Square root Irrational

**I-Think** · October 25th 2010, 03:55 PM

Quick proof veracity test for this question

Let $d\in{N}$ that is not the square of another natural number. Prove that $\sqrt{d}$ is not a rational number

My proof
Assume $\sqrt{d}\in{Q}$ , hence $\sqrt{d}=\frac{a}{b}$ , where $a,b\in{N}$ and $gcd(a,b)=1$

So $d=\frac{a^2}{b^2}$ , so $db^2=a^2$ . Note $d|a^2$ , so d|a
Hence $a=dn$
and $a^2=d^2n^2$

Therefore
$d=\frac{d^{2}n^{2}}{b^2}$ , divide by b

$1=\frac{dn^2}{b^2}$

So $b^2=n^{2}d$ and $d=\frac{b^2}{n^2}$
So $\sqrt{d}=\frac{b}{n}$
Thus $\sqrt{d}=\frac{a}{b}=\frac{b}{n}$
Hence
$\frac{a}{b}=\frac{b}{n}\rightarrow{b^2=an}\rightar row{\frac{b^2}{a}=n}$
Hence $a|b$
But this gives a contradiction as $gcd(a,b)=1$
Hence our initial assumption is wrong and $\sqrt{d}$ is irrational
QED
Is this proof 100% correct?

**Pim** · October 26th 2010, 02:51 AM

To me it seems correct. Except for one thing: you say "divide by b" , but you're actually dividing by d. All the steps are correct though.

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