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Show that if $n > 1$ is not a $k$-th power, then its $k$-th root is irrational, i.e. not of the form $u/v$...
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This question already has an answer here:
How to prove: if $a,b in mathbb N$, then $a^{1/b}$ is an integer or an irrational number?
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any solutions or suggestions on this one?
An integer $n$ is called a $k$-th power if $n = m^k$ for some $m ∈ Z$.
Show that if $n > 1$ is not a $k$-th power, then its $k$-th root is irrational, i.e. not of the form $u/v$ for $u, v ∈ N$.
discrete-mathematics
asked Jan 30 at 22:38
SomethingSomething
817
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marked as duplicate by Robert Israel, Lord Shark the Unknown, Lee David Chung Lin, Leucippus, Martin Argerami Jan 31 at 4:35
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
How to prove: if $a,b in mathbb N$, then $a^{1/b}$ is an integer or an irrational number?
12 answers
any solutions or suggestions on this one?
An integer $n$ is called a $k$-th power if $n = m^k$ for some $m ∈ Z$.
Show that if $n > 1$ is not a $k$-th power, then its $k$-th root is irrational, i.e. not of the form $u/v$ for $u, v ∈ N$.
discrete-mathematics
asked Jan 30 at 22:38
SomethingSomething
817
$endgroup$
marked as duplicate by Robert Israel, Lord Shark the Unknown, Lee David Chung Lin, Leucippus, Martin Argerami Jan 31 at 4:35
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
How to prove: if $a,b in mathbb N$, then $a^{1/b}$ is an integer or an irrational number?
12 answers
any solutions or suggestions on this one?
An integer $n$ is called a $k$-th power if $n = m^k$ for some $m ∈ Z$.
Show that if $n > 1$ is not a $k$-th power, then its $k$-th root is irrational, i.e. not of the form $u/v$ for $u, v ∈ N$.
discrete-mathematics
asked Jan 30 at 22:38
SomethingSomething
817
$endgroup$
This question already has an answer here:
How to prove: if $a,b in mathbb N$, then $a^{1/b}$ is an integer or an irrational number?
12 answers
any solutions or suggestions on this one?
An integer $n$ is called a $k$-th power if $n = m^k$ for some $m ∈ Z$.
Show that if $n > 1$ is not a $k$-th power, then its $k$-th root is irrational, i.e. not of the form $u/v$ for $u, v ∈ N$.
This question already has an answer here:
How to prove: if $a,b in mathbb N$, then $a^{1/b}$ is an integer or an irrational number?
12 answers
discrete-mathematics
discrete-mathematics
asked Jan 30 at 22:38
SomethingSomething
817
asked Jan 30 at 22:38
SomethingSomething
817
asked Jan 30 at 22:38
SomethingSomething
817
asked Jan 30 at 22:38
SomethingSomething
817
asked Jan 30 at 22:38
SomethingSomething
817
817
marked as duplicate by Robert Israel, Lord Shark the Unknown, Lee David Chung Lin, Leucippus, Martin Argerami Jan 31 at 4:35
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Robert Israel, Lord Shark the Unknown, Lee David Chung Lin, Leucippus, Martin Argerami Jan 31 at 4:35
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
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6 Answers
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Assume $frac{u}{v}$ is in its lowest terms, i.e. $u$ and $v$ have no common factors, then $frac{u^k}{v^k}$ is in its lowest terms, and will not be an integer unless $v=1$.
answered Jan 30 at 22:49
herb steinbergherb steinberg
3,1132311
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If $sqrt[k]{n} = u/v$, then $v^k n = u^k$. Consider now the power of each prime $p$ that appears in the factorization on both sides. You get a number of the form $ak+c$ on the LHS and a number of the form $bk$ on the $RHS$. Therefore, $ak+c=bk$ and so $c$ is a multiple of $k$. This means that $n$ is a $k$-th power.
answered Jan 30 at 22:47
lhflhf
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Claim 1: Let $c$ and $d$ be integers. If there is a prime $p ge 2$ that divides $d$ but not $c$, then $frac{c}{d}$ is not an integer.
