Introduction to Calculus and Analysis I Section1 Answer

The answer of Problem Section1, page 106

SECTION 1.1a, page 2

Q1. (a)

​ Assume that is rational, then

$$a+x=\frac{p_{ax}}{q_{ax}}$$

where and have no common factors.

Since is rational, then

$$x = \frac{p_{ax}}{q_ax} - \frac{p_a}{q_a}$$

however, this contradicts to the statement of problem.

The prove is same to .

Q1.(b)

​ Assume that and are two rational numbers, we find that

$$a<a+\frac{b-a}{\sqrt2}<b$$

clearly, $a+\frac{b-a}{\sqrt2}$ is irrational.

Q2.(b)

​ Assume, for the sake of contradicion, that $\sqrt n$ is rational. This means it can be expressed as the ratio of two integers, $p$ and $q$, where $p\ne 0$ and the fraction is in its simplest form:

$$\sqrt n= \frac{p}{q}$$

Squaring the both sides, we get:

$$n = \frac{p^2}{q^2} \\ p^2 = n q^2$$

which implies that is a factor of , then must have factor . We say , substituing for gives us

$$n^2p'^2 = nq^2\\ q^2 = np'^2$$

hence, and share a common factor , this contradicts our assumption that is in its simplest form.

Therefore, our initial assumption that is is rational must be false. Thus, we conclude that is irrational.