Prove that for any prime positive integer p, \sqrt{p} is an irrational number.

Foundation

Mathematics Foundation

Properties of Real Number

Irrational Number

Playing with Numbers

Question

Prove that for any prime positive integer

p

\sqrt{p}

is an irrational number.

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Let us assume on the contrary that

\sqrt{p}

is rational. Then, there exist positive co-primes

a

and

b

such that

\sqrt{p} = \frac{a}{b}

∴ p = \frac{a^{2}}{b^{2}}

∴ b^{2} p = a^{2}

\Rightarrow p ∣ a^{2} [∵ p ∣ b^{2} p]

∴ p ∣ a

∴ a = p c

for some positive integer

c

Now,

b^{2} p = a^{2}

∴ b^{2} p = p^{2} c^{2} [∵ a = p c]

∴ b^{2} = p c^{2}

\Rightarrow p ∣ b^{2} [∵ p ∣ p c^{2}]

\Rightarrow p ∣ b

∴ p ∣ a

and

p ∣ b

This contradicts that

a

and

b

are co-primes.

Hence,

\sqrt{p}

is irrational