Suppose the $k$-th power of $n$ is rational but nonintegral. Then let $a$ and $b$ be relatively prime integers satisfying $(frac{a}{b})^k = m$. Then as $a$ and $b$ are relatively prime, there is a prime $p ge 2$ that divides $b$ but not $a$. Thus letting $c=a^k$ and $d=b^k$, that prime $p$ divides $d$ but not $c$ [Note that the prime factors of $m^k$ are the same as that of $m$ for every integer $m$ and every positive integer $k$.] Thus $frac{c}{d} = frac{a^k}{b^k} = (frac{a}{b})^k$ cannot be the integer $n$ after all by Claim 1.
answered Jan 30 at 22:47
MikeMike
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How about proving by the contrapositive. We only care about $min Bbb Q / {Bbb Z}$, so let:
$$m=frac ab, a,b text{ coprime }, bneq 1$$
Then $$m^k =frac{a^k}{b^k}$$
Which is also irreducible. In other words.
$$min Bbb Q / {Bbb Z} implies m^kin Bbb Q / {Bbb Z}$$
We thusly rule out the possibility $m$ is a rational non-integer given integer $m^k$, so it must either be an integer or irrational.
answered Jan 30 at 22:50
Rhys HughesRhys Hughes
7,0501630
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Your claim is equivalent to this:
If the $k$-th root of $n$ is rational, then $n$ is a $k$-th power.
So let $(u/v)^k = n$, i.e. $u^k = n v^k$. We may assume that $text{gcd}(u,v) = 1$. Then $u^k$ must divide $n$ which means $n = m u^k$ for some $m in mathbb{N}$. This yields $u^k = m u^k v^k$, i.e. $1 = m v^k$ which is possible only when $m = 1$ and $v = 1$. Hence $u/v = u in mathbb{N}$.
answered Jan 30 at 22:56
Paul FrostPaul Frost
12.2k3935
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By contrapositive, without using prime factorisation:
Set $x = sqrt[uproot{1}leftroot{-1}n]{a}$, and denote $y$ its decimal part. Suppose $x$ is rational.
Claim: $y = 0$ (so that $sqrt[uproot{1}leftroot{-1}n]{a}$ is an integer).
Let $q$ be the smallest positive integer such that each $qx^i$, for $i = 1, dots, n-1$, is an integer. Note that $q' = qy$ is an integer, since $qx$ is.
On the other hand,
$$ q'x^i = q( x-lfloor xrfloor)x^i = qx^{i + 1} - lfloor xrfloor qx^i.$$
By hypothesis, in this formula, $qx^i$ and $qx^{i + 1}$ are integers if $i < n-1$ – and if $i = n-1$, $qx^{i + 1} = qa$ is an integer too. There results that each $q'x^i$ is an integer.
However $q$ is the smallest positive integer which has this property, and $0 ≤ q' < q$,so necessarily $q' = 0$, whence $y = 0$.
answered Jan 30 at 23:39
BernardBernard
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Assume $frac{u}{v}$ is in its lowest terms, i.e. $u$ and $v$ have no common factors, then $frac{u^k}{v^k}$ is in its lowest terms, and will not be an integer unless $v=1$.
answered Jan 30 at 22:49
herb steinbergherb steinberg
3,1132311
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add a comment |
$begingroup$
Assume $frac{u}{v}$ is in its lowest terms, i.e. $u$ and $v$ have no common factors, then $frac{u^k}{v^k}$ is in its lowest terms, and will not be an integer unless $v=1$.
answered Jan 30 at 22:49
herb steinbergherb steinberg
3,1132311
$endgroup$
add a comment |
$begingroup$
Assume $frac{u}{v}$ is in its lowest terms, i.e. $u$ and $v$ have no common factors, then $frac{u^k}{v^k}$ is in its lowest terms, and will not be an integer unless $v=1$.
answered Jan 30 at 22:49
herb steinbergherb steinberg
3,1132311
$endgroup$
Assume $frac{u}{v}$ is in its lowest terms, i.e. $u$ and $v$ have no common factors, then $frac{u^k}{v^k}$ is in its lowest terms, and will not be an integer unless $v=1$.
