Mixing
Square root of 6 proof rationality
I was proving $sqrt 6 notin Bbb Q$, by assuming its negation and stating that: $exists (p,q) in Bbb Z times Bbb Z^*/ gcd(p,q) = 1$, and $sqrt 6 = (p/q)$.
$implies p^2 = 2 times 3q^2 implies exists k in Bbb Z; p = 2k implies 2k^2 = 3q^2$ and found two possible cases, either $q$ is even or odd, if even we get contradiction that $gcd(p, q) neq 1$, if odd we get contradiction that $2k^2 = 3q^2$.
Is it a right path for reasoning it?
rational-numbers
edited Nov 17 '18 at 23:00
Mr. Brooks
36111237
asked Nov 15 '18 at 1:12
J.Moh
395
add a comment |
I was proving $sqrt 6 notin Bbb Q$, by assuming its negation and stating that: $exists (p,q) in Bbb Z times Bbb Z^*/ gcd(p,q) = 1$, and $sqrt 6 = (p/q)$.
$implies p^2 = 2 times 3q^2 implies exists k in Bbb Z; p = 2k implies 2k^2 = 3q^2$ and found two possible cases, either $q$ is even or odd, if even we get contradiction that $gcd(p, q) neq 1$, if odd we get contradiction that $2k^2 = 3q^2$.
Is it a right path for reasoning it?
rational-numbers
edited Nov 17 '18 at 23:00
Mr. Brooks
36111237
asked Nov 15 '18 at 1:12
J.Moh
395
Is the $k$ a typo? What is $p, k$ and $q$? Why is $q$ being even a condradiction with $gcd(p,q) = 1$? And why is $2k^2 = 3q^2$ a contradiction? You just said $2k^2 = 3q^2$ so that isn't a contradiction? You are on the right track but you haven't explained any of your arguments.
– fleablood
Nov 15 '18 at 2:13
Why is $q$ even a contradiction? What if $k$ or $p$ (whats the difference) is odd?
– fleablood
Nov 15 '18 at 2:20
I rearranged it!
– J.Moh
Nov 15 '18 at 3:51
add a comment |
I was proving $sqrt 6 notin Bbb Q$, by assuming its negation and stating that: $exists (p,q) in Bbb Z times Bbb Z^*/ gcd(p,q) = 1$, and $sqrt 6 = (p/q)$.
$implies p^2 = 2 times 3q^2 implies exists k in Bbb Z; p = 2k implies 2k^2 = 3q^2$ and found two possible cases, either $q$ is even or odd, if even we get contradiction that $gcd(p, q) neq 1$, if odd we get contradiction that $2k^2 = 3q^2$.
Is it a right path for reasoning it?
rational-numbers
edited Nov 17 '18 at 23:00
Mr. Brooks
36111237
asked Nov 15 '18 at 1:12
J.Moh
395
I was proving $sqrt 6 notin Bbb Q$, by assuming its negation and stating that: $exists (p,q) in Bbb Z times Bbb Z^*/ gcd(p,q) = 1$, and $sqrt 6 = (p/q)$.
$implies p^2 = 2 times 3q^2 implies exists k in Bbb Z; p = 2k implies 2k^2 = 3q^2$ and found two possible cases, either $q$ is even or odd, if even we get contradiction that $gcd(p, q) neq 1$, if odd we get contradiction that $2k^2 = 3q^2$.
Is it a right path for reasoning it?
rational-numbers
rational-numbers
edited Nov 17 '18 at 23:00
Mr. Brooks
36111237
asked Nov 15 '18 at 1:12
J.Moh
395
edited Nov 17 '18 at 23:00
Mr. Brooks
36111237
asked Nov 15 '18 at 1:12
J.Moh
395
edited Nov 17 '18 at 23:00
Mr. Brooks
36111237
edited Nov 17 '18 at 23:00
Mr. Brooks
36111237
edited Nov 17 '18 at 23:00
Mr. Brooks
36111237
36111237
asked Nov 15 '18 at 1:12
J.Moh
395
asked Nov 15 '18 at 1:12
J.Moh
395
asked Nov 15 '18 at 1:12
J.Moh
395
395
Is the $k$ a typo? What is $p, k$ and $q$? Why is $q$ being even a condradiction with $gcd(p,q) = 1$? And why is $2k^2 = 3q^2$ a contradiction? You just said $2k^2 = 3q^2$ so that isn't a contradiction? You are on the right track but you haven't explained any of your arguments.
– fleablood
Nov 15 '18 at 2:13
Why is $q$ even a contradiction? What if $k$ or $p$ (whats the difference) is odd?
– fleablood
Nov 15 '18 at 2:20
I rearranged it!
