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mrwall-e
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Two problems from chapter 3 of "How to Prove It" I just cannot get my head around
The thing I hate about proving things is that I can always think it through in my head, but don't know how to express it mathematically for a sound proof.
7) Suppose a, b, c, and d are real numbers, 0 < a < b, and d > 0. Prove that if ac >= bd then c > d.
8) Suppose x and y are real numbers, and 3x + 2y >= 5. Prove that if x > 1 then y < 1.
none.
7) We will prove the contrapositive. Suppose d > c. Thus, multiplying the inequality by any constant a will result in ad > ac. This is where I get stuck - something so simple, but I can't think of a sound way to prove it.
8) Suppose x > 1. Thus, 3x > 3. Then, by the transitive property (?) 3x > 5 - 2y. Therefore, 5 - 2y < 3. Subtracting 5 from both sides yields -2y < -2. This is equivalent to y < 1. Therefore, assuming 3x >= 5 - 2y, if x > 1 then y < 1. But this is not sound, because there is no transitive property?
Thanks
PS. This actually isn't a homework question.. but it's in the form of one so I figured put it here.
The thing I hate about proving things is that I can always think it through in my head, but don't know how to express it mathematically for a sound proof.
Homework Statement
7) Suppose a, b, c, and d are real numbers, 0 < a < b, and d > 0. Prove that if ac >= bd then c > d.
8) Suppose x and y are real numbers, and 3x + 2y >= 5. Prove that if x > 1 then y < 1.
Homework Equations
none.
The Attempt at a Solution
7) We will prove the contrapositive. Suppose d > c. Thus, multiplying the inequality by any constant a will result in ad > ac. This is where I get stuck - something so simple, but I can't think of a sound way to prove it.
8) Suppose x > 1. Thus, 3x > 3. Then, by the transitive property (?) 3x > 5 - 2y. Therefore, 5 - 2y < 3. Subtracting 5 from both sides yields -2y < -2. This is equivalent to y < 1. Therefore, assuming 3x >= 5 - 2y, if x > 1 then y < 1. But this is not sound, because there is no transitive property?
Thanks
PS. This actually isn't a homework question.. but it's in the form of one so I figured put it here.
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