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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 "nonnegative" integers so Hoempa's counterexample with b= 2 does not apply. I would do this: If a< b are positive integers then, multiplying both sides by a gives while multiplying both sides by b gives . The "transitive property" of "<" then gives . If a= 0 b> 0 is positive so while is positive so we still have ("Integer" is not really relevant here. If a and b are any nonnegative real numbers, the statement is still true.) 
March 27th, 2014, 06:04 AM  #5  
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361  Re: Prove if a^2 < b^2 then a < b Quote:
I like Pero's suggestion. a < b gives 0 < b  a. Since a and b are nonnegative 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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