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: CA Posts: 1,112 Thanks: 580 
looks basically like the standard proof that $\sqrt{2}$ is irrational so it should be fine.

April 7th, 2017, 11:58 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 16,792 Thanks: 1238  
April 8th, 2017, 11:55 AM  #4 
Member Joined: Aug 2016 From: Used to be Earth Posts: 64 Thanks: 14  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  
Member Joined: Aug 2016 From: Used to be Earth Posts: 64 Thanks: 14  Quote:
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  
Senior Member Joined: Sep 2015 From: CA Posts: 1,112 Thanks: 580  Quote:
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  
Senior Member Joined: May 2016 From: USA Posts: 577 Thanks: 248  Quote:
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: 16,792 Thanks: 1238 
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  
Member Joined: Aug 2016 From: Used to be Earth Posts: 64 Thanks: 14  Quote:
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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