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Snezhnost [94]
4 years ago
15

Proof by Contradiction : Show that √ 2 is irrational.

Mathematics
1 answer:
myrzilka [38]4 years ago
5 0

Answer:

\sqrt2 is irrational

Step-by-step explanation:

Let us assume that \sqrt2 is rational. Thus, it can be expressed in the form of fraction \frac{x}{y}, where x and y are co-prime to each other.

\sqrt2 = \frac{x}{y}

Squaring both sides,

2 = \frac{x^2}{y^2}

Now, it is clear that x is an even number. So, let us substitute x = 2u

Thus,

2 = \frac{(2u)^2}{y^2}\\y^2 = 2u^2

Thus, y^2is 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 \frac{x}{y}  was in lowest form.

Hence, \sqrt2 is not a rational number. But \sqrt2 is a an irrational number.

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