answered Jan 30 at 22:49
herb steinbergherb steinberg
3,1132311
answered Jan 30 at 22:49
herb steinbergherb steinberg
3,1132311
answered Jan 30 at 22:49
herb steinbergherb steinberg
3,1132311
answered Jan 30 at 22:49
herb steinbergherb steinberg
3,1132311
3,1132311
add a comment |
add a comment |
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If $sqrt[k]{n} = u/v$, then $v^k n = u^k$. Consider now the power of each prime $p$ that appears in the factorization on both sides. You get a number of the form $ak+c$ on the LHS and a number of the form $bk$ on the $RHS$. Therefore, $ak+c=bk$ and so $c$ is a multiple of $k$. This means that $n$ is a $k$-th power.
answered Jan 30 at 22:47
lhflhf
167k11172404
$endgroup$
add a comment |
$begingroup$
If $sqrt[k]{n} = u/v$, then $v^k n = u^k$. Consider now the power of each prime $p$ that appears in the factorization on both sides. You get a number of the form $ak+c$ on the LHS and a number of the form $bk$ on the $RHS$. Therefore, $ak+c=bk$ and so $c$ is a multiple of $k$. This means that $n$ is a $k$-th power.
answered Jan 30 at 22:47
lhflhf
167k11172404
$endgroup$
add a comment |
$begingroup$
If $sqrt[k]{n} = u/v$, then $v^k n = u^k$. Consider now the power of each prime $p$ that appears in the factorization on both sides. You get a number of the form $ak+c$ on the LHS and a number of the form $bk$ on the $RHS$. Therefore, $ak+c=bk$ and so $c$ is a multiple of $k$. This means that $n$ is a $k$-th power.
answered Jan 30 at 22:47
lhflhf
167k11172404
$endgroup$
If $sqrt[k]{n} = u/v$, then $v^k n = u^k$. Consider now the power of each prime $p$ that appears in the factorization on both sides. You get a number of the form $ak+c$ on the LHS and a number of the form $bk$ on the $RHS$. Therefore, $ak+c=bk$ and so $c$ is a multiple of $k$. This means that $n$ is a $k$-th power.
answered Jan 30 at 22:47
lhflhf
167k11172404
answered Jan 30 at 22:47
lhflhf
167k11172404
answered Jan 30 at 22:47
lhflhf
167k11172404
answered Jan 30 at 22:47
lhflhf
167k11172404
167k11172404
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Claim 1: Let $c$ and $d$ be integers. If there is a prime $p ge 2$ that divides $d$ but not $c$, then $frac{c}{d}$ is not an integer.
Suppose the $k$-th power of $n$ is rational but nonintegral. Then let $a$ and $b$ be relatively prime integers satisfying $(frac{a}{b})^k = m$. Then as $a$ and $b$ are relatively prime, there is a prime $p ge 2$ that divides $b$ but not $a$. Thus letting $c=a^k$ and $d=b^k$, that prime $p$ divides $d$ but not $c$ [Note that the prime factors of $m^k$ are the same as that of $m$ for every integer $m$ and every positive integer $k$.] Thus $frac{c}{d} = frac{a^k}{b^k} = (frac{a}{b})^k$ cannot be the integer $n$ after all by Claim 1.
answered Jan 30 at 22:47
MikeMike
4,611512
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Claim 1: Let $c$ and $d$ be integers. If there is a prime $p ge 2$ that divides $d$ but not $c$, then $frac{c}{d}$ is not an integer.
Suppose the $k$-th power of $n$ is rational but nonintegral. Then let $a$ and $b$ be relatively prime integers satisfying $(frac{a}{b})^k = m$. Then as $a$ and $b$ are relatively prime, there is a prime $p ge 2$ that divides $b$ but not $a$. Thus letting $c=a^k$ and $d=b^k$, that prime $p$ divides $d$ but not $c$ [Note that the prime factors of $m^k$ are the same as that of $m$ for every integer $m$ and every positive integer $k$.] Thus $frac{c}{d} = frac{a^k}{b^k} = (frac{a}{b})^k$ cannot be the integer $n$ after all by Claim 1.
answered Jan 30 at 22:47
MikeMike
4,611512
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Claim 1: Let $c$ and $d$ be integers. If there is a prime $p ge 2$ that divides $d$ but not $c$, then $frac{c}{d}$ is not an integer.