– J.Moh
Nov 15 '18 at 3:51
add a comment |
Is the $k$ a typo? What is $p, k$ and $q$? Why is $q$ being even a condradiction with $gcd(p,q) = 1$? And why is $2k^2 = 3q^2$ a contradiction? You just said $2k^2 = 3q^2$ so that isn't a contradiction? You are on the right track but you haven't explained any of your arguments.
– fleablood
Nov 15 '18 at 2:13
Why is $q$ even a contradiction? What if $k$ or $p$ (whats the difference) is odd?
– fleablood
Nov 15 '18 at 2:20
I rearranged it!
– J.Moh
Nov 15 '18 at 3:51
Is the $k$ a typo? What is $p, k$ and $q$? Why is $q$ being even a condradiction with $gcd(p,q) = 1$? And why is $2k^2 = 3q^2$ a contradiction? You just said $2k^2 = 3q^2$ so that isn't a contradiction? You are on the right track but you haven't explained any of your arguments.
– fleablood
Nov 15 '18 at 2:13
Is the $k$ a typo? What is $p, k$ and $q$? Why is $q$ being even a condradiction with $gcd(p,q) = 1$? And why is $2k^2 = 3q^2$ a contradiction? You just said $2k^2 = 3q^2$ so that isn't a contradiction? You are on the right track but you haven't explained any of your arguments.
– fleablood
Nov 15 '18 at 2:13
Why is $q$ even a contradiction? What if $k$ or $p$ (whats the difference) is odd?
– fleablood
Nov 15 '18 at 2:20
Why is $q$ even a contradiction? What if $k$ or $p$ (whats the difference) is odd?
– fleablood
Nov 15 '18 at 2:20
I rearranged it!
– J.Moh
Nov 15 '18 at 3:51
I rearranged it!
– J.Moh
Nov 15 '18 at 3:51
add a comment |
4 Answers
4
active
oldest
votes
How about a proof by descent?
First show that $2^2<6<3^2$. Then if $sqrt{6}$ is to be rational it must have a form $p/q$ where $p,qin mathbb{Z}, q>0, 2q<p<3q$. By simple algebra the square root is also equal to $6q/p$, thus
$p/q=6q/ptext{.....Eq. 1}$.
Now if $a/b=c/d$ then also
$a/b=(ma+nc)/(mb+nd)$
for any coefficients $m,n$ where the denominator is nonzero. In particular, Eq. 1 implies
$p/q=(3p-6q)/(3q-p)text{.....Eq. 2}$
where we already have $2q<p<3q$ and thus $0<3q-p<q$. So the proposed rational fraction $p/q$ must be equal to an alternative rational fraction with a smaller positive denominator. This causes an infinite descent contradiction forcing the assumption of a rational value to be false.
We can form a similar proof for the square root of any natural number that is not a squared integer. "Not a squared integer" is needed because the square root must be strictly between two adjacent integers to obtain a descent of positive denominators.
answered Nov 18 '18 at 2:40
Oscar Lanzi
12.1k12036
add a comment |
Assume $sqrt{6} = {a over b}$ with $a, b in mathbb{Z}$ . This implies
$$2 cdot 3 b^2 = a^2$$
Use the fundamental theorem of algebra to decompose both sides into a unique prime factorization. There are an even number of factors of $2$ on the rhs and an odd number of factors of $2$ on the lhs... a contradition. Same for factors of $3$.
Thus $sqrt{6} neq {a over b}$, i.e., is irrational.
answered Nov 15 '18 at 2:32
David G. Stork
9,87021232
Yeah, I tried to reach it without prime factorisation. A second look?
– J.Moh
Nov 15 '18 at 2:42
You are considering the value of the rhs and lhs of the equation. I am instead looking at the number of prime factors. Frankly, I think that is so much more elegant... but I suppose this is a matter of taste. Mine applies to $sqrt{5/3}$ whereas yours doesn't... at least not as directly.
– David G. Stork
Nov 15 '18 at 2:47
Our professor used the value way, so I am following his steps for this, though, thanks!
– J.Moh
Nov 15 '18 at 2:50
Show him/her the better way! And stand out as a student!
– David G. Stork
Nov 15 '18 at 2:51
Well, that doesn t apply in Morocco. Pray for me I win visa lottery so I can join a better faculty in the US :)
– J.Moh
Nov 15 '18 at 3:49
|
show 7 more comments
I'll play more generally
and see what happens.
Suppose
$sqrt{m}
=dfrac{u}{v}
$
where
$m = prod_{p in P} p^{m_i},
u = prod_{p in P} p^{u_i},
v = prod_{p in P} p^{v_i}
$.
Then
$prod_{p in P} p^{m_i}
=dfrac{u^2}{v^2}
=dfrac{ prod_{p in P} p^{2u_i}}{prod_{p in P} p^{2v_i}}
$
so
$prod_{p in P} p^{m_i}prod_{p in P} p^{2v_i}
=prod_{p in P} p^{2u_i}
$
or
$prod_{p in P} p^{m_i+2v_i}
=prod_{p in P} p^{2u_i}
$.
