Question 1017551
I can think of 2 ways
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1) You can do a proof by contradiction
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assume that there are integers p and q such that
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square root(3) = p/q which implies that
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3 = p^2 / q^2 and p/q in lowest terms(no common factor except 1)
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3q^2 = p^2
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so p^2 is divisible by 3 and p is as well
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this means that there exists an integer k, such that, p = 3k, therefore
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3q^2 = p^2 implies 3q^2 = (3k)^2 implies 3q^2 = 9k^2
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divide both sides by 3 and we get
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q^2 = 3k^2
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which implies q is divisible by 3
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but we also have p is divisible by 3 and
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this means that p and q are divisible by 3, so they were not in lowest terms - the contradiction that p and q only have 1 as a divisor
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2) consider x^2 - p = 0 where p is prime
then apply the Rational Root Theorem