anonymous
  • anonymous
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?
Mathematics
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schrodinger
  • schrodinger
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KingGeorge
  • KingGeorge
It would seem so, but perhaps your teacher wants something more fundamental?
KingGeorge
  • KingGeorge
What class is this for btw?
anonymous
  • anonymous
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.

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jim_thompson5910
  • jim_thompson5910
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
  • anonymous
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
  • KingGeorge
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
KingGeorge
  • KingGeorge
Strictly increasing meaning: If \(a
anonymous
  • anonymous
That is good too KingGeorge, but this is just the first chapter. So you aren't suppose to know what a "derivative" is.
KingGeorge
  • KingGeorge
Well then, jim's way is a good formal way of doing it using only simple algebra.
anonymous
  • anonymous
Thanks everybody.

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