understand including both in the problem. Isn't the problem statement the same if you delete one of them?
$endgroup$
– David G. Stork
Jan 21 at 5:30
add a comment |
$begingroup$
Prove or disprove:
$forall{x}inmathbb{Z}$, if $forall{x}inmathbb{N}$, $a^xmid{x}$, then $a=1$ or $a=-1$.
I tried proving the positive but I'm not sure if it is the right approach.
proof-writing divisibility
edited Jan 21 at 5:31
David G. Stork
11k41432
asked Jan 21 at 5:26
macymacy
305
$endgroup$
Prove or disprove:
$forall{x}inmathbb{Z}$, if $forall{x}inmathbb{N}$, $a^xmid{x}$, then $a=1$ or $a=-1$.
I tried proving the positive but I'm not sure if it is the right approach.
proof-writing divisibility
proof-writing divisibility
edited Jan 21 at 5:31
David G. Stork
11k41432
asked Jan 21 at 5:26
macymacy
305
edited Jan 21 at 5:31
David G. Stork
11k41432
asked Jan 21 at 5:26
macymacy
305
edited Jan 21 at 5:31
David G. Stork
11k41432
edited Jan 21 at 5:31
David G. Stork
11k41432
edited Jan 21 at 5:31
David G. Stork
11k41432
11k41432
asked Jan 21 at 5:26
macymacy
305
asked Jan 21 at 5:26
macymacy
305
asked Jan 21 at 5:26
macymacy
305
305
3
$begingroup$
Why do you have both $forall x in mathbb{Z}$ and $forall x in mathbb{N}$? I don't understand including both in the problem. Isn't the problem statement the same if you delete one of them?
$endgroup$
– David G. Stork
Jan 21 at 5:30
add a comment |
3
$begingroup$
Why do you have both $forall x in mathbb{Z}$ and $forall x in mathbb{N}$? I don't understand including both in the problem. Isn't the problem statement the same if you delete one of them?
$endgroup$
– David G. Stork
Jan 21 at 5:30
3
3
$begingroup$
Why do you have both $forall x in mathbb{Z}$ and $forall x in mathbb{N}$? I don't understand including both in the problem. Isn't the problem statement the same if you delete one of them?
$endgroup$
– David G. Stork
Jan 21 at 5:30
$begingroup$
Why do you have both $forall x in mathbb{Z}$ and $forall x in mathbb{N}$? I don't understand including both in the problem. Isn't the problem statement the same if you delete one of them?
$endgroup$
– David G. Stork
Jan 21 at 5:30
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
We have: if $a ge 2 implies a^x ge 2^x > ximplies a^x > ximplies a^x nmid x$. Thus if $a^x mid x implies a = 1$. Also, if $a le -2 implies |a^x| ge 2^x > x implies |a^x| > ximplies |a^x| nmid ximplies a^x nmid x$. Thus $a^x mid x implies a = -1$. Together we have: $a = pm 1$.
answered Jan 21 at 5:35
DeepSeaDeepSea
71.3k54488
$endgroup$
add a comment |
$begingroup$
Note that if $a^x | x,$ then $ a^x leq x.$ Now consider when this is possible and try to take it from there.
answered Jan 21 at 5:35
Jack PfaffingerJack Pfaffinger
374112
$endgroup$
add a comment |
$begingroup$
If $a > 1$ is should be become apparent that $a^m > m$. After all $a^m = a*a*a.... $ and as $a*a = a+a+a+a... ge a+a$ we have $a*a*a* ge a+a+a+.... > 1+1+1+.... = m$.
(To make this clearer consider $5^3 = 5*5^2 = 5^2 + 5^2 + 5^2 + 5^2 +5^2 ge 5^2 + 5^2 > 1 + 5^2 = 1 + 5*5 = 1 + 5+5 +5 +5+5 ge 1+ 5+5 > 1+1+1 = 3$)
We can try to prove that more formally.
If $a > 1$ then $a ge 2$ and $a^m ge 2^m =(1+1)^m =_{text{by binomial theorem}} sum_{k=0}^m {mchoose k}1^k ge sum_{k=0}^m ge m+1$.
Now for positive numbers if $a|b$ then $ale b$ so $a^m|m$ means $a^m le m$ but that's a contradiction if $a > 1$.
So if $a$ is a positive integer and $a^m|m$ then $a =1$.
But what about $0$ and negative values?
Well $0 not mid k$ if $kne 0$ so that $a ne 0$.
