Let a, b be positive numbers such that a

Question

Let a, b be positive numbers such that a < b. Show a^2 < b^2. Is it really as easy as taking a < b and squaring both sides?

kinggeorge · Answer

It would seem so, but perhaps your teacher wants something more fundamental?

kinggeorge · Answer

What class is this for btw?

anonymous · Answer

Its in a book titled "A First Course in Calculus" 5th Edition by Serge Lang. Its in the first chapter. There are 7 proof questions at the end. I got the first 3, but the rest are problems similar to this. Such questions would probably mostly fall in the category of real analysis, but I've never taken such a course.

jim_thompson5910 · Answer

Here is how I would do it a < b a - b < 0 (a+b)(a - b) < (a+b)*0 ... multiply both sides by (a+b) ... see note below (a+b)(a - b) < 0 a^2 - b^2 < 0 a^2 < b^2 Note: since a > 0 and b > 0, we know that a+b > 0. Multiplying both sides of an inequality by a positive number will not change the inequality sign.

anonymous · Answer

Ah!. Nice jim_thompson. But would it really be that necessary to do that rather than just squaring both sides? I guess I'm just confused what the question is trying to entail.

kinggeorge · Answer

The other way I would consider doing it, especially since this is an introductory calculus book, is show that $x^2$ is a strictly increasing function.

Note that $$\frac{d}{dx}\,x^2=2x$$Since, we are assuming $a,b$ positive, if $x=a$ or $b$, then the slope is positive. Hence, it's strictly increasing, and if $a<b$, then $a^2<b^2$.

kinggeorge · Answer

Strictly increasing meaning: If $a<b$, then $f(a)<f(b)$.

anonymous · Answer

That is good too KingGeorge, but this is just the first chapter. So you aren't suppose to know what a "derivative" is.

kinggeorge · Answer

Well then, jim's way is a good formal way of doing it using only simple algebra.

anonymous · Answer

Thanks everybody.