Question

prove that
$\sqrt{2}$
is an irrational number pls pls pls​

Answer 1

Answer:

Let us assume that √2 is a rational number.

then, as we know a rational number should be in the form of p/q

where p and q are co- prime number.

So,

√2 = p/q { where p and q are co- prime}

√2q = p

Now, by squaring both the side

we get,

(√2q)² = p²

2q² = p² ........ ( i )

So,

if 2 is the factor of p²

then, 2 is also a factor of p ..... ( ii )

=> Let p = 2m { where m is any integer }

squaring both sides

p² = (2m)²

p² = 4m²

putting the value of p² in equation ( i )

2q² = p²

2q² = 4m²

q² = 2m²

So,

if 2 is factor of q²

Then 2 is also factor of q

Since

2 is factor of p & q both

So, our assumption that p & q are co- prime is wrong

Hence √2 is an irrational number

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Answer 2

Step-by-step explanation:

Let us assume to the contrary that $\sqrt{2 }$ is a rational number.

It can be expressed in the form of p/q

where p and q are co-primes and q≠ 0.

⇒ √2 = p/q

⇒ 2 = p2/q2 (Squaring on both the sides)

⇒ 2q2 = p2………………………………..(1)

This means that 2 divides p2. This means that 2 divides p because each factor should appear two times for the square to exist.

So we have p = 2r

where r is some integer.

⇒ p2 = 4r2………………………………..(2)

from equation (1) and (2)

⇒ 2q2 = 4r2

⇒ q2 = 2r2

Where q2 is multiply of 2 and also q is multiple of 2

Then p, q have a common factor of 2. This runs contrary to their being co-primes. Consequently, p / q is not a rational number. This demonstrates that √2 is an irrational number.

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