Question

prove 2/3 irrational​

Answer 1

m/n = √2 + √3

Square both sides:

m^{2} / n^{2} = 5 + 2√6

"Solve" for √6:

√6 = (m^2 - 5n^2) / (2n^2)

so if √2 + √3 is rational, then so is √6.

Let a and b be the integers with gcd(a,b) = 1 such that

a/b = √6

Square both sides and multiply by b^2:

a^2 = 6b^2

Now, the right side is divisible by 2, so a^2 is divisible by 2, which then implies that a is divisible by 2 (since 2 is prime). Therefore we can write a=2k for some integer k:

4k^2 = (2k)^2 = 6b^2

Divide by 2:

2k^2 = 3b^2

Now the left side is divisible by 2, so 3b^2 is divisible by 2, from which it follows that b is divisible by 2.

However, this would mean that 2 divides gcd (a,b) = 1. Contradiction.

Therefore √6 is irrational, and therefore √2 + √3 is irrational also.

Hence, proved...

Thanks..

Answer 2

$\huge\fcolorbox{red}{yellow}{HEY MATE}$

An $\bold {irrational\:number}$ cannot be expressed as a ratio between two numbers and cannot be written as a simple fraction.

But, 2/3 can be expressed as a ratio between two numbers and can be written as a simple fraction. So, $\bold{2/3 \:is \:rational.}$

Since 2/3 is rational, we cannot prove it as irrational.

$\large\underline\bold\pink {HOPE\:\: THAT\: \:HELPS \:\:U}$

$<marquee> ---PLZZ MARK AS BRAINLIEST ---$