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 = 2u
Thus,
[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.