Suppose the $k$-th power of $n$ is rational but nonintegral. Then let $a$ and $b$ be relatively prime integers satisfying $(frac{a}{b})^k = m$. Then as $a$ and $b$ are relatively prime, there is a prime $p ge 2$ that divides $b$ but not $a$. Thus letting $c=a^k$ and $d=b^k$, that prime $p$ divides $d$ but not $c$ [Note that the prime factors of $m^k$ are the same as that of $m$ for every integer $m$ and every positive integer $k$.] Thus $frac{c}{d} = frac{a^k}{b^k} = (frac{a}{b})^k$ cannot be the integer $n$ after all by Claim 1.
answered Jan 30 at 22:47
MikeMike
4,611512
$endgroup$
Claim 1: Let $c$ and $d$ be integers. If there is a prime $p ge 2$ that divides $d$ but not $c$, then $frac{c}{d}$ is not an integer.
Suppose the $k$-th power of $n$ is rational but nonintegral. Then let $a$ and $b$ be relatively prime integers satisfying $(frac{a}{b})^k = m$. Then as $a$ and $b$ are relatively prime, there is a prime $p ge 2$ that divides $b$ but not $a$. Thus letting $c=a^k$ and $d=b^k$, that prime $p$ divides $d$ but not $c$ [Note that the prime factors of $m^k$ are the same as that of $m$ for every integer $m$ and every positive integer $k$.] Thus $frac{c}{d} = frac{a^k}{b^k} = (frac{a}{b})^k$ cannot be the integer $n$ after all by Claim 1.
answered Jan 30 at 22:47
MikeMike
4,611512
answered Jan 30 at 22:47
MikeMike
4,611512
answered Jan 30 at 22:47
MikeMike
4,611512
answered Jan 30 at 22:47
MikeMike
4,611512
4,611512
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How about proving by the contrapositive. We only care about $min Bbb Q / {Bbb Z}$, so let:
$$m=frac ab, a,b text{ coprime }, bneq 1$$
Then $$m^k =frac{a^k}{b^k}$$
Which is also irreducible. In other words.
$$min Bbb Q / {Bbb Z} implies m^kin Bbb Q / {Bbb Z}$$
We thusly rule out the possibility $m$ is a rational non-integer given integer $m^k$, so it must either be an integer or irrational.
answered Jan 30 at 22:50
Rhys HughesRhys Hughes
7,0501630
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add a comment |
$begingroup$
How about proving by the contrapositive. We only care about $min Bbb Q / {Bbb Z}$, so let:
$$m=frac ab, a,b text{ coprime }, bneq 1$$
Then $$m^k =frac{a^k}{b^k}$$
Which is also irreducible. In other words.
$$min Bbb Q / {Bbb Z} implies m^kin Bbb Q / {Bbb Z}$$
We thusly rule out the possibility $m$ is a rational non-integer given integer $m^k$, so it must either be an integer or irrational.
answered Jan 30 at 22:50
Rhys HughesRhys Hughes
7,0501630
$endgroup$
add a comment |
$begingroup$
How about proving by the contrapositive. We only care about $min Bbb Q / {Bbb Z}$, so let:
$$m=frac ab, a,b text{ coprime }, bneq 1$$
Then $$m^k =frac{a^k}{b^k}$$
Which is also irreducible. In other words.
$$min Bbb Q / {Bbb Z} implies m^kin Bbb Q / {Bbb Z}$$
We thusly rule out the possibility $m$ is a rational non-integer given integer $m^k$, so it must either be an integer or irrational.
answered Jan 30 at 22:50
Rhys HughesRhys Hughes
7,0501630
$endgroup$
How about proving by the contrapositive. We only care about $min Bbb Q / {Bbb Z}$, so let:
$$m=frac ab, a,b text{ coprime }, bneq 1$$
Then $$m^k =frac{a^k}{b^k}$$
Which is also irreducible. In other words.
