(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

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