prove that 7-3√5 is irrational
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⇒ Assume that 7 - 3√5 is rational. ... ⇒ In x = 7 - 3√5, both sides are rational as assumed. ⇒ Here, as both sides of x = 7 - 3√5 are rational, then so should be √5 = (7 - x)/3. But this contradicts our earlier assumption that 7 - 3√5 is rational, because the RHS of √5 = (7 - x)/3 is rational while the LHS is irrational.
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