By unique factorization,
$m_i+2v_i
=2u_i
$,
so
$m_i
=2u_i-2v_i
=2(u_i-v_i)
$.
Therefore
$m
=prod_{p in P} p^{m_i}
=prod_{p in P} p^{2(u_i-v_i)}
=left(prod_{p in P} p^{u_i-v_i}right)^2
$
so $m$ is a perfect square.
Therefore
the square root of an integer
is rational
only if
it is a square.
I'll now try to generalize this
to $k$-th roots,
with as much cut-and-paste
as possible.
Suppose
$sqrt[k]{m}
=dfrac{u}{v}
$
where
$m = prod_{p in P} p^{m_i},
u = prod_{p in P} p^{u_i},
v = prod_{p in P} p^{v_i}
$.
Then
$prod_{p in P} p^{m_i}
=dfrac{u^k}{v^k}
=dfrac{ prod_{p in P} p^{ku_i}}{prod_{p in P} p^{kv_i}}
$
so
$prod_{p in P} p^{m_i}prod_{p in P} p^{kv_i}
=prod_{p in P} p^{ku_i}
$
or
$prod_{p in P} p^{m_i+kv_i}
=prod_{p in P} p^{ku_i}
$.
By unique factorization,
$m_i+kv_i
=ku_i
$,
so
$m_i
=ku_i-kv_i
=k(u_i-v_i)
$.
Therefore
$m
=prod_{p in P} p^{m_i}
=prod_{p in P} p^{k(u_i-v_i)}
=left(prod_{p in P} p^{u_i-v_i}right)^k
$
so $m$ is a perfect $k$-th power.
Therefore
the $k$-th root of an integer
is rational
only if
it is a $k$-th power.
answered Nov 15 '18 at 1:37
marty cohen
72.6k549127
add a comment |
When you get $2p^2 = 3q^2$ that means that $2|q^2$ so $2|q$ and that $3|p^2$ so $3|p$.
Then if you replace $p = 3p'$ for some integer $p'$ and $q=2q'$ for some integer $q'$ you get
$2(3p')^2 = 3(2q')^2$
$18p'^2 = 12q'^2$
$3p'^3 = 2q'^2$.
Can you finish from there?
This assumes you know Euclid's lemma that 1) prime numbers exist and 2) if $p$ is prime and $p|a*b$ then either $p|a$ or $p|b$ (or both).
answered Nov 15 '18 at 2:28
fleablood
68.3k22685
I tried to reach it without prime factorisation, can you have a second look?
– J.Moh
Nov 15 '18 at 2:42
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
How about a proof by descent?
First show that $2^2<6<3^2$. Then if $sqrt{6}$ is to be rational it must have a form $p/q$ where $p,qin mathbb{Z}, q>0, 2q<p<3q$. By simple algebra the square root is also equal to $6q/p$, thus
$p/q=6q/ptext{.....Eq. 1}$.
Now if $a/b=c/d$ then also
$a/b=(ma+nc)/(mb+nd)$
for any coefficients $m,n$ where the denominator is nonzero. In particular, Eq. 1 implies
$p/q=(3p-6q)/(3q-p)text{.....Eq. 2}$
where we already have $2q<p<3q$ and thus $0<3q-p<q$. So the proposed rational fraction $p/q$ must be equal to an alternative rational fraction with a smaller positive denominator. This causes an infinite descent contradiction forcing the assumption of a rational value to be false.
We can form a similar proof for the square root of any natural number that is not a squared integer. "Not a squared integer" is needed because the square root must be strictly between two adjacent integers to obtain a descent of positive denominators.
answered Nov 18 '18 at 2:40
Oscar Lanzi
12.1k12036
add a comment |
How about a proof by descent?
First show that $2^2<6<3^2$. Then if $sqrt{6}$ is to be rational it must have a form $p/q$ where $p,qin mathbb{Z}, q>0, 2q<p<3q$. By simple algebra the square root is also equal to $6q/p$, thus
$p/q=6q/ptext{.....Eq. 1}$.
Now if $a/b=c/d$ then also
$a/b=(ma+nc)/(mb+nd)$
for any coefficients $m,n$ where the denominator is nonzero. In particular, Eq. 1 implies
$p/q=(3p-6q)/(3q-p)text{.....Eq. 2}$
where we already have $2q<p<3q$ and thus $0<3q-p<q$. So the proposed rational fraction $p/q$ must be equal to an alternative rational fraction with a smaller positive denominator. This causes an infinite descent contradiction forcing the assumption of a rational value to be false.
We can form a similar proof for the square root of any natural number that is not a squared integer. "Not a squared integer" is needed because the square root must be strictly between two adjacent integers to obtain a descent of positive denominators.
answered Nov 18 '18 at 2:40
Oscar Lanzi
12.1k12036
add a comment |
How about a proof by descent?
