Show that there is no rational number whose square is $2$ or $8$
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4Possible duplicate of Square root of $2$ is irrational. Note that since $8$ is a square times $2$, whether $2$ has a rational square root is equivalent to whether $8$ does. So, this is a duplicate. – Rushabh Mehta Nov 30 '19 at 07:24
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@DonThousand A better target in my opinion: https://math.stackexchange.com/questions/5/ – Maximilian Janisch Nov 30 '19 at 09:48
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Suppose there existes a rational $\frac{a}{b}$, with b nonzero, in reduced form, with a and b coprime such that its square was 2.
By beeing coprime you have that if 2 is a factor of a, it mustn't be a factor of b. And vice versa.
Then $a^2=2b^2$. Notice that if a number has any power (also 0) of 2 as a factor then its square has an even power of 2 as a factor. Thus, $a^2$ and $b^2$ have an even power of 2 as factors. But $a^2=2b^2$ and $2b^2$ has an odd power of 2 as factor. This is a contradiction for the prime decomposition of numbers is unique by the F.T.Arithmetic.
The proof for 8 is very similar.
N. Pullbacki
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