Question

Writing and Proof: If true prove it, if false give a counterexample. Use contradiction when proving.
(a) For each positive real number x, if x is irrational, then \(\displaystyle{x}^{{2}}\) is irrational.
(b) For every pair of real numbers a nd y, if x+y is irrational, then x if irrational or y is irrational

Answer 1

(a) The given statement is false. Because for each positive real number x, if x is irrational number, then it is not necessary to their square \(\displaystyle{x}^{{2}}\) is also irrational. For example:
consider \(\displaystyle{x}=\sqrt{{5}}\) ((it is positive real number) Then,
\(\displaystyle{x}^{{2}}={\left(\sqrt{{5}}\right)}^{{2}}\)
\(\displaystyle{x}^{{2}}={5}\) {since, \(\displaystyle{\left(\sqrt{{a}}\right)}^{{2}}={a}\)}
Since, 5 is not an irrational number because it can be written in the form of \(\displaystyle\frac{{p}}{{q}}\) {where p and q are integer}.
That is \(\displaystyle{5}=\frac{{5}}{{1}}\).
Hence, given statement is false.
(b)The given statement is true.To prove this, contrary assume that x and y both are rational number. So, \(\displaystyle{x}=\frac{{a}}{{b}}\) {where a and b are the integers}
\(\displaystyle{y}=\frac{{c}}{{d}}\) {where c and d are the integers}
According to the given statement:
x+y=irrational number
\(\displaystyle\frac{{a}}{{b}}+\frac{{c}}{{d}}\)=ZSK irrational number
\(\displaystyle\frac{{{a}{d}+{c}{d}}}{{{b}{d}}}=\) irrational number
Let, ad+cd=p and bd=q.
Then,
\(\displaystyle\frac{{p}}{{q}}=\) irrational number
rational number=irrational number
Which is the contradiction. Therefore, our consideration values {x and y both are rational number} are wrong. Hence, the given statement is true.
For example:
\(\displaystyle{x}={2},{y}=\sqrt{{3}}\), then \(\displaystyle{x}+{y}={2}+\sqrt{{3}}=\) irrational