First show that $2^2<6<3^2$. Then if $sqrt{6}$ is to be rational it must have a form $p/q$ where $p,qin mathbb{Z}, q>0, 2q<p<3q$. By simple algebra the square root is also equal to $6q/p$, thus
$p/q=6q/ptext{.....Eq. 1}$.
Now if $a/b=c/d$ then also
$a/b=(ma+nc)/(mb+nd)$
for any coefficients $m,n$ where the denominator is nonzero. In particular, Eq. 1 implies
$p/q=(3p-6q)/(3q-p)text{.....Eq. 2}$
where we already have $2q<p<3q$ and thus $0<3q-p<q$. So the proposed rational fraction $p/q$ must be equal to an alternative rational fraction with a smaller positive denominator. This causes an infinite descent contradiction forcing the assumption of a rational value to be false.
We can form a similar proof for the square root of any natural number that is not a squared integer. "Not a squared integer" is needed because the square root must be strictly between two adjacent integers to obtain a descent of positive denominators.
answered Nov 18 '18 at 2:40
Oscar Lanzi
12.1k12036
How about a proof by descent?
First show that $2^2<6<3^2$. Then if $sqrt{6}$ is to be rational it must have a form $p/q$ where $p,qin mathbb{Z}, q>0, 2q<p<3q$. By simple algebra the square root is also equal to $6q/p$, thus
$p/q=6q/ptext{.....Eq. 1}$.
Now if $a/b=c/d$ then also
$a/b=(ma+nc)/(mb+nd)$
for any coefficients $m,n$ where the denominator is nonzero. In particular, Eq. 1 implies
$p/q=(3p-6q)/(3q-p)text{.....Eq. 2}$
where we already have $2q<p<3q$ and thus $0<3q-p<q$. So the proposed rational fraction $p/q$ must be equal to an alternative rational fraction with a smaller positive denominator. This causes an infinite descent contradiction forcing the assumption of a rational value to be false.
We can form a similar proof for the square root of any natural number that is not a squared integer. "Not a squared integer" is needed because the square root must be strictly between two adjacent integers to obtain a descent of positive denominators.
answered Nov 18 '18 at 2:40
Oscar Lanzi
12.1k12036
edited Nov 21 '18 at 0:03
answered Nov 18 '18 at 2:40
Oscar Lanzi
12.1k12036
answered Nov 18 '18 at 2:40
Oscar Lanzi
12.1k12036
answered Nov 18 '18 at 2:40
Oscar Lanzi
12.1k12036
12.1k12036
add a comment |
add a comment |
Assume $sqrt{6} = {a over b}$ with $a, b in mathbb{Z}$ . This implies
$$2 cdot 3 b^2 = a^2$$
Use the fundamental theorem of algebra to decompose both sides into a unique prime factorization. There are an even number of factors of $2$ on the rhs and an odd number of factors of $2$ on the lhs... a contradition. Same for factors of $3$.
Thus $sqrt{6} neq {a over b}$, i.e., is irrational.
answered Nov 15 '18 at 2:32
David G. Stork
9,87021232
Yeah, I tried to reach it without prime factorisation. A second look?
– J.Moh
Nov 15 '18 at 2:42
You are considering the value of the rhs and lhs of the equation. I am instead looking at the number of prime factors. Frankly, I think that is so much more elegant... but I suppose this is a matter of taste. Mine applies to $sqrt{5/3}$ whereas yours doesn't... at least not as directly.
– David G. Stork
Nov 15 '18 at 2:47
Our professor used the value way, so I am following his steps for this, though, thanks!
– J.Moh
Nov 15 '18 at 2:50
Show him/her the better way! And stand out as a student!
– David G. Stork
Nov 15 '18 at 2:51
Well, that doesn t apply in Morocco. Pray for me I win visa lottery so I can join a better faculty in the US :)
– J.Moh
Nov 15 '18 at 3:49
|
show 7 more comments
Assume $sqrt{6} = {a over b}$ with $a, b in mathbb{Z}$ . This implies
$$2 cdot 3 b^2 = a^2$$
Use the fundamental theorem of algebra to decompose both sides into a unique prime factorization. There are an even number of factors of $2$ on the rhs and an odd number of factors of $2$ on the lhs... a contradition. Same for factors of $3$.
Thus $sqrt{6} neq {a over b}$, i.e., is irrational.
answered Nov 15 '18 at 2:32
David G. Stork
9,87021232
Yeah, I tried to reach it without prime factorisation. A second look?
– J.Moh
Nov 15 '18 at 2:42
You are considering the value of the rhs and lhs of the equation. I am instead looking at the number of prime factors. Frankly, I think that is so much more elegant... but I suppose this is a matter of taste. Mine applies to $sqrt{5/3}$ whereas yours doesn't... at least not as directly.
– David G. Stork
Nov 15 '18 at 2:47
Our professor used the value way, so I am following his steps for this, though, thanks!
