Proof by Contradiction : Show that √ 2 is irrational.

Answer: [tex]\sqrt2[/tex] is irrational Step-by-step explanation:Let us assume that [tex]\sqrt2[/tex] is rational. Thus, it can be expressed in the form of fraction [tex]\frac{x}{y}[/tex], where x and y are co-prime to each other.[tex]\sqrt2[/tex] = [tex]\frac{x}{y}[/tex]Squaring both sides,[tex]2 = \frac{x^2}{y^2}[/tex]Now, it is clear that x is an even number. So, let us substitute x = 2uThus,[tex]2 = \frac{(2u)^2}{y^2}\\y^2 = 2u^2[/tex]Thus, [tex]y^2[/tex]is even, which follows the fact that y is also an even number. But this is a contradiction as x and y have a common factor that is 2 but we assumed that the fraction [tex]\frac{x}{y}[/tex]  was in lowest form.Hence, [tex]\sqrt2[/tex] is not a rational number. But [tex]\sqrt2[/tex] is a an irrational number.