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 April 7th, 2017, 09:40 AM #1 Member   Joined: Aug 2016 From: Used to be Earth Posts: 64 Thanks: 14 Is This a Correct Proof? Hello, Is the below proof attempt correct? -- Let m, n and q be natural numbers, but q is not a perfect square, and m and n have no common divisors. Let's assume m/n=√q. Then, m²/n²=q. So, m² is divisible by n², which implies m is divisible by n. But this is impossible since m and n have no common divisors. Therefore, m/n≠√q, which shows that the √ of an integer is either another integer or irrational. -- Regards
 April 7th, 2017, 04:17 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,125 Thanks: 1103 looks basically like the standard proof that $\sqrt{2}$ is irrational so it should be fine. Thanks from sKebess
April 7th, 2017, 11:58 PM   #3
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Quote:
 Originally Posted by sKebess m² is divisible by n², which implies m is divisible by n.
Do you know why?

April 8th, 2017, 11:55 AM   #4
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Quote:
 Originally Posted by skipjack Do you know why?
Let $\displaystyle m=p_1p_2...p_n$, where $\displaystyle p_i$ is a prime factor of m. Then $\displaystyle m^2=p_1^2p_2^2...p_n^2$.

The same argument holds for $\displaystyle n$.

If m² is divisible by n² then they must share at least one squared prime factors, since we would then have m²=qn², where q is an integer.

The $\displaystyle \sqrt{ }$ of this(these) squared common prime factor(s) must also be shared by m and n, because of the first argument.

etc...

Last edited by sKebess; April 8th, 2017 at 11:58 AM.

 April 8th, 2017, 12:17 PM #5 Member   Joined: Aug 2016 From: Used to be Earth Posts: 64 Thanks: 14 Just realized there's a simpler way to show this. $\displaystyle m=q\times p_1p_2...p_n$ $\displaystyle \leftrightarrow$ $\displaystyle m^2=q^2\times p_1^2p_2^2...p_n^2$ where $\displaystyle p_i$ is a prime factor of $\displaystyle m$, and $\displaystyle q$ an integer. $\displaystyle n=p_1p_2...p_n$ $\displaystyle \leftrightarrow$ $\displaystyle n^2=p_1^2p_2^2...p_n^2$. So $\displaystyle \frac{m^2}{n^2}=\frac{q^2\times p_1^2p_2^2...p_n^2}{p_1^2p_2^2...p_n^2}=q^2$ $\displaystyle \leftrightarrow$ $\displaystyle \frac{m}{n}=\frac{q\times p_1p_2...p_n}{p_1p_2...p_n}=q$ Last edited by sKebess; April 8th, 2017 at 12:24 PM.
April 8th, 2017, 12:34 PM   #6
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Quote:
 Originally Posted by romsek looks basically like the standard proof that $\displaystyle \sqrt2$ is irrational so it should be fine.
Yes, what I found strange is the proof for $\displaystyle \sqrt2$ given in my textbook is 2, 3 times longer than this.
I was then asked to prove $\displaystyle \sqrt5$ was irrational, but what's the point really if the general proof is so simple...? which is why I wasn't sure whether I missed something or not.

Last edited by sKebess; April 8th, 2017 at 12:41 PM.

April 8th, 2017, 01:13 PM   #7
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Quote:
 Originally Posted by sKebess but what's the point really if the general proof is so simple...? which is why I wasn't sure whether I missed something or not.
historical reasons, that's all

most likely $\sqrt{2}$ being the hypotenuse of the simplest isosceles right triangle was the first irrational number the ancients came across and so was of particular interest.

April 8th, 2017, 01:45 PM   #8
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Quote:
 Originally Posted by romsek historical reasons, that's all most likely $\sqrt{2}$ being the hypotenuse of the simplest isosceles right triangle was the first irrational number the ancients came across and so was of particular interest.
This is exactly correct.

Greek geometry was initially based upon the premise that there existed some unit of length that made every linear measurement in a rectilinear figure a whole number of units. Discovering that something as simple as a right isosceles triangle disproved that premise represented a logical crisis.

Furthermore, I suspect the proof that the square root of 2 was irrational was older than the proof that the factorization of a number into powers of primes is unique. You do not need any theorem about unique factorization to show irrationality.

 April 9th, 2017, 12:39 AM #9 Global Moderator   Joined: Dec 2006 Posts: 19,731 Thanks: 1808 Although the general result is true, its proof ultimately relies on the fundamental theorem of arithmetic, which was established by C. F. Gauss by 1801, but √2 was proved to be irrational a long time before that.
April 9th, 2017, 08:27 AM   #10
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Quote:
 Originally Posted by skipjack Although the general result is true, its proof ultimately relies on the fundamental theorem of arithmetic, which was established by C. F. Gauss by 1801, but √2 was proved to be irrational a long time before that.
Perhaps I'm missing something, but couldn't we arrive at the same conclusion with the following:

If m is divisible by n, then we have
m/n=k (by the definition of divisibility) <=> m²/n²=k²

So, m² is also divisible by n², since k² is an integer.

Since each step was equivalent to the previous one doesn't this show that m² is divisible by n² iff m is divisible n?

The fundamental theorem of arithmetic was not needed here, only the definition of divisibility was needed. I'd think even Pythagoras knew of this.

What did I miss?

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