Chapter 1.7, Problem 11E

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A.
(a) Claim. There is a unique three-digit number whose digits have sum 8 and product 10.
“Proof.” Let x, y, and z be the digits. Then $x+y+z=\text{8}$ and $xyz=\text{1}0$ . The only factors of 10 are 1, 2, 5, and 10, but since 10 is not a digit, the digits must be 1, 2, and 5. The sum of these digits is 8. Therefore, 125 is the only three-digit number whose digits have sum 8 and product 10.
(b) Claim. There is a unique set of three consecutive odd numbers that are all prime.
“Proof.” The consecutive odd numbers 3, 5, and 7 are all prime. Suppose that x, y, and z are consecutive odd numbers, all prime, and $x\ne 3$ . Then $y=x+\text{2}$ and $z=x+\text{4}$ . Since x is prime, when x is divided by 3, the remainder is 1 or 2. In case the remainder is 1, then $x=\text{3}k+\text{1}$ for some integer $k\ge \text{1}$ . But then $y=x+\text{2}=\text{3}k+\text{3}=\text{3}(k+\text{1})$ , so y is not prime. In case the remainder is 2, then $x=\text{3}k+\text{2}$ for some
integer $k\ge \text{1}$ . But then $z=x+\text{4}=\text{3}k+\text{2}+\text{4}=\text{3}(k+\text{2})$ , so z is not prime. In either case we reach the contradiction that y or z is not prime. Thus $x=\text{3}$ and so $y=\text{5}$ and $z=\text{7}$ . Therefore, the only three consecutive odd primes are 3, 5, and 7.
(c) Claim. If x is any real number, then either $\pi -x$ is irrational or $\pi +x$ is irrational.
“Proof.” It is known that p is an irrational number; that is, p cannot be written in the form $\frac{a}{b}$ for integers a and b. Consider $x=\pi$ . Then $\pi -x=0$ , which is rational, but $\pi +x=\text{2}\pi$ . If 2p were rational, then $\text{2}\pi =\frac{a}{b}$ for some integers a and b. Then $\pi =\frac{a}{2b}$ , so p is rational. This is impossible, so 2p is irrational. Therefore, either $\pi -x$ or $\pi +x$ is irrational.
(d) Claim. If x is any real number, then either $\pi -x$ is irrational or $\pi +x$ is irrational.
“Proof.” It is known that p is an irrational number; that is, p cannot be written in the form $\frac{a}{b}$ for integers a and b. Let x be any real number. Suppose both $\pi -x$ and $\pi +x$ are rational. Then, since the sum of two rational numbers is always rational, $(\pi -x)+(\pi +x)=\text{2}\pi$ is rational. Then $\text{2}\pi =\frac{a}{b}$ for some integers a and b. Then $\pi =\frac{a}{2b}$ , so p is rational. This is impossible. Therefore, at least one of $\pi -x$ or $\pi +x$ is irrational.
(e) Claim. For all real numbers x and y, $x^{\text{2}}-\text{3}x=y^{\text{2}}-\text{3}{y}^{\text{2}}$ if and only if $x=y$ or $x+y=\text{3}$ .
“Proof.” Suppose that $x^{\text{2}}-\text{3}x=y^{\text{2}}-\text{3}y$ . Then $x^{\text{2}}+xy-\text{3}x-xy-y^{\text{2}}+\text{3}y=0$ , so $(x+y-\text{3})(x-y)=0$ . Therefore, $x=y$ or $x+y=\text{3}$ .
(f) Claim. For all real numbers x and y, the equality $xy=\frac{1}{2}{(x+y)}^{\text{2}}$ holds if and only if $x=y=0$ .
“Proof.”
Part (i) Suppose that $x=y=0$ . Then $xy=0=\frac{1}{2}{(x+y)}^{\text{2}}$ , so the equality holds.
Part (ii) Suppose that x and y are real numbers and $xy=\frac{1}{2}{(x+y)}^{\text{2}}$ . Then $\text{2}xy=x^{\text{2}}+\text{2}xy+y^{\text{2}}$ , so $x^{\text{2}}+y^{\text{2}}=0$ . Since the square of a real number is never negative, $x^{\text{2}}=y^{\text{2}}=0$ , so $x=y=0$ .
(g) Claim. If n is prime and $n+\text{5}$ or $n+\text{12}$ is prime, then $n=\text{2}$ .
“Proof.” Assume that n is a prime number. Then $n\ge \text{2}$ , so if $n+\text{5}$ is prime, then $n+\text{5}$ must be odd. Therefore, n must be even. Since 2 is the only even prime, $n=\text{2}$ .
(h) Claim. Let a, b, and c be real numbers with $a\ne 0$ . If $a{x}^{\text{2}}+bx+c=0$ has no rational roots, then $c{x}^{\text{2}}+bx+a=0$ has no rational roots.
“Proof.” Suppose that $c{x}^{\text{2}}+bx+a=0$ has a rational root p/q. Then $c{(p/q)}^{\text{2}}+b(p/q)+a=0$ . Then $c+b(q/p)+a{(q/p)}^{\text{2}}=0$ , so q/p is a rational root of the equation $a{x}^{\text{2}}+bx+c=0$ .