Chapter 1.6, Problem 9E

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A.
(a) Claim. Every polynomial of degree 3 with real coefficients has a real zero.
“Proof.” The polynomial $p(x)=x^{\text{3}}-\text{8}$ has degree 3, real coefficients, and a real zero $(x=\text{2})$ . Thus the statement “Every polynomial of degree 3 with real coefficients does not have a real zero” is false, and hence its denial, “Every polynomial of degree 3 with real coefficients has a real zero,” is true.
(b) Claim. There is a unique polynomial whose first derivative is $\text{2}x+\text{3}$ and which has a zero at $x=\text{1}$ .
“Proof.” The antiderivative of $\text{2}x+\text{3}$ is $x^{\text{2}}+\text{3}x+C$ . If we let $p(x)=x^{\text{2}}+\text{3}x-\text{4}$ , then $p^{\prime }(x)=\text{2}x+\text{3}$ and $p\left(\text{1}\right)=0$ . So p(x) is the desired polynomial.
(c) Claim. Every prime number greater than 2 is odd.
“Proof.” The prime numbers greater than 2 are $\text{3},\text{ 5},\text{ 7},\text{ 11},\text{ 13},\text{ 17},\text{ 19},\dots$ . None of these is even, so all of them are odd.
(d) Claim. There exists an irrational number r such that $r^{\sqrt{2}}$ is rational.
“Proof.” If ${\sqrt{3}}^{\sqrt{2}}$ is rational, then $r=\sqrt{3}$ s the desired example. Otherwise, ${\sqrt{3}}^{\sqrt{2}}$ is irrational and ${\left({\sqrt{3}}^{\sqrt{2}}\right)}^{\sqrt{2}}={\left(\sqrt{3}\right)}^2=3$ , which is rational. Therefore, either $\sqrt{3}$ or ${\sqrt{3}}^{\sqrt{2}}$ is an irrational number r such that $r^{\sqrt{2}}$ is rational.
(e) Claim. For every real number x, $\left|x\right|\ge 0$ .
“Proof.” We proceed by three cases: $x>0$ , $x=0$ , and $x<0$ .
Case 1. $x>0$ . Choose, for example, $x=\text{4}$ . Then $\left|\text{4}\right|=\text{4}$ . Thus $\left|x\right|\ge 0$ .
Case 2. $x=0$ . Then $\left|0\right|=0$ . Thus $\left|x\right|\ge 0$ .
Case 3. $x<0$ . Choose, for example, $x=-\text{5}$ . Then $|-\text{5}|=\text{5}$ . Thus $\left|x\right|\ge 0$ .
(f ) Claim. If x is prime, then $x+\text{7}$ is composite.
“Proof.” Let x be a prime number. If $x=\text{2}$ , then $x+\text{7}=\text{9}$ , which is composite. If $x\ne \text{2}$ , then x is odd, so $x+\text{7}$ is even and greater than 2. In this case, too, $x+\text{7}$ is composite. Therefore, if x is prime, then $x+\text{7}$ is composite.
(g) Claim. If t is an irrational number, then $t-\text{8}$ is irrational.
“Proof.” Suppose there exists an irrational number t such that $t-\text{8}$ is rational. Then $t-\text{8}=\frac{p}{q}$ , where p and q are integers and $q\ne 0$ . Then $t=\frac{p}{q}+8=\frac{p+8q}{q}$ , with $p+\text{8}q$ and q integers and $q\ne 0$ . This is a contradiction because t is irrational. Therefore, for all irrational numbers t, $t-\text{8}$ is irrational.
(h) Claim. For real numbers x and y, if $xy=0$ , then $x=0$ or $y=0$ .
“Proof.”
Case 1. If $x=0$ , then $xy=0y=0$ .
Case 2. If $y=0$ , then $xy=x0=0$ .
In either case, $xy=0$ .
(i) Claim. For every real number x in the interval $\left(\text{3},\text{6}\right)$ , there is a natural number K such that for every real number y, if $y>K$ , then $\frac{1}{y}<\frac{1}{10}$ .
“Proof.” Assume that x is in the interval $\left(\text{3},\text{6}\right)$ . Then $x>\text{3}$ . Let K be $x+\text{7}$ . Then $K>\text{1}0$ . Suppose that y is a real number and $y>K$ . Then $y>\text{1}0$ , so $\frac{1}{y}<\frac{1}{10}$ .
( j) Claim. For every natural number n, $n\le n^{\text{2}}$ .
“Proof.” Let n be a natural number. Since n is a natural number, $\text{1}\le n$ . Since n is positive, $n·\text{1}\le n·n$ . Therefore, $n\le n^{\text{2}}$ for all natural numbers n.