My Math Forum Prove if a^2 < b^2 then a < b

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 March 27th, 2014, 12:55 AM #1 Senior Member     Joined: Oct 2013 From: Far far away Posts: 422 Thanks: 18 Prove if a^2 < b^2 then a < b Prove that if a^2 < b^2 then a < b for all whole numbers Well, I tried very hard to prove this but somehow I just can't do it. In the mean time I tried proving the converse: If a < b then a^2 < b^2 Here's my proof for (If a < b then a^2 < b^2) Definition: a < b iff a + c = b where a, b, c are whole numbers and c is not zero Then (a + c)^2 = b^2 implies a^2 + 2ac + c^2 = b^2 implies a^2 + (2ac + c^2) = b^2 implies a^2 < b^2...by (the definition of "less than" above) and the (fact that 2ac + c^2 is a nonzero whole number) Is this proof sound? Proving the original statement (if a^2 < b^2 then a < b for all whole numbers) is very difficult for me. Thanks
 March 27th, 2014, 01:02 AM #2 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: Prove if a^2 < b^2 then a < b Yes, but it's generally simpler to use a > b iff a -b > 0. Try this for a^2 and b^2. Also, hint: think about factorising.
 March 27th, 2014, 04:45 AM #3 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: Prove if a^2 < b^2 then a < b Careful, (a, b) = (0, -1) gives 0 = a^2 < b^2 = 1 but a > b instead of a < b.
 March 27th, 2014, 05:44 AM #4 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Prove if a^2 < b^2 then a < b The set of "whole numbers" is the set of "non-negative" integers so Hoempa's counter-example with b= -2 does not apply. I would do this: If a< b are positive integers then, multiplying both sides by a gives $a^2< ab$ while multiplying both sides by b gives $ab< b^2$. The "transitive property" of "<" then gives $a^2< b^2$. If a= 0 b> 0 is positive so $a^2= 0$ while $b^2$ is positive so we still have $a^2< b^2$ ("Integer" is not really relevant here. If a and b are any non-negative real numbers, the statement is still true.)
March 27th, 2014, 06:04 AM   #5
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Re: Prove if a^2 < b^2 then a < b

Quote:
 Originally Posted by HallsofIvy The set of "whole numbers" is the set of "non-negative" integers...
Not necessarily, switching to non-negative numbers is a good idea. (My counterexample has b = -1).

I like Pero's suggestion. a < b gives 0 < b - a. Since a and b are non-negative and b > a, a + b > 0 so 0 < (b - a)(a + b) = b^2 - a^2 which gives a^2 < b^2.

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