– J.Moh
Nov 15 '18 at 2:50
Show him/her the better way! And stand out as a student!
– David G. Stork
Nov 15 '18 at 2:51
Well, that doesn t apply in Morocco. Pray for me I win visa lottery so I can join a better faculty in the US :)
– J.Moh
Nov 15 '18 at 3:49
|
show 7 more comments
Assume $sqrt{6} = {a over b}$ with $a, b in mathbb{Z}$ . This implies
$$2 cdot 3 b^2 = a^2$$
Use the fundamental theorem of algebra to decompose both sides into a unique prime factorization. There are an even number of factors of $2$ on the rhs and an odd number of factors of $2$ on the lhs... a contradition. Same for factors of $3$.
Thus $sqrt{6} neq {a over b}$, i.e., is irrational.
answered Nov 15 '18 at 2:32
David G. Stork
9,87021232
Assume $sqrt{6} = {a over b}$ with $a, b in mathbb{Z}$ . This implies
$$2 cdot 3 b^2 = a^2$$
Use the fundamental theorem of algebra to decompose both sides into a unique prime factorization. There are an even number of factors of $2$ on the rhs and an odd number of factors of $2$ on the lhs... a contradition. Same for factors of $3$.
Thus $sqrt{6} neq {a over b}$, i.e., is irrational.
answered Nov 15 '18 at 2:32
David G. Stork
9,87021232
edited Nov 16 '18 at 5:11
answered Nov 15 '18 at 2:32
David G. Stork
9,87021232
answered Nov 15 '18 at 2:32
David G. Stork
9,87021232
answered Nov 15 '18 at 2:32
David G. Stork
9,87021232
9,87021232
Yeah, I tried to reach it without prime factorisation. A second look?
– J.Moh
Nov 15 '18 at 2:42
You are considering the value of the rhs and lhs of the equation. I am instead looking at the number of prime factors. Frankly, I think that is so much more elegant... but I suppose this is a matter of taste. Mine applies to $sqrt{5/3}$ whereas yours doesn't... at least not as directly.
– David G. Stork
Nov 15 '18 at 2:47
Our professor used the value way, so I am following his steps for this, though, thanks!
– J.Moh
Nov 15 '18 at 2:50
Show him/her the better way! And stand out as a student!
– David G. Stork
Nov 15 '18 at 2:51
Well, that doesn t apply in Morocco. Pray for me I win visa lottery so I can join a better faculty in the US :)
– J.Moh
Nov 15 '18 at 3:49
|
show 7 more comments
Yeah, I tried to reach it without prime factorisation. A second look?
– J.Moh
Nov 15 '18 at 2:42
You are considering the value of the rhs and lhs of the equation. I am instead looking at the number of prime factors. Frankly, I think that is so much more elegant... but I suppose this is a matter of taste. Mine applies to $sqrt{5/3}$ whereas yours doesn't... at least not as directly.
– David G. Stork
Nov 15 '18 at 2:47
Our professor used the value way, so I am following his steps for this, though, thanks!
– J.Moh
Nov 15 '18 at 2:50
Show him/her the better way! And stand out as a student!
– David G. Stork
Nov 15 '18 at 2:51
Well, that doesn t apply in Morocco. Pray for me I win visa lottery so I can join a better faculty in the US :)
– J.Moh
Nov 15 '18 at 3:49
Yeah, I tried to reach it without prime factorisation. A second look?
– J.Moh
Nov 15 '18 at 2:42
Yeah, I tried to reach it without prime factorisation. A second look?
– J.Moh
Nov 15 '18 at 2:42
You are considering the value of the rhs and lhs of the equation. I am instead looking at the number of prime factors. Frankly, I think that is so much more elegant... but I suppose this is a matter of taste. Mine applies to $sqrt{5/3}$ whereas yours doesn't... at least not as directly.
– David G. Stork
Nov 15 '18 at 2:47
You are considering the value of the rhs and lhs of the equation. I am instead looking at the number of prime factors. Frankly, I think that is so much more elegant... but I suppose this is a matter of taste. Mine applies to $sqrt{5/3}$ whereas yours doesn't... at least not as directly.
– David G. Stork
Nov 15 '18 at 2:47
Our professor used the value way, so I am following his steps for this, though, thanks!
– J.Moh
Nov 15 '18 at 2:50
Our professor used the value way, so I am following his steps for this, though, thanks!
– J.Moh
Nov 15 '18 at 2:50
Show him/her the better way! And stand out as a student!
– David G. Stork
Nov 15 '18 at 2:51
Show him/her the better way! And stand out as a student!
– David G. Stork
Nov 15 '18 at 2:51
Well, that doesn t apply in Morocco. Pray for me I win visa lottery so I can join a better faculty in the US :)
– J.Moh
Nov 15 '18 at 3:49
Well, that doesn t apply in Morocco. Pray for me I win visa lottery so I can join a better faculty in the US :)
– J.Moh
Nov 15 '18 at 3:49
|
show 7 more comments
I'll play more generally
and see what happens.
