Question

prove that the following are irrational 1/√2​

Answer 1

About The Topic:

➜ What are $\bf \purple {Rational~Numbers}$?

$\implies$ A number 'x' is called Irrational if it cannot be written in the form of ${\bf{p/q}}$, where p and q are Integers and q ≠ 0 .

➜ There are Infinitely many rational numbers.

➜ Examples- ${\bf{{\sqrt{2} , \: \: \sqrt{3} , \: \: \sqrt{15} \: \: etc}}}$

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Question:

• Prove that 1/√2 is an Irrational Number.

Solution:

• Let's Assume 1/√2 is a Rational Number!

So,

• We can write this as $\frac{1}{ \sqrt{2} } = \frac{a}{b} -- (1)$

Here,

— A and B are two Co-Prime numbers, and B is not equal to Zero.

Now,

• Simplifying the equation (1) multiplying by $\sqrt{2}$ both sides, we get:

• $\textbf{1 = a√2/b}$

Now,

• Dividing by 'b', we get:

⇢${\bf{a√2}}$

$~~~~~~~~~~~~$ ${\bf{Or}}$

⇢ ${\bf{b/a = √2}}$

Here,

• a and b are Integers

• So, $\frac{b}{a}$ is a rational Number!

So,

• $\sqrt{2}$ should be a rational number.

Therefore:

• ${\bf{\frac{1}{ \sqrt{2} }}}$ is an Irrational number.

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*Note: When we use the symbol √ , we assume that it is the positive square root of the number.

Answer 2

GIVEN:

1/√2

TO FIND:

Prove that 1/√2 is irrational.

SOLUTION:

Let us assume, to the contrary, that 1/√2 is rational. That is, we can find co - prime integers p and q (q ≠ 0) such that

$\rm{\dashrightarrow \dfrac{1}{\sqrt{2}} = \dfrac{p}{q}}$

$\rm{\dashrightarrow \dfrac{ 1 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \dfrac{p}{q}}$

$\rm{\dashrightarrow \dfrac{\sqrt{2}}{2} = \dfrac{p}{q}}$

☆$\rm{\dashrightarrow \sqrt{2} = \dfrac{2p}{q} }$☆

• Since, p and q are integers so 2p/q is rational, and so √2 is rational.

But this contradicts the fact that √2 is irrational.

So, we conclude that √2 is an irrational.

• Hence, 1/√2 is an irrational number.

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