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A question in my book says to prove that pi is irrational, I found a proof which I'm happy with and found a similar one on the web however on the solutions they have done:
assume √π is rational i.e [tex] \sqrt{\pi} = \frac{p}{q} p,q \in \mathbb{Z} [/tex]
[tex] \pi = \frac{p^2}{q^2}, p^2,q^2 \in \mathbb{Z} ∴ \pi \mathrm{is\ rational} [/tex]
∴ contradiction √π irrational,
could anyone explain how it's a contradiction? I've pasted exactly what they have in the solution
assume √π is rational i.e [tex] \sqrt{\pi} = \frac{p}{q} p,q \in \mathbb{Z} [/tex]
[tex] \pi = \frac{p^2}{q^2}, p^2,q^2 \in \mathbb{Z} ∴ \pi \mathrm{is\ rational} [/tex]
∴ contradiction √π irrational,
could anyone explain how it's a contradiction? I've pasted exactly what they have in the solution