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I want to prove for $a,b \in \mathbb{Q}$

\begin{align} a+b\sqrt{2}=0 \quad \Leftrightarrow \quad a=b=0 \end{align} For $\Leftarrow$ is trivial, now I am in trouble of showing $\Rightarrow$ Simply I can say $\sqrt{2}$ is irrational and go proceed, but I want some formal mathematical proof.

Martin Sleziak
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phy_math
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  • This boils down to the fact that $\sqrt 2$ is irrational. Any mathematical proof you want for the above fact will also prove that. More specifically, prove from the above that if the left hand side is true, then $a,b$ can be chosen to be integers. Then, prove that if $a$ and $b$ are the smallest possible solutions in absolute value, you can find solutions smaller than that. – Sarvesh Ravichandran Iyer Sep 08 '17 at 03:48

3 Answers3

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Suppose $a+b\sqrt{2}=0$, with $b\neq 0$. Then $a=p/q \space(q\neq 0), b=r/s \space (r \neq 0)$ Then $\sqrt{2}=\frac{-ps}{qr}$ so square both sides and proceed with your regular 'square root 2 is irrational' proof.

Wen
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If $b=0$ or $a=0$, we are done (!). So assume $b\neq 0, a\neq 0$. Then we have $$\sqrt{2} = -\frac{a}{b}.$$

Can you take it from here?

Krish
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Assume:($a\neq b\neq 0$) $$a+b\sqrt{2}=0$$ $$(a+b\sqrt{2})(\frac{a-b\sqrt{2}}{a-b\sqrt{2}})=0$$ $$\frac{a^2-2b^2}{a-b\sqrt{2}}=0$$ $$\left\{\begin{matrix} a^2-2b^2=0 & & \\ a-b\sqrt{2}\neq 0& & \end{matrix}\right.$$ $$\left\{\begin{matrix} \frac{a}{b}=\sqrt{2} & & \\ \frac{a}{b}\neq \sqrt{2}& & \end{matrix}\right.$$ Contradiction

So:($a=b=0$) for $a+b\sqrt{2}=0$

ProblemSolver1
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