Question

a) Show that V2 is an irrational number.

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Answer 1

$\LARGE{ \underline{ \blue{ \sf{Required \: answer:}}}}$

This proof starts with the assumption that √2 is equal to a rational number a / b. Let us assume on the contrary, and here a and b are coprimes.

➝ √2 = a/b

Squaring both sides,

➝ 2 = a² / b²

Cross multiplying,

➝ 2b² = a²

From here, we can conclude that a² is divisible by 2. Since a² is divisible by 2, it can be inferred that a is also divisible by 2.

Since a is even, it can be written as 2p where p is some other whole number.

Substituting a = 2p in the above equation:

➝ 2b² = a²

➝ 2b² = (2p)²

➝ 2b² = 4p²

Dividing both sides of the equation by 2:

➝ b² = 2p²

By the same logic as before, b² is also divisible by 2. So q is an even number. This contradicts our assumption that a and b are coprimes because they have also a common factor 2.

Thus, √2 is a irrational number !!!

Answer 2

$\large\green\bf{ANSWER :-}$

✳️ Given ✳️

• √2 is an irrational number.

✳️ To Prove ✳️

• We have to prove that√2 is an irrational number.

✳️ Solution ✳️

✒️ Let√2 is an irrational number.

So, √2 = $\dfrac{p}{q}$ (where p and q are co - prime number and q is not equal to 0)

➕ Squaring both sides we get,

$\implies$ 2 = $\dfrac{p²}{q²}$

$\red\bigstar$ Cross multiplication :-

$\implies$ q² = 2p² ........ Equation no (1)

So, we can say that q² is divisible by 2 and q is also divisible by 2.

Again,

✒️ Let, q = 2r

➕ Squaring both sides we get,

$\implies$ q² = 4r² ........ Equation no (2)

From equation no (1) and (2) we get

$\implies$ 2p² = 4r²

➕ Divided both sides by 2 we get,

$\implies$ p² = 2r²

$\therefore$ So, we can say that p² is divisible by 2 and p is also divisible by 2.

And, from this we have reach a conclusion that p and q both divisible by 2.

It contradict the fact that p and q are co- prime number.

Hence, we can say that√2 is an irrational number.

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Step-by-step explanation: