Question

show that 3-5Root2 is irrational
​

Answer 1

Answer:

hope it helps you

Step-by-step explanation:

soo min here

Answer 2

$\large{\red{\underline{To\:prove:}}}$

3-5√2 is irrational.

$\large{\red{\underline{Given:}}}$

√5 is irrational.

$\large{\red{\underline{Assumption:}}}$

Let (3-2√5) be a rational number.

$\large{\red{\underline{Solution:}}}$

So, it can be written in the form of p/q :-

$3 - 2 \sqrt{5} = \frac{p}{q} \\ - 2 \sqrt{5} = \frac{p - 3q}{q} \\ - \sqrt{5} = - ( \frac{p + 3q}{2q} ) \\ \sqrt{5} = \frac{p + 3q}{2q}$

Let

$\frac{p + 3q}{2q}$

be a rational number.

But we know,

√5 is irrational.

So, this contradiction arises dur to our wrong assumption that (3-2√5) is rational.

$\large{\boxed{Thus,\: (3-2√5) \:is \:irrational.}}$