Question

prove that root 2 is irrational​

Answer 1

Answer:

Step-by-step explanation:

✓2=a/b

✓2b=a

2b=a^2

b=a^2/2

If a number is equal a^2 then it is also equal to a

✓2=b/c

✓2c=b

2c=b^2

c=b^2/2

If a number is equal to b^2 then it is also equal to b

We know that a&b are Co prime hence our assumption that ✓2 is rational is wrong

Hence ✓2 is irrational

Answer 2

Solution:

Let us assume that, √2 is a rational number of simplest form $\frac{a}{b}$, having no common factor other than 1.

√2 = $\frac{a}{b}$

On squaring both sides, we get ;

2 = $\frac{a^{2}}{b^{2}}$

⇒ a² = 2b²

Clearly, a² is divisible by 2.

So, a is also divisible by 2.

Now, let some integer be c.

⇒ a = 2c

Substituting for a, we get ;

⇒ 2b² = 2c

Squaring both sides,

⇒ 2b² = 4c²

⇒ b² = 2c²

This means that, 2 divides b², and so 2 divides b.

Therefore, a and b have at least 2 as a common factor. But this contradicts the fact that a and b have no common factor other than 1.

This contradiction has arises because of our assumption that √2 is rational.

So, we conclude that √2 is irrational.