And for negative numbers $a|b iff small{|}asmall{|}large{mid}small{|}bsmall{|}$ and if $a < 0$ then $a^m = pm |a|^m$. So this holds for negative $a$ as well.
answered Jan 21 at 5:57
fleabloodfleablood
71.9k22686
$endgroup$
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
$begingroup$
We have: if $a ge 2 implies a^x ge 2^x > ximplies a^x > ximplies a^x nmid x$. Thus if $a^x mid x implies a = 1$. Also, if $a le -2 implies |a^x| ge 2^x > x implies |a^x| > ximplies |a^x| nmid ximplies a^x nmid x$. Thus $a^x mid x implies a = -1$. Together we have: $a = pm 1$.
answered Jan 21 at 5:35
DeepSeaDeepSea
71.3k54488
$endgroup$
add a comment |
$begingroup$
We have: if $a ge 2 implies a^x ge 2^x > ximplies a^x > ximplies a^x nmid x$. Thus if $a^x mid x implies a = 1$. Also, if $a le -2 implies |a^x| ge 2^x > x implies |a^x| > ximplies |a^x| nmid ximplies a^x nmid x$. Thus $a^x mid x implies a = -1$. Together we have: $a = pm 1$.
answered Jan 21 at 5:35
DeepSeaDeepSea
71.3k54488
$endgroup$
add a comment |
$begingroup$
We have: if $a ge 2 implies a^x ge 2^x > ximplies a^x > ximplies a^x nmid x$. Thus if $a^x mid x implies a = 1$. Also, if $a le -2 implies |a^x| ge 2^x > x implies |a^x| > ximplies |a^x| nmid ximplies a^x nmid x$. Thus $a^x mid x implies a = -1$. Together we have: $a = pm 1$.
answered Jan 21 at 5:35
DeepSeaDeepSea
71.3k54488
$endgroup$
We have: if $a ge 2 implies a^x ge 2^x > ximplies a^x > ximplies a^x nmid x$. Thus if $a^x mid x implies a = 1$. Also, if $a le -2 implies |a^x| ge 2^x > x implies |a^x| > ximplies |a^x| nmid ximplies a^x nmid x$. Thus $a^x mid x implies a = -1$. Together we have: $a = pm 1$.
answered Jan 21 at 5:35
DeepSeaDeepSea
71.3k54488
answered Jan 21 at 5:35
DeepSeaDeepSea
71.3k54488
answered Jan 21 at 5:35
DeepSeaDeepSea
71.3k54488
answered Jan 21 at 5:35
DeepSeaDeepSea
71.3k54488
71.3k54488
add a comment |
add a comment |
$begingroup$
Note that if $a^x | x,$ then $ a^x leq x.$ Now consider when this is possible and try to take it from there.
answered Jan 21 at 5:35
Jack PfaffingerJack Pfaffinger
374112
$endgroup$
add a comment |
$begingroup$
Note that if $a^x | x,$ then $ a^x leq x.$ Now consider when this is possible and try to take it from there.
answered Jan 21 at 5:35
Jack PfaffingerJack Pfaffinger
374112
$endgroup$
add a comment |
$begingroup$
Note that if $a^x | x,$ then $ a^x leq x.$ Now consider when this is possible and try to take it from there.
answered Jan 21 at 5:35
Jack PfaffingerJack Pfaffinger
374112
$endgroup$
Note that if $a^x | x,$ then $ a^x leq x.$ Now consider when this is possible and try to take it from there.
answered Jan 21 at 5:35
Jack PfaffingerJack Pfaffinger
374112
answered Jan 21 at 5:35
Jack PfaffingerJack Pfaffinger
374112
answered Jan 21 at 5:35
Jack PfaffingerJack Pfaffinger
374112
answered Jan 21 at 5:35
Jack PfaffingerJack Pfaffinger
374112
374112
add a comment |
add a comment |
$begingroup$
If $a > 1$ is should be become apparent that $a^m > m$. After all $a^m = a*a*a.... $ and as $a*a = a+a+a+a... ge a+a$ we have $a*a*a* ge a+a+a+.... > 1+1+1+.... = m$.
(To make this clearer consider $5^3 = 5*5^2 = 5^2 + 5^2 + 5^2 + 5^2 +5^2 ge 5^2 + 5^2 > 1 + 5^2 = 1 + 5*5 = 1 + 5+5 +5 +5+5 ge 1+ 5+5 > 1+1+1 = 3$)
We can try to prove that more formally.
If $a > 1$ then $a ge 2$ and $a^m ge 2^m =(1+1)^m =_{text{by binomial theorem}} sum_{k=0}^m {mchoose k}1^k ge sum_{k=0}^m ge m+1$.
Now for positive numbers if $a|b$ then $ale b$ so $a^m|m$ means $a^m le m$ but that's a contradiction if $a > 1$.
So if $a$ is a positive integer and $a^m|m$ then $a =1$.
But what about $0$ and negative values?
Well $0 not mid k$ if $kne 0$ so that $a ne 0$.
And for negative numbers $a|b iff small{|}asmall{|}large{mid}small{|}bsmall{|}$ and if $a < 0$ then $a^m = pm |a|^m$. So this holds for negative $a$ as well.
answered Jan 21 at 5:57
fleabloodfleablood
71.9k22686
$endgroup$
add a comment |
$begingroup$
If $a > 1$ is should be become apparent that $a^m > m$. After all $a^m = a*a*a.... $ and as $a*a = a+a+a+a... ge a+a$ we have $a*a*a* ge a+a+a+.... > 1+1+1+.... = m$.
