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Problem 8

For a right triangle, suppose that the hypotenuse has length \(c\) feet and the lengths of the sides are \(a\) feet and \(b\) feet. (a) What is a formula for the area of this right triangle? What is an isosceles triangle? (b) State the Pythagorean Theorem for right triangles. \(\star\) (c) Prove that the right triangle described above is an isosceles triangle if and only if the area of the right triangle is \(\frac{1}{4} c^{2}\).

Problem 8

(a) Prove that for each real number \(x,(x+\sqrt{2})\) is irrational or \((-x+\sqrt{2})\) is irrational. (b) Generalize the proposition in Part (a) for any irrational number (instead of just \(\sqrt{2}\) ) and then prove the new proposition.

Problem 9

Prove that for each natural number \(n, \sqrt{3 n+2}\) is not a natural number.

Problem 9

Are the following propositions true or false? Justify each conclusion with a counterexample or a proof. (a) For all integers \(a\) and \(b\) with \(a \neq 0,\) the equation \(a x+b=0\) has a rational number solution. (b) For all integers \(a, b,\) and \(c,\) if \(a, b,\) and \(c\) are odd, then the equation \(a x^{2}+b x+c=0\) has no solution that is a rational number. Hint: Do not use the quadratic formula. Use a proof by contradiction and recall that any rational number can be written in the form \(\frac{p}{q},\) where \(p\) and \(q\) are integers, \(q>0\), and \(p\) and \(q\) have no common factor greater than \(1 .\) (c) For all integers \(a, b, c,\) and \(d,\) if \(a, b, c,\) and \(d\) are odd, then the equation \(a x^{3}+b x^{2}+c x+d=0\) has no solution that is a rational number.

Problem 9

A real number \(x\) is defined to be a rational number provided there exist integers \(m\) and \(n\) with \(n \neq 0\) such that \(x=\frac{m}{n}\). A real number that is not a rational number is called an irrational number. It is known that if \(x\) is a positive rational number, then there exist positive integers \(m\) and \(n\) with \(n \neq 0\) such that \(x=\frac{m}{n}\). Is the following proposition true or false? Explain. For each positive real number \(x,\) if \(x\) is irrational, then \(\sqrt{x}\) is irrational.

Problem 10

Extending the idea in Exercise (1) of Section \(3.4,\) we can represent three consecutive integers as \(m, m+1,\) and \(m+2,\) where \(m\) is an integer. (a) Explain why we can also represent three consecutive integers as \(k-1\), \(k,\) and \(k+1,\) where \(k\) is an integer. \- (b) Explain why Proposition 3.27 proves that the product of any three consecutive integers is divisible by 3 . \- (c) Prove that the product of three consecutive integers is divisible by 6 .

Problem 10

One of the most famous unsolved problems in mathematics is a conjecture made by Christian Goldbach in a letter to Leonhard Euler in 1742. The conjecture made in this letter is now known as Goldbach's Conjecture. The conjecture is as follows: Every even integer greater than 2 can be expressed as the sum of two (not necessarily distinct) prime numbers. Currently, it is not known if this conjecture is true or false. (a) Write \(50,142,\) and 150 as a sum of two prime numbers. (b) Prove the following: If Goldbach's Conjecture is true, then every integer greater than 5 can be written as a sum of three prime numbers. (c) Prove the following: If Goldbach's Conjecture is true, then every odd integer greater than 7 can be written as a sum of three odd prime numbers.

Problem 10

(a) Prove Part (2) of Proposition 3.23 . For each \(x \in \mathbb{R},|-x|=|x|\). (b) Prove Part (2) of Theorem 3.25 . For all real numbers \(x\) and \(y,|x y|=|x||y|\).

Problem 10

Is the following proposition true or false? Justify your conclusion. For each real number \(x, x(1-x) \leq \frac{1}{4}\).

Problem 11

Two prime numbers that differ by 2 are called twin primes. For example, 3 and 5 are twin primes, 5 and 7 are twin primes, and 11 and 13 are twin primes. Determine at least two other pairs of twin primes. Is the following proposition true or false? Justify your conclusion. For all natural numbers \(p\) and \(q\) if \(p\) and \(q\) are twin primes other than 3 and \(5,\) then \(p q+1\) is a perfect square and 36 divides \(p q+1\).

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