Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $c$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.