Show that $\sqrt{10}$ is irrational using the Fundamental theorem of arithmetic. I know I can prove it by way of contradiction. But I also want to know how to do this with this method.

By contradiction, $\sqrt{10}$ is rational. Then we can write $\frac{a}{b} = \sqrt{10}$ where a and b are integers and b is not 0 and the gcd(a,b) = 1. Then $10 = \frac{a^2}{b^2} => 10b^2 = a^2$. Because $a^2$ is even, a is even, and then we can write a = 2k where k is an integer. Then $10b^2 = (2k)^2 => 10b^2 = 4k^2 => 10b^2 = 2(2k^2) => 5b^2 = 2k^2$

We do not have a gcd(a,b) = 1 Because our assumption was wrong, $\sqrt{10}$ is irrational.

I want to know how to use the theorem. I looked at it and I got confused. Thanks


From $10b^2=a^2$: look at exponent of $2$ in both sides. LHS has odd power of $2$ while RHS has even power of $2$ - contradiction.

Note that this argument holds by fundamental theorem of arithmetic, which enable to speak about powers of $2$ and ensure the unicity of the exponent.

  • $\begingroup$ I am still quite confused. How does the left side have an odd power of 2? $\endgroup$ – DezTop Oct 16 '13 at 10:13
  • $\begingroup$ Because $b^2$ is square and thus has an even power of $2$ (possible 0), while $10$ has a odd power of $2$. Consequently theire product has an odd power of $2$. $\endgroup$ – Fabio Lucchini Oct 16 '13 at 11:59

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