Mathematica 2014

21

√2 ൌ ܽ ܾ 2 = ௔ మ ௕ మ ܾ ଶ = ܽ ଶ

2

As ܽ ଶ is double another number (a and b are integers), it must be even. Therefore, let us call it 2n. We now have 2 ܾ ଶ = ܽ ଶ , or : 2 ܾ ଶ = ሺ2݊ሻ ଶ 2 ܾ ଶ = 4݊ ଶ ܾ ଶ = 2݊ ଶ This tells us that ܾ ଶ is an even number, as it is another number doubled, and even squares are always the product of the same even number. As both a and b are even, ௔ ௕ can be further cancelled, but this is a contradiction with the original √2 is irrational. In conclusion, both methods are very powerful for complicated or simple proofs, and the logic behind them cannot be questioned or doubted, making them extremely useful for proving more difficult formulae and theorem. the fraction statement. Thus, by disproving this statement, we can prove that

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