$$min Bbb Q / {Bbb Z} implies m^kin Bbb Q / {Bbb Z}$$
We thusly rule out the possibility $m$ is a rational non-integer given integer $m^k$, so it must either be an integer or irrational.
answered Jan 30 at 22:50
Rhys HughesRhys Hughes
7,0501630
answered Jan 30 at 22:50
Rhys HughesRhys Hughes
7,0501630
answered Jan 30 at 22:50
Rhys HughesRhys Hughes
7,0501630
answered Jan 30 at 22:50
Rhys HughesRhys Hughes
7,0501630
7,0501630
add a comment |
add a comment |
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Your claim is equivalent to this:
If the $k$-th root of $n$ is rational, then $n$ is a $k$-th power.
So let $(u/v)^k = n$, i.e. $u^k = n v^k$. We may assume that $text{gcd}(u,v) = 1$. Then $u^k$ must divide $n$ which means $n = m u^k$ for some $m in mathbb{N}$. This yields $u^k = m u^k v^k$, i.e. $1 = m v^k$ which is possible only when $m = 1$ and $v = 1$. Hence $u/v = u in mathbb{N}$.
answered Jan 30 at 22:56
Paul FrostPaul Frost
12.2k3935
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Your claim is equivalent to this:
If the $k$-th root of $n$ is rational, then $n$ is a $k$-th power.
So let $(u/v)^k = n$, i.e. $u^k = n v^k$. We may assume that $text{gcd}(u,v) = 1$. Then $u^k$ must divide $n$ which means $n = m u^k$ for some $m in mathbb{N}$. This yields $u^k = m u^k v^k$, i.e. $1 = m v^k$ which is possible only when $m = 1$ and $v = 1$. Hence $u/v = u in mathbb{N}$.
answered Jan 30 at 22:56
Paul FrostPaul Frost
12.2k3935
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add a comment |
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Your claim is equivalent to this:
If the $k$-th root of $n$ is rational, then $n$ is a $k$-th power.
So let $(u/v)^k = n$, i.e. $u^k = n v^k$. We may assume that $text{gcd}(u,v) = 1$. Then $u^k$ must divide $n$ which means $n = m u^k$ for some $m in mathbb{N}$. This yields $u^k = m u^k v^k$, i.e. $1 = m v^k$ which is possible only when $m = 1$ and $v = 1$. Hence $u/v = u in mathbb{N}$.
answered Jan 30 at 22:56
Paul FrostPaul Frost
12.2k3935
$endgroup$
Your claim is equivalent to this:
If the $k$-th root of $n$ is rational, then $n$ is a $k$-th power.
So let $(u/v)^k = n$, i.e. $u^k = n v^k$. We may assume that $text{gcd}(u,v) = 1$. Then $u^k$ must divide $n$ which means $n = m u^k$ for some $m in mathbb{N}$. This yields $u^k = m u^k v^k$, i.e. $1 = m v^k$ which is possible only when $m = 1$ and $v = 1$. Hence $u/v = u in mathbb{N}$.
answered Jan 30 at 22:56
Paul FrostPaul Frost
12.2k3935
answered Jan 30 at 22:56
Paul FrostPaul Frost
12.2k3935
answered Jan 30 at 22:56
Paul FrostPaul Frost
12.2k3935
answered Jan 30 at 22:56
Paul FrostPaul Frost
12.2k3935
12.2k3935
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By contrapositive, without using prime factorisation:
Set $x = sqrt[uproot{1}leftroot{-1}n]{a}$, and denote $y$ its decimal part. Suppose $x$ is rational.
Claim: $y = 0$ (so that $sqrt[uproot{1}leftroot{-1}n]{a}$ is an integer).
Let $q$ be the smallest positive integer such that each $qx^i$, for $i = 1, dots, n-1$, is an integer. Note that $q' = qy$ is an integer, since $qx$ is.
