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In my Discrete Math class, we need to prove there aren't nonzero rational numbers $a$ and $b$ such that $5a^2 = 3b^2$.

I thought about using The Fundamental Theorem of Arithmetic but can't think of a way to apply it. How can I approach this proof in a good manner?

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  • $\begingroup$ Hint: You can suppose $a$ and $b$ are non-zero integers. $\endgroup$ – Jakobian Jul 3 '19 at 14:23
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    $\begingroup$ Hint: Write $a=\frac{p}{q}$ and $b = \frac{m}{n}$ to reduce the problem to when $a$ and $b$ are integers. $\endgroup$ – AHusain Jul 3 '19 at 14:23
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Have you proved that $\sqrt{2}$ is irrational? I believe the proof of this is much the same.

To begin, clear fractions and assume without loss of generality that $a$ and $b$ are integers that are co-prime. (If they share a factor, the square of that factor divides both sides and you can cancel it.)

Next, notice that $3b^2 = 5a^2$ would imply that $3$ must divide $a$. However, this implies that $3^2$ divides $a^2$ and you'll find that $3$ must also divide $b$ a contradiction to our coprime assumption.

Exactly the same could have be done with respect to $5$ instead to get the same result.

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  • $\begingroup$ How do you conclude that 3 must divide a? $\endgroup$ – F. Zer Jul 3 '19 at 14:29
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    $\begingroup$ @F.Zer Since we require everything to be an integer, the left-hand side is an integer and so $3$ divides $5a^2$. However, $3$ certainly doesn't divide $5$ and so it must divide $a^2$. i.e. It divides $a$. $\endgroup$ – Laarz Jul 3 '19 at 14:36
  • $\begingroup$ Thank you. I understood, now. So, $3^2$ divides $a^2$. How do you know 3 must also divide $b$? $\endgroup$ – F. Zer Jul 3 '19 at 16:30
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    $\begingroup$ Since $3^2$ divides $5a^2$, $3^2$ must also divide $3b^2$, so $3$ must divide $b^2$. $\endgroup$ – Martin Kochanski Jul 3 '19 at 17:36
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Let $a=a_1/a_2$, $b=b_1/b_2$, then $5b_2^2a_1^2=3b_1^2a_2^2$. Now the number of fives in the left side's prime factorisation is odd but even in the right side's factorisation.

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    $\begingroup$ Worth emphasis: such sketches implicitly use the Fundamental Theorem of Arithmetic (or a simpler form that every integer can be written uniquely in the form $\, 5^{\large k} n\,$ where $\,5\nmid n).\,$ To obtain a rigorous proof it is essential to be explicit about invoking these basic Theorems (which are crucial to the proof). $\endgroup$ – Bill Dubuque Jul 3 '19 at 19:12
  • $\begingroup$ Above I meant "every nonzero integer". $\endgroup$ – Bill Dubuque Jul 4 '19 at 0:16
  • $\begingroup$ See my alternative solution, please. $\endgroup$ – Michael Hoppe Jul 5 '19 at 16:19
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Yet a complete other solution. Essentially we want to show that $\sqrt{5/3}$ is irrational. Now suppose that $$\frac{a^2}{b^2}=\frac53\quad\text{with positive integers $a$, $b$.}$$ It's easy to show that under this assumption the square of $$\frac{\frac53b-a}{a-b}= \frac{5b-3a}{3(a-b)}$$ equals $5/3$, too. Now verify by an easy calculation that both numerator and denominator are positive and $$5b-3a<a \quad\text{and}\quad3(a-b)<b,$$ which immediately leads to a contradiction.

Motivation: First, let $k$ be natural number which is no square. Hence there exists a unique natural number $k$ satisfying $n^2<k<(n+1)^2$.

Now $k$ can be approximated be continued fractions $f_1$, $f_2$, and so on. If $a/b$ is $f_m$, the prior fraction is $$f_{m-1}=\frac{kb-na}{a-nb}.$$ It's not quite "messy" to show that assuming $(a/b)^2=k$ we get $$f_{m-1}^2=k,\quad0<kb-na<a\quad\text{and}\quad0<a-nb<b.$$

The procedure is easily extended if $k$ is a fraction of positive integers, none of which is square (otherwise the positivity will fail), provided $k>1$.

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  • $\begingroup$ +1 I like this answer, and I agree with the statement that $(5/3b-a)/(a-b)$ squares to $5/3$, but the proof that I can see relies on some messy algebra. Your exposition also leaves open the question of how you came up with this proof, which would be nice. $\endgroup$ – Cheerful Parsnip Jul 16 '19 at 18:33
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    $\begingroup$ See my expanded answer, please. $\endgroup$ – Michael Hoppe Jul 17 '19 at 11:56

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