Question

prove that 3-2√5 is an irrational number
please answer me quickly

Answer 1

Answer:

Let us assume that 3 + 2 root 5 is a rational number.  

So we can write this number as  

3 + 2 root 5 = a/b  

Here a and b are two integers and  co prime numbers and b is not equal to 0

Subtract 3 both sides we get  

2 root 5 = a/b - 3

2 root 5 = (a-3b)/b

Now divide by 2 we get  

root 5 = (a-3b)/2b

Here a and b are integer so (a-3b)/2b is a rational number so root 5 should be a rational number But root 5 is a irrational number so it contradict the fact  

Hence result is 3 + 2 root 5 is a irrational number

hope it helps you

plzzzz mark it brainliest

and practice it thoroughly as it can come in the cbse final exam of class 10 ok

have a nice day

Answer 2

$\huge \underline{ \underline {\mathfrak{answer = }}}$

Let us assume that $3 + 2\sqrt{5}$ is a rational no.

$3 + 2 \sqrt{5} = \frac{a}{b}$

(Where a and b are co- prime numbers and b≠0)
$3 - \frac{a}{b} = 2 \sqrt{5}$

$\frac{3b - a}{b} = 2 \sqrt{5}$

$\frac{3b - a}{2b} = \sqrt{5}$

$\frac{3b - a}{2b}$It is a rational no. Where a and b are integers.

But, $\sqrt{5}$is a irrational no. But above states that it is rational, but it is a contradiction.

So, we can conclude that $3 + 2\sqrt{5}$ is a irrational no.

$<marquee>$
.... MAY IT WILL HELP YOU...