My Math Forum  

Go Back   My Math Forum > High School Math Forum > Algebra

Algebra Pre-Algebra and Basic Algebra Math Forum


Reply
 
LinkBack Thread Tools Display Modes
March 27th, 2014, 12:55 AM   #1
Senior Member
 
shunya's Avatar
 
Joined: Oct 2013
From: Far far away

Posts: 429
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
shunya is offline  
 
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.
Pero is offline  
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.
Hoempa is offline  
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 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 non-negative real numbers, the statement is still true.)
HallsofIvy is offline  
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:
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.
Hoempa is offline  
Reply

  My Math Forum > High School Math Forum > Algebra

Tags
<, prove



Search tags for this page
Click on a term to search for related topics.
Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
prove 7>(2^0.5+5^0.5+11^0.5) Albert.Teng Algebra 8 February 7th, 2013 08:14 PM
Goldbach's conjecture (to prove or not to prove) octaveous Number Theory 13 September 23rd, 2010 04:36 AM
prove: rose3 Number Theory 1 February 16th, 2010 09:23 AM
prove prove prove. currently dont know where to post qweiop90 Algebra 1 July 31st, 2008 06:27 AM
prove prove prove. currently dont know where to post qweiop90 New Users 1 December 31st, 1969 04:00 PM





Copyright © 2019 My Math Forum. All rights reserved.