Suppose
$sqrt{m}
=dfrac{u}{v}
$
where
$m = prod_{p in P} p^{m_i},
u = prod_{p in P} p^{u_i},
v = prod_{p in P} p^{v_i}
$.
Then
$prod_{p in P} p^{m_i}
=dfrac{u^2}{v^2}
=dfrac{ prod_{p in P} p^{2u_i}}{prod_{p in P} p^{2v_i}}
$
so
$prod_{p in P} p^{m_i}prod_{p in P} p^{2v_i}
=prod_{p in P} p^{2u_i}
$
or
$prod_{p in P} p^{m_i+2v_i}
=prod_{p in P} p^{2u_i}
$.
By unique factorization,
$m_i+2v_i
=2u_i
$,
so
$m_i
=2u_i-2v_i
=2(u_i-v_i)
$.
Therefore
$m
=prod_{p in P} p^{m_i}
=prod_{p in P} p^{2(u_i-v_i)}
=left(prod_{p in P} p^{u_i-v_i}right)^2
$
so $m$ is a perfect square.
Therefore
the square root of an integer
is rational
only if
it is a square.
I'll now try to generalize this
to $k$-th roots,
with as much cut-and-paste
as possible.
Suppose
$sqrt[k]{m}
=dfrac{u}{v}
$
where
$m = prod_{p in P} p^{m_i},
u = prod_{p in P} p^{u_i},
v = prod_{p in P} p^{v_i}
$.
Then
$prod_{p in P} p^{m_i}
=dfrac{u^k}{v^k}
=dfrac{ prod_{p in P} p^{ku_i}}{prod_{p in P} p^{kv_i}}
$
so
$prod_{p in P} p^{m_i}prod_{p in P} p^{kv_i}
=prod_{p in P} p^{ku_i}
$
or
$prod_{p in P} p^{m_i+kv_i}
=prod_{p in P} p^{ku_i}
$.
By unique factorization,
$m_i+kv_i
=ku_i
$,
so
$m_i
=ku_i-kv_i
=k(u_i-v_i)
$.
Therefore
$m
=prod_{p in P} p^{m_i}
=prod_{p in P} p^{k(u_i-v_i)}
=left(prod_{p in P} p^{u_i-v_i}right)^k
$
so $m$ is a perfect $k$-th power.
Therefore
the $k$-th root of an integer
is rational
only if
it is a $k$-th power.
answered Nov 15 '18 at 1:37
marty cohen
72.6k549127
add a comment |
I'll play more generally
and see what happens.
Suppose
$sqrt{m}
=dfrac{u}{v}
$
where
$m = prod_{p in P} p^{m_i},
u = prod_{p in P} p^{u_i},
v = prod_{p in P} p^{v_i}
$.
Then
$prod_{p in P} p^{m_i}
=dfrac{u^2}{v^2}
=dfrac{ prod_{p in P} p^{2u_i}}{prod_{p in P} p^{2v_i}}
$
so
$prod_{p in P} p^{m_i}prod_{p in P} p^{2v_i}
=prod_{p in P} p^{2u_i}
$
or
$prod_{p in P} p^{m_i+2v_i}
=prod_{p in P} p^{2u_i}
$.
By unique factorization,
$m_i+2v_i
=2u_i
$,
so
$m_i
=2u_i-2v_i
=2(u_i-v_i)
$.
Therefore
$m
=prod_{p in P} p^{m_i}
=prod_{p in P} p^{2(u_i-v_i)}
=left(prod_{p in P} p^{u_i-v_i}right)^2
$
so $m$ is a perfect square.
Therefore
the square root of an integer
is rational
only if
it is a square.
I'll now try to generalize this
to $k$-th roots,
with as much cut-and-paste
as possible.
Suppose
$sqrt[k]{m}
=dfrac{u}{v}
$
where
$m = prod_{p in P} p^{m_i},
u = prod_{p in P} p^{u_i},
v = prod_{p in P} p^{v_i}
$.
Then
$prod_{p in P} p^{m_i}
=dfrac{u^k}{v^k}
=dfrac{ prod_{p in P} p^{ku_i}}{prod_{p in P} p^{kv_i}}
$
so
$prod_{p in P} p^{m_i}prod_{p in P} p^{kv_i}
=prod_{p in P} p^{ku_i}
$
or
$prod_{p in P} p^{m_i+kv_i}
=prod_{p in P} p^{ku_i}
$.
By unique factorization,
$m_i+kv_i
=ku_i
$,
so
$m_i
=ku_i-kv_i
=k(u_i-v_i)
$.
