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 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

(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
${\displaystyle \frac{a}{b}+\frac{c}{d}}$= irrational number
${\displaystyle \frac{ad+cd}{bd}=}$ 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