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Is it sufficient to say that providing the shorter two sides of a right triangle can be expressed as integers that work out to equal the value of the hypotenuse, then the value of the hypotenuse must be irrational? For example, suppose I wish to prove that $\sqrt 5$ is irrational, if the shorter sides are 1 and 2 then the length of the hypotenuse $\sqrt 5$ must be irrational. Please tell me if I've made a mistake.

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    $\begingroup$ Have you heard of Pythagorean triples? Such as 3,4,5 $\endgroup$ – A.Riesen Jun 1 '16 at 16:52
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    $\begingroup$ Can't say I have. $\endgroup$ – Michael Lee Jun 1 '16 at 16:53
  • $\begingroup$ @MichaelLee Well, it should answer your question. $3^2+4^2=5^2$. $\endgroup$ – Clement C. Jun 1 '16 at 16:57
  • $\begingroup$ $$5^2+12^2=13^2, \;8^2+15^2=17^2,\;161^2+240^2=289^2,...$$ $\endgroup$ – DonAntonio Jun 1 '16 at 17:01
  • $\begingroup$ Oh! Now I follow you! If the sides consist of 3,4,5 then the hypotenuse is indeed a rational number. I can't thank you all enough for helping me here, it would have taken me ages to figure it out by myself. $\endgroup$ – Michael Lee Jun 1 '16 at 23:59
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Hint:

take two integer numbers $m>n>0$, than: $$ a=m^2-n^2 \qquad b=2mn \qquad c=m^2+n^2 $$ are integer such that $a^2+b^2=c^2$ (easy to prove). These are called Pythagorean triples.

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