명제의 증명과 절대부등식 연습문제 — Sundstrom, Mathematical Reasoning · 3.2절 · 9번

Sundstrom, Mathematical Reasoning · 3.2절 · 9번영어 원문

A real number $x$ is defined to be a rational number

provided

$\text{ there exist integers } m \text{ and } n \text{ with } n \ne 0 \text{ such that } x = \frac{m}{n}$

. 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 $n \ne 0$ such that $x = \dfrac{m}{n}$ . Is the following proposition true or false? Explain.

For each positive real number

, if

is irrational, then

\sqrt x

is irrational.

정답·풀이 보기 (영어 원문)

The statement is true. It is easier to prove the contrapositive, which is:

For each positive real number

, if

\sqrt{x}

is rational, then

is rational.

Let $x$ be a positive real number. If there exist positive integers $m$ and $n$ such that $\sqrt{x} = \dfrac{m}{n}$ , then $x = \dfrac{m^2}{n^2}$ .

한 번 더 — 비슷한 문제Sundstrom, Mathematical Reasoning · 3.1절영어 원문

Use a counterexample to prove the following statement is false.

For all integers

and

, if 5 divides

or 5 divides

, then 5 divides

(5a + b)

다 풀었으면 정답 확인

A counterexample for this statement will be values of $a$ and $b$ for which 5 divides $a$ or 5 divides $b$ , and 5 does not divide $5a + b$ . One counterexample for the statement is $a = 5$ and $b = 1$ . For these values, the hyothesis is true since 5 divides $a$ and the conclusion is false since $5a + b = 26$ and 5 does not divide 26.

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