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

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SUMMARY

This discussion focuses on proving inequalities using exercises from Daniel J. Velleman's "How to Prove It." The three exercises include proving that if \( a < b < 0 \) then \( a^2 > b^2 \), that if \( 0 < a < b \) then \( \frac{1}{b} < \frac{1}{a} \), and that if \( a < b \) then \( \frac{(a+b)}{2} < b \). The participants provide step-by-step solutions, emphasizing the importance of manipulating inequalities correctly, such as dividing instead of multiplying in certain contexts. The discussion highlights the significance of understanding the properties of inequalities in mathematical proofs.

PREREQUISITES
  • Understanding of real numbers and their properties
  • Familiarity with basic algebraic manipulation
  • Knowledge of inequalities and their implications
  • Experience with mathematical proofs and logical reasoning
NEXT STEPS
  • Study the properties of inequalities in real analysis
  • Learn about the concept of mathematical induction for proving statements
  • Explore the implications of the Arithmetic Mean-Geometric Mean Inequality
  • Practice additional exercises from "How to Prove It" by Daniel J. Velleman
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Students of mathematics, educators teaching proof techniques, and anyone interested in enhancing their understanding of inequalities and mathematical reasoning.

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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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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