 # 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 Chesley 2020-12-14 Answered
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
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(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 ${x}^{2}$ is also irrational. For example:
consider $x=\sqrt{5}$ ((it is positive real number) Then,
${x}^{2}={\left(\sqrt{5}\right)}^{2}$
${x}^{2}=5$ {since, ${\left(\sqrt{a}\right)}^{2}=a$}
Since, 5 is not an irrational number because it can be written in the form of $\frac{p}{q}$ {where p and q are integer}.
That is $5=\frac{5}{1}$.
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, $x=\frac{a}{b}$ {where a and b are the integers}
$y=\frac{c}{d}$ {where c and d are the integers}
According to the given statement:
x+y=irrational number
$\frac{a}{b}+\frac{c}{d}$= irrational number
$\frac{ad+cd}{bd}=$ irrational number
Then,
$\frac{p}{q}=$ 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:
$x=2,y=\sqrt{3}$, then $x+y=2+\sqrt{3}=$ irrational