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I am doing exercises form Velleman's How to Prove It
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
Homework Statement
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