Therefore
$m
=prod_{p in P} p^{m_i}
=prod_{p in P} p^{k(u_i-v_i)}
=left(prod_{p in P} p^{u_i-v_i}right)^k
$
so $m$ is a perfect $k$-th power.
Therefore
the $k$-th root of an integer
is rational
only if
it is a $k$-th power.
answered Nov 15 '18 at 1:37
marty cohen
72.6k549127
add a comment |
I'll play more generally
and see what happens.
Suppose
$sqrt{m}
=dfrac{u}{v}
$
where
$m = prod_{p in P} p^{m_i},
u = prod_{p in P} p^{u_i},
v = prod_{p in P} p^{v_i}
$.
Then
$prod_{p in P} p^{m_i}
=dfrac{u^2}{v^2}
=dfrac{ prod_{p in P} p^{2u_i}}{prod_{p in P} p^{2v_i}}
$
so
$prod_{p in P} p^{m_i}prod_{p in P} p^{2v_i}
=prod_{p in P} p^{2u_i}
$
or
$prod_{p in P} p^{m_i+2v_i}
=prod_{p in P} p^{2u_i}
$.
By unique factorization,
$m_i+2v_i
=2u_i
$,
so
$m_i
=2u_i-2v_i
=2(u_i-v_i)
$.
Therefore
$m
=prod_{p in P} p^{m_i}
=prod_{p in P} p^{2(u_i-v_i)}
=left(prod_{p in P} p^{u_i-v_i}right)^2
$
so $m$ is a perfect square.
Therefore
the square root of an integer
is rational
only if
it is a square.
I'll now try to generalize this
to $k$-th roots,
with as much cut-and-paste
as possible.
Suppose
$sqrt[k]{m}
=dfrac{u}{v}
$
where
$m = prod_{p in P} p^{m_i},
u = prod_{p in P} p^{u_i},
v = prod_{p in P} p^{v_i}
$.
Then
$prod_{p in P} p^{m_i}
=dfrac{u^k}{v^k}
=dfrac{ prod_{p in P} p^{ku_i}}{prod_{p in P} p^{kv_i}}
$
so
$prod_{p in P} p^{m_i}prod_{p in P} p^{kv_i}
=prod_{p in P} p^{ku_i}
$
or
$prod_{p in P} p^{m_i+kv_i}
=prod_{p in P} p^{ku_i}
$.
By unique factorization,
$m_i+kv_i
=ku_i
$,
so
$m_i
=ku_i-kv_i
=k(u_i-v_i)
$.
Therefore
$m
=prod_{p in P} p^{m_i}
=prod_{p in P} p^{k(u_i-v_i)}
=left(prod_{p in P} p^{u_i-v_i}right)^k
$
so $m$ is a perfect $k$-th power.
Therefore
the $k$-th root of an integer
is rational
only if
it is a $k$-th power.
answered Nov 15 '18 at 1:37
marty cohen
72.6k549127
I'll play more generally
and see what happens.
Suppose
$sqrt{m}
=dfrac{u}{v}
$
where
$m = prod_{p in P} p^{m_i},
u = prod_{p in P} p^{u_i},
v = prod_{p in P} p^{v_i}
$.
Then
$prod_{p in P} p^{m_i}
=dfrac{u^2}{v^2}
=dfrac{ prod_{p in P} p^{2u_i}}{prod_{p in P} p^{2v_i}}
$
so
$prod_{p in P} p^{m_i}prod_{p in P} p^{2v_i}
=prod_{p in P} p^{2u_i}
$
or
$prod_{p in P} p^{m_i+2v_i}
=prod_{p in P} p^{2u_i}
$.
By unique factorization,
$m_i+2v_i
=2u_i
$,
so
$m_i
=2u_i-2v_i
=2(u_i-v_i)
$.
Therefore
$m
=prod_{p in P} p^{m_i}
=prod_{p in P} p^{2(u_i-v_i)}
=left(prod_{p in P} p^{u_i-v_i}right)^2
$
so $m$ is a perfect square.
Therefore
the square root of an integer
is rational
only if
it is a square.
I'll now try to generalize this
to $k$-th roots,
with as much cut-and-paste
as possible.
Suppose
$sqrt[k]{m}
=dfrac{u}{v}
$
where
$m = prod_{p in P} p^{m_i},
u = prod_{p in P} p^{u_i},
v = prod_{p in P} p^{v_i}
$.
Then
$prod_{p in P} p^{m_i}
=dfrac{u^k}{v^k}
=dfrac{ prod_{p in P} p^{ku_i}}{prod_{p in P} p^{kv_i}}
$
so
$prod_{p in P} p^{m_i}prod_{p in P} p^{kv_i}
=prod_{p in P} p^{ku_i}
$
or
$prod_{p in P} p^{m_i+kv_i}
=prod_{p in P} p^{ku_i}
$.
By unique factorization,
$m_i+kv_i
=ku_i
$,
so
$m_i
=ku_i-kv_i
=k(u_i-v_i)
$.
