명제의 증명과 절대부등식

This exercise is intended to provide another rationale as to why a proof by contradiction works. Suppose that we are trying to prove that a statement $P$ is true. Instead of proving this statement, assume that we prove that the conditional statement “If $\neg P$, then $C$” is true, where $C$ is some contradiction. Recall that a contradiction is a statement that is always false.

In symbols, write a statement that is a disjunction and that is logically equivalent to $\neg P \to C$.

This exercise is intended to provide another rationale as to why a proof by contradiction works. Suppose that we are trying to prove that a statement $P$ is true. Instead of proving this statement, assume that we prove that the conditional statement “If $\neg P$, then $C$” is true, where $C$ is some contradiction. Recall that a contradiction is a statement that is always false.

Now explain why $P$ must be true if we prove that the negation of $P$ implies a contradiction.

Prove that the cube root of 2 is an irrational number. That is, prove that if $r$ is a real number such that $r^3 = 2$, then $r$ is an irrational number.

Prove the following propositions:

For all real numbers $x$ and $y$, if $x$ is rational and $y$ is irrational, then $x + y$ is irrational.

Prove the following propositions:

For all nonzero real numbers $x$ and $y$, if $x$ is rational and $y$ is irrational, then $xy$ is irrational.

Complete the following.

Is the real number $\sqrt 2 + \sqrt 5$ a rational number or an irrational number? Justify your conclusion.

Complete the following.

Prove that for each real number $x$, $\left(x + \sqrt{2} \right)$ is irrational or $\left(-x + \sqrt{2} \right)$ is irrational.

Complete the following.

Generalize the proposition in Part (a) for any irrational number (instead of just $\sqrt{2}$) and then prove the new proposition.

Justify your conclusion for each of the following.

Is the base 2 logarithm of 32, $\log_2 (32)$, a rational number or an irrational number?

Justify your conclusion for each of the following.

Is the base 2 logarithm of 3, $\log _2 (3)$, a rational number or an irrational number?

Prove each of the following propositions:

For each real number $\theta$, if $0 \lt \theta \lt \dfrac{\textstyle \pi }{\textstyle 2}$, then $\left[ {\sin (\theta) + \cos (\theta) } \right] > 1$.

Hint: What happens if you expand $[ {\sin (\theta) + \cos (\theta) } ]^2$? Don't forget your trigonometric identities.

Prove each of the following propositions:

For all real numbers $a$ and $b$, if $a \ne 0$ and $b \ne 0$, then $\sqrt{a^2 + b^2} \ne a + b$.

Prove each of the following propositions:

If $n$ is an integer greater than 2, then for all integers $m$, $n$ does not divide $m$ or $n + m \ne nm$.

Prove each of the following propositions:

For all real numbers $a$ and $b$, if $a > 0$ and $b > 0$, then

$$\frac{2}{a} + \frac{2}{b} \ne \frac{4}{a + b}$$

.

Prove that there do not exist three consecutive natural numbers such that the cube of the largest is equal to the sum of the cubes of the other two.

Hint: Three consecutive natural numbers can be represented by $n$, $n + 1$, and $n + 2$, where $n \in \mathbb{N}$, or three consecutive natural numbers can be represented by $m - 1$, $m$, and $m + 1$, where $m \in \mathbb{N}$.

Three natural numbers $a$, $b$, and $c$ with $a \lt b \lt c$ are called a Pythagorean triple

provided that $a^2 + b^2 = c^2$. For example, the numbers 3, 4, and 5 form a Pythagorean triple, and the numbers 5, 12, and 13 form a Pythagorean triple.

Verify that if $a = 20$, $b = 21$, and $c = 29$, then $a^2 + b^2 = c^2$, and hence, 20, 21, and 29 form a Pythagorean triple.

A magic square

is a square array of natural numbers whose rows, columns, and diagonals all sum to the same number. For example, the following is a 3 by 3 magic square since the sum of the 3 numbers in each row is equal to 15, the sum of the 3 numbers in each column is equal to 15, and the sum of the 3 numbers in each diagonal is equal to 15.

| 8 | 3 | 4 | | --- | --- | --- | | 1 | 5 | 9 | | 6 | 7 | 2 |

Prove that the following 4 by 4 square cannot be completed to form a magic square.

| | 1 | | 2 | | --- | --- | --- | --- | | 3 | 4 | 5 | | | 6 | 7 | | 8 | | 9 | | 10 | |

Using only the digits 1 through 9 one time each, is it possible to construct a 3 by 3 magic square with the digit 3 in the center square? That is, is it possible to construct a magic square of the form

| $a$ | $b$ | $c$ | | --- | --- | --- | | $d$ | 3 | $e$ | | $f$ | $g$ | $h$ |

where $a, b, c, d, e, f, g, h$ are all distinct digits, none of which is equal to 3? Either construct such a magic square or prove that it is not possible.