# Chapter 1: Real Variables: examples II

Examples II.

1) Show that no rational number can have its cube equal to 2.

Proof 1.

Proof by contradiction. Let $x=p/q$. $q \neq 0, p, q \in Z.$ (p, q have no common factors).

Let $x^{3}=2$. Hence, $\frac{p^{3}}{q^{3}}=2$. Hence, $p^{3}=2q^{3}$. Hence, p^{3} is even because we know that even times even is even and even times odd i also even and odd times odd is odd. Hence, p ought to be even. Let $p=2m$. Then, again $q^{3}=4m^{3}$. Hence, $q^{3}$ is even. Hence, q is even. But, this means that p and q have a common factor 2 which contradicits our hypothesis. Hence, the proof.. QED.

Proof 2)

Let given rational fraction be $\frac{p}{q}$, $q \neq 0, p, q \in Z$.

Let $\frac{p}{q}=\frac{m^{3}}{n^{3}}$, $n \neq 0, m,n \in Z$.

Since p and q do not have any common factors, m and n also do not have any common factors.

Case I: p is even, q is odd so clearly, they do not have any common factors.

Case IIL p is odd, q is odd but with no common factors.

Case I: since m and n are without any common factors, and $m^{3}, n^{3}$ are also in its lowest terms, we have $p=m^{3}, q=n^{3}$.

Case II: similar to case I above.

Proof 3.

A more general proposition, due to Gauss, includes those two above problems as special cases. Consider the following algebraic equation; $x^{n}+p_{1}x^{n-1}+p_{2}x^{n-2}+\ldots +p_{n}=0$.

with integral coefficients,, cannot have a rational root but non integral root.

Proof 3:

For suppose that the equation has a root a/b, where a and b are integers without a common factor, and b is positive. Writing a/b for x, and multiply both the sides of the equation $b^{n-1}$, e obtain $-\frac{a^{n}}{b}=p_{1}a^{n-1}+p_{2}a^{n-2}b+\ldots +p_{n}b^{n-1}$,

a fraction in the lowest terms equal to an integer, which is absurd. thus, b=1, and the root is a. It is clear that a must be a divisor of $p_{n}$

Proof 4.

Show that if $p_{n}=1$ and neither of $1+p_{1}+p_{2}+p_{3}+\ldots$,, $1-p_{1}+p_{2}-p_{3}+\ldots$ is zero, then the equation cannot have a rational root.

I will put the proof later.

Problem 5.

Find the rational toots, if any of $x^{4}-4x^{3}-8x^{2}+13x+10=0$.’

Solution.

The roots can only be integral and so $\pm 1, \pm 2, \pm 3, \pm 5 pm 10$ are the only possibilities: whether these are roots can be determined by tiral. it is clear that can in this way determine the rational roots of any equation.

More later,

Nalin Pithwa

# Analysis: Chapter 1: part 10: algebraic operations with real numbers

Algebraic operations with real numbers.

We now proceed to meaning of the elementary algebraic operations such as addition, as applied to real numbers in general.

(i),  Addition. In order to define the sum of two numbers $\alpha$ and $\beta$, we consider the following two classes: (i) the class (c) formed by all sums $c=a+b$, (ii) the class (C) formed by all sums $C=A+B$. Clearly, $c < C$ in all cases.

Again, there cannot be more than one rational number which does not belong either to (c) or to (C). For suppose there were two, say r and s, and let s be the greater. Then, both r and s must be greater than every c and less than every C; and so $C-c$ cannot be less than $s-r$. But, $C-c=(A-a)+(B-b)$;

and we can choose a, b, A, B so that both $A-a$ and $B-b$ are as small as we like; and this plainly contradicts our hypothesis.

If every rational number belongs to (c) or to (C), the classes (c), (C) form a section of the rational numbers, that is to say, a number $\gamma$. If there is one which does not, we add it to (C). We have now a section or real number $\gamma$, which must clearly be rational, since it corresponds to the least member of (C). In any case we call $\gamma$ the sum of $\alpha$ and $\beta$and write $\gamma=\alpha + \beta$.

If both $\alpha$ and $\beta$ are rational, they are the least members of the upper classes (A) and (B). In this case it is clear that $\alpha + \beta$ is the least member of (C), so that our definition agrees with our previous ideas of addition.

(ii) Subtraction.

We define $\alpha - \beta$ by the equation $\alpha-\beta=\alpha +(-\beta)$.

The idea of subtraction accordingly presents no fresh difficulties.

More later,

Nalin Pithwa