# Check my three proofs

1. Jul 31, 2012

### Government$I am doing exercises form Velleman's How to Prove It 1. The problem statement, all variables and given/known data 1. Suppose a and b are real numbers. Prove that if a < b < 0 then a^2 > b^2. 2. Suppose a and b are real numbers. Prove that if 0 < a < b then 1/b < 1/a. 3. Suppose a and b are real numbers. Prove that if a < b then (a+b)/2 < b. 2. The attempt at a solution 1. Since a < b < 0 , this means that both a and b are negative numbers. Then i multiply a < b by a and get a^2 > ab and then again multiply a < b by b and get ab > b^2. Then since, a^2 > ab and ab > b^2 therefore a^2 > b^2. 2. I multiply a < b by ab and get 1/b < 1/a. 3. I add b to a < b and get a + b < 2b and divide this by 2 and get (a+b)/2 < b 2. Jul 31, 2012 ### HallsofIvy Staff Emeritus Yes, that's excellent. Well, you meandivide by ab, not multiply. And you should specifically say that "because 0< a< b, ab> 0". Yes. Notice that the same kind of argument shows that a< (a+b)/2 so that the "mean" of two distinct numbers always lies between them. 3. Jul 31, 2012 ### Government$

Thank you for response.

2. Yes i meant divide not multiply.
3. I haven't noticed that but it's a nice little insight.

So, is the key in proofs to manipulate equation to get for A to B without putting in numbers since that isn't proof?