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I've been working through The Four Pillars of Geometry by John Stillwell. In exercise 2.5.3 he asks,

How can we be sure that lengths $a,b,c>0$ with $a^2+b^2=c^2$ actually fit together to make a triangle? (Hint: Show that $a+b>c$)

Lemma: $a+b>c$. Suppose not that $a+b\leq c$ then $a^2+b^2+2ab\leq c^2$ and $2ab\leq 0$ $\Rightarrow\Leftarrow$ a contradiction is reached since $a,b>0$.

So my two questions are, (1) is my proof of the hint correct, and (2) how does that hint lead to the statement that $a,b$ and $c$ do in fact form a triangle.

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up vote 5 down vote accepted

Your proof of your lemma looks correct. To answer (2), you have $a+b>c$. Also, from $a^2+b^2=c^2$, you have $a^2\lt c^2$ and $b^2\lt c^2$, so $a\lt c$ and $b\lt c$ as all lengths are positive. So $a+c\gt b$ and $b+c\gt a$. Since any two lengths are greater than the third, you know that in a Euclidean plane, by the method of Euclid I.22 that you can construct a triangle with sides $a$, $b$, and $c$.

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Thanks for the fast answer and pointer to an excellent source! – ttt Jun 8 '11 at 0:08
@Tony, no problem. I used that online version of the Elements quite a bit when I also self-studied some modern geometry. – yunone Jun 8 '11 at 0:11

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