This is the following question:

Suppose that $a, b, c$ are integers such that

$a\sqrt{2} + b = c\sqrt{3}$

(i) By squaring both sides of the equation, show that $a = b = c = 0$

The answer says that you put the equation into the following form:

if $ab \neq 0$

$\sqrt{2} = \frac{3c^2 − 2a^2 − b^2}{2ab}$

is rational — a contradiction and so a = 0 or b = 0.

Why would $a$ or $b$ be 0? (I get that you cannot express an irrational number as the quotient of two rational numbers).

  • 1
    $\begingroup$ BEcause the last step you took was of form $0(\sqrt 2)=0.$ Which is true but then you can't say $\sqrt 2 = {0 \over 0}$. $\endgroup$ – user103816 Oct 26 '15 at 7:08
  • $\begingroup$ because the expression for square root 2 is rational which is impossible. whenever ab not equals 0 you can get such exprssion.this forces ab=0 $\endgroup$ – Neel Oct 26 '15 at 7:14
  • $\begingroup$ I am not sure about this but try seeing this $2*√2ab = 3c^2 - 2a^2 - b^2$ now since a,b,c are all integers they can't give something irrational as above so I guess it's justified that a = b = c = 0 $\endgroup$ – Arnav Das Oct 26 '15 at 7:20
  • $\begingroup$ Since supposing $ab\ne0$ leads to a contradiction, you have to conclude that $ab=0$. Now, a product of two integers is zero if and only if one of the factors is zero. $\endgroup$ – egreg Oct 26 '15 at 8:58

If $a,b \in \mathbb{Z}$, then $$ 2a^{2} + b^{2} + 2ab\sqrt{2} = 3c^{2}, $$ and then \begin{align} (*)\ \ \ \ ab\sqrt{2} = \frac{3c^{2}-2a^{2}-b^{2}}{2} . \end{align} The number $\sqrt{2}$ is irrational; so $ab \neq 0$ leads to a contradiction, and hence $ab = 0$. We claim that $a=b=c = 0$; without loss of generality, let $a = 0$ and let $b \neq 0$. Then from $(*)$ we have $|b| = \sqrt{3}|c|$, which, by the fact that $\sqrt{3}$ is irrational, shows that $bc \neq 0$ is absurd; hence $bc = 0$. But either $b \neq 0$ or $c \neq 0$ also gives absurdity, so $b=c=0$.


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