Algebra Solutions: The square root of a prime integer is not rational

If $p$ is a prime, prove that there do not exist nonzero integers $a$ and $b$ such that $a^2 = pb^2$ . (I.e., $\sqrt{p}$ is not rational.)

Suppose to the contrary that $a^2 = pb^2$ for some integers $a$ and $b$ . Then by the fundamental theorem of arithmetic we may factor $a^2$ and $pb^2$ as products of primes, and that these factorizations must be the same up to a reordering of the factors. Note, however, that $p$ must appear an even number of times in the factorization of $a^2$ and an odd number of times in the factorization of $pb^2$ . Since no integer is both even and odd, we have a contradiction.

Algebra Solutions

The square root of a prime integer is not rational

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