On the other hand,
$$ q'x^i = q( x-lfloor xrfloor)x^i = qx^{i + 1} - lfloor xrfloor qx^i.$$
By hypothesis, in this formula, $qx^i$ and $qx^{i + 1}$ are integers if $i < n-1$ – and if $i = n-1$, $qx^{i + 1} = qa$ is an integer too. There results that each $q'x^i$ is an integer.
However $q$ is the smallest positive integer which has this property, and $0 ≤ q' < q$,so necessarily $q' = 0$, whence $y = 0$.
answered Jan 30 at 23:39
BernardBernard
124k741116
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add a comment |
$begingroup$
By contrapositive, without using prime factorisation:
Set $x = sqrt[uproot{1}leftroot{-1}n]{a}$, and denote $y$ its decimal part. Suppose $x$ is rational.
Claim: $y = 0$ (so that $sqrt[uproot{1}leftroot{-1}n]{a}$ is an integer).
Let $q$ be the smallest positive integer such that each $qx^i$, for $i = 1, dots, n-1$, is an integer. Note that $q' = qy$ is an integer, since $qx$ is.
On the other hand,
$$ q'x^i = q( x-lfloor xrfloor)x^i = qx^{i + 1} - lfloor xrfloor qx^i.$$
By hypothesis, in this formula, $qx^i$ and $qx^{i + 1}$ are integers if $i < n-1$ – and if $i = n-1$, $qx^{i + 1} = qa$ is an integer too. There results that each $q'x^i$ is an integer.
However $q$ is the smallest positive integer which has this property, and $0 ≤ q' < q$,so necessarily $q' = 0$, whence $y = 0$.
answered Jan 30 at 23:39
BernardBernard
124k741116
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add a comment |
$begingroup$
By contrapositive, without using prime factorisation:
Set $x = sqrt[uproot{1}leftroot{-1}n]{a}$, and denote $y$ its decimal part. Suppose $x$ is rational.
Claim: $y = 0$ (so that $sqrt[uproot{1}leftroot{-1}n]{a}$ is an integer).
Let $q$ be the smallest positive integer such that each $qx^i$, for $i = 1, dots, n-1$, is an integer. Note that $q' = qy$ is an integer, since $qx$ is.
On the other hand,
$$ q'x^i = q( x-lfloor xrfloor)x^i = qx^{i + 1} - lfloor xrfloor qx^i.$$
By hypothesis, in this formula, $qx^i$ and $qx^{i + 1}$ are integers if $i < n-1$ – and if $i = n-1$, $qx^{i + 1} = qa$ is an integer too. There results that each $q'x^i$ is an integer.
However $q$ is the smallest positive integer which has this property, and $0 ≤ q' < q$,so necessarily $q' = 0$, whence $y = 0$.
answered Jan 30 at 23:39
BernardBernard
124k741116
$endgroup$
By contrapositive, without using prime factorisation:
Set $x = sqrt[uproot{1}leftroot{-1}n]{a}$, and denote $y$ its decimal part. Suppose $x$ is rational.
Claim: $y = 0$ (so that $sqrt[uproot{1}leftroot{-1}n]{a}$ is an integer).
Let $q$ be the smallest positive integer such that each $qx^i$, for $i = 1, dots, n-1$, is an integer. Note that $q' = qy$ is an integer, since $qx$ is.
On the other hand,
$$ q'x^i = q( x-lfloor xrfloor)x^i = qx^{i + 1} - lfloor xrfloor qx^i.$$
By hypothesis, in this formula, $qx^i$ and $qx^{i + 1}$ are integers if $i < n-1$ – and if $i = n-1$, $qx^{i + 1} = qa$ is an integer too. There results that each $q'x^i$ is an integer.
However $q$ is the smallest positive integer which has this property, and $0 ≤ q' < q$,so necessarily $q' = 0$, whence $y = 0$.
answered Jan 30 at 23:39
BernardBernard
124k741116
answered Jan 30 at 23:39
BernardBernard
124k741116
answered Jan 30 at 23:39
BernardBernard
124k741116
answered Jan 30 at 23:39
BernardBernard
124k741116
124k741116
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