Therefore
$m
=prod_{p in P} p^{m_i}
=prod_{p in P} p^{k(u_i-v_i)}
=left(prod_{p in P} p^{u_i-v_i}right)^k
$
so $m$ is a perfect $k$-th power.
Therefore
the $k$-th root of an integer
is rational
only if
it is a $k$-th power.
answered Nov 15 '18 at 1:37
marty cohen
72.6k549127
answered Nov 15 '18 at 1:37
marty cohen
72.6k549127
answered Nov 15 '18 at 1:37
marty cohen
72.6k549127
answered Nov 15 '18 at 1:37
marty cohen
72.6k549127
72.6k549127
add a comment |
add a comment |
When you get $2p^2 = 3q^2$ that means that $2|q^2$ so $2|q$ and that $3|p^2$ so $3|p$.
Then if you replace $p = 3p'$ for some integer $p'$ and $q=2q'$ for some integer $q'$ you get
$2(3p')^2 = 3(2q')^2$
$18p'^2 = 12q'^2$
$3p'^3 = 2q'^2$.
Can you finish from there?
This assumes you know Euclid's lemma that 1) prime numbers exist and 2) if $p$ is prime and $p|a*b$ then either $p|a$ or $p|b$ (or both).
answered Nov 15 '18 at 2:28
fleablood
68.3k22685
I tried to reach it without prime factorisation, can you have a second look?
– J.Moh
Nov 15 '18 at 2:42
add a comment |
When you get $2p^2 = 3q^2$ that means that $2|q^2$ so $2|q$ and that $3|p^2$ so $3|p$.
Then if you replace $p = 3p'$ for some integer $p'$ and $q=2q'$ for some integer $q'$ you get
$2(3p')^2 = 3(2q')^2$
$18p'^2 = 12q'^2$
$3p'^3 = 2q'^2$.
Can you finish from there?
This assumes you know Euclid's lemma that 1) prime numbers exist and 2) if $p$ is prime and $p|a*b$ then either $p|a$ or $p|b$ (or both).
answered Nov 15 '18 at 2:28
fleablood
68.3k22685
I tried to reach it without prime factorisation, can you have a second look?
– J.Moh
Nov 15 '18 at 2:42
add a comment |
When you get $2p^2 = 3q^2$ that means that $2|q^2$ so $2|q$ and that $3|p^2$ so $3|p$.
Then if you replace $p = 3p'$ for some integer $p'$ and $q=2q'$ for some integer $q'$ you get
$2(3p')^2 = 3(2q')^2$
$18p'^2 = 12q'^2$
$3p'^3 = 2q'^2$.
Can you finish from there?
This assumes you know Euclid's lemma that 1) prime numbers exist and 2) if $p$ is prime and $p|a*b$ then either $p|a$ or $p|b$ (or both).
answered Nov 15 '18 at 2:28
fleablood
68.3k22685
When you get $2p^2 = 3q^2$ that means that $2|q^2$ so $2|q$ and that $3|p^2$ so $3|p$.
Then if you replace $p = 3p'$ for some integer $p'$ and $q=2q'$ for some integer $q'$ you get
$2(3p')^2 = 3(2q')^2$
$18p'^2 = 12q'^2$
$3p'^3 = 2q'^2$.
Can you finish from there?
This assumes you know Euclid's lemma that 1) prime numbers exist and 2) if $p$ is prime and $p|a*b$ then either $p|a$ or $p|b$ (or both).
answered Nov 15 '18 at 2:28
fleablood
68.3k22685
answered Nov 15 '18 at 2:28
fleablood
68.3k22685
answered Nov 15 '18 at 2:28
fleablood
68.3k22685
answered Nov 15 '18 at 2:28
fleablood
68.3k22685
68.3k22685
I tried to reach it without prime factorisation, can you have a second look?
– J.Moh
Nov 15 '18 at 2:42
add a comment |
I tried to reach it without prime factorisation, can you have a second look?
– J.Moh
Nov 15 '18 at 2:42
I tried to reach it without prime factorisation, can you have a second look?
– J.Moh
Nov 15 '18 at 2:42
I tried to reach it without prime factorisation, can you have a second look?
– J.Moh
Nov 15 '18 at 2:42
add a comment |
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Is the $k$ a typo? What is $p, k$ and $q$? Why is $q$ being even a condradiction with $gcd(p,q) = 1$? And why is $2k^2 = 3q^2$ a contradiction? You just said $2k^2 = 3q^2$ so that isn't a contradiction? You are on the right track but you haven't explained any of your arguments.
– fleablood
Nov 15 '18 at 2:13
Why is $q$ even a contradiction? What if $k$ or $p$ (whats the difference) is odd?
– fleablood
Nov 15 '18 at 2:20
I rearranged it!
– J.Moh
Nov 15 '18 at 3:51