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

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In summary, the conversation discusses three proofs involving real numbers and their relationships. The first proof shows that if a and b are real numbers and a < b < 0, then a^2 > b^2. The second proof shows that if 0 < a < b, then 1/b < 1/a. The third proof shows that if a < b, then (a+b)/2 < b. The expert summarizer also points out an additional insight in the third proof, stating that the mean of two distinct numbers always lies between them. The conversation ends with a question about the key to proofs, to which the expert summarizer responds by stating that the key is to manipulate equations without using specific numbers.
  • #1
Government$
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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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  • #2
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.
 
  • #3
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?
 

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

1. How do you prove an inequality using mathematical induction?

First, you must establish a base case by showing that the inequality holds for the first value. Next, assume that the inequality holds for some arbitrary value, and then use mathematical operations to show that it also holds for the next value. Finally, use the principle of mathematical induction to prove that the inequality holds for all values.

2. When is it appropriate to use the contrapositive to prove an inequality?

The contrapositive is a useful tool when proving inequalities involving logical statements, as it allows you to flip the statement and use logical equivalences to prove the inequality. It is especially helpful when the original statement is complex, as it allows you to reduce the statement to simpler terms.

3. Can a counterexample be used to prove an inequality?

No, a counterexample cannot be used to prove an inequality. A counterexample is a specific case that disproves a general statement, but it does not provide a valid proof. In order to prove an inequality, you must use logical reasoning and mathematical operations to show that the statement holds for all cases.

4. How do you know if an inequality is true for all real numbers?

In order to prove that an inequality holds for all real numbers, you must use mathematical operations and logical reasoning to show that the statement holds for any arbitrary real number. This can be done through methods such as substitution, algebraic manipulation, or proof by contradiction.

5. Are there any shortcuts or tricks for proving inequalities?

While there may be some specific techniques for certain types of inequalities, there is no general shortcut or trick for proving inequalities. In order to prove an inequality, you must use logical reasoning and mathematical operations to show that the statement holds for all cases. It may also be helpful to break the inequality into smaller parts or to use known properties and theorems to simplify the statement.

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