(To make this clearer consider $5^3 = 5*5^2 = 5^2 + 5^2 + 5^2 + 5^2 +5^2 ge 5^2 + 5^2 > 1 + 5^2 = 1 + 5*5 = 1 + 5+5 +5 +5+5 ge 1+ 5+5 > 1+1+1 = 3$)
We can try to prove that more formally.
If $a > 1$ then $a ge 2$ and $a^m ge 2^m =(1+1)^m =_{text{by binomial theorem}} sum_{k=0}^m {mchoose k}1^k ge sum_{k=0}^m ge m+1$.
Now for positive numbers if $a|b$ then $ale b$ so $a^m|m$ means $a^m le m$ but that's a contradiction if $a > 1$.
So if $a$ is a positive integer and $a^m|m$ then $a =1$.
But what about $0$ and negative values?
Well $0 not mid k$ if $kne 0$ so that $a ne 0$.
And for negative numbers $a|b iff small{|}asmall{|}large{mid}small{|}bsmall{|}$ and if $a < 0$ then $a^m = pm |a|^m$. So this holds for negative $a$ as well.
answered Jan 21 at 5:57
fleabloodfleablood
71.9k22686
$endgroup$
add a comment |
$begingroup$
If $a > 1$ is should be become apparent that $a^m > m$. After all $a^m = a*a*a.... $ and as $a*a = a+a+a+a... ge a+a$ we have $a*a*a* ge a+a+a+.... > 1+1+1+.... = m$.
(To make this clearer consider $5^3 = 5*5^2 = 5^2 + 5^2 + 5^2 + 5^2 +5^2 ge 5^2 + 5^2 > 1 + 5^2 = 1 + 5*5 = 1 + 5+5 +5 +5+5 ge 1+ 5+5 > 1+1+1 = 3$)
We can try to prove that more formally.
If $a > 1$ then $a ge 2$ and $a^m ge 2^m =(1+1)^m =_{text{by binomial theorem}} sum_{k=0}^m {mchoose k}1^k ge sum_{k=0}^m ge m+1$.
Now for positive numbers if $a|b$ then $ale b$ so $a^m|m$ means $a^m le m$ but that's a contradiction if $a > 1$.
So if $a$ is a positive integer and $a^m|m$ then $a =1$.
But what about $0$ and negative values?
Well $0 not mid k$ if $kne 0$ so that $a ne 0$.
And for negative numbers $a|b iff small{|}asmall{|}large{mid}small{|}bsmall{|}$ and if $a < 0$ then $a^m = pm |a|^m$. So this holds for negative $a$ as well.
answered Jan 21 at 5:57
fleabloodfleablood
71.9k22686
$endgroup$
If $a > 1$ is should be become apparent that $a^m > m$. After all $a^m = a*a*a.... $ and as $a*a = a+a+a+a... ge a+a$ we have $a*a*a* ge a+a+a+.... > 1+1+1+.... = m$.
(To make this clearer consider $5^3 = 5*5^2 = 5^2 + 5^2 + 5^2 + 5^2 +5^2 ge 5^2 + 5^2 > 1 + 5^2 = 1 + 5*5 = 1 + 5+5 +5 +5+5 ge 1+ 5+5 > 1+1+1 = 3$)
We can try to prove that more formally.
If $a > 1$ then $a ge 2$ and $a^m ge 2^m =(1+1)^m =_{text{by binomial theorem}} sum_{k=0}^m {mchoose k}1^k ge sum_{k=0}^m ge m+1$.
Now for positive numbers if $a|b$ then $ale b$ so $a^m|m$ means $a^m le m$ but that's a contradiction if $a > 1$.
So if $a$ is a positive integer and $a^m|m$ then $a =1$.
But what about $0$ and negative values?
Well $0 not mid k$ if $kne 0$ so that $a ne 0$.
And for negative numbers $a|b iff small{|}asmall{|}large{mid}small{|}bsmall{|}$ and if $a < 0$ then $a^m = pm |a|^m$. So this holds for negative $a$ as well.
answered Jan 21 at 5:57
fleabloodfleablood
71.9k22686
answered Jan 21 at 5:57
fleabloodfleablood
71.9k22686
answered Jan 21 at 5:57
fleabloodfleablood
71.9k22686
answered Jan 21 at 5:57
fleabloodfleablood
71.9k22686
71.9k22686
add a comment |
add a comment |
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3
$begingroup$
Why do you have both $forall x in mathbb{Z}$ and $forall x in mathbb{N}$? I don't understand including both in the problem. Isn't the problem statement the same if you delete one of them?
$endgroup$
– David G. Stork
Jan 21 at 5:30