Proving Inequalities: Three Exercises from Velleman's How to Prove It"

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The discussion focuses on proving three inequalities from Velleman's "How to Prove It." The first proof demonstrates that if a and b are negative, then a^2 is greater than b^2. The second proof shows that for positive a and b, the inequality 1/b < 1/a holds true by dividing rather than multiplying. The third proof confirms that the average of two distinct numbers lies between them. Participants emphasize the importance of manipulating equations to establish proofs without relying on specific numerical examples.
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I am doing exercises form Velleman's How to Prove It

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
 
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Government$ said:
I am doing exercises form Velleman's How to Prove It

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.
Yes, that's excellent.

2. I multiply a < b by ab and get 1/b < 1/a.
Well, you meandivide by ab, not multiply. And you should specifically say that "because 0< a< b, ab> 0".

3. I add b to a < b and get a + b < 2b and divide this by 2 and get (a+b)/2 < b
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.
 
Thank you for response.

HallsofIvy said:
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.

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?
 
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