Question

Prove that√2+√4is irrational

Answer 1

Answer:

Let us assume on the contrary at root 2+ root 4 is rational

Then, there exist co-prime positive integer a and b

such that.

root 2+root 4=a/b

root 2+a/b=root 4

root 2b-a/b= root 4

root 4 is rational [a, b are integers root 2b-a/b is a rational number.

our assumption is incorrect.

Answer 2

Answer:

$R = QUQ'$

If possible let $\sqrt{2}$ is a rational number (assume)

$\sqrt{2} = \frac{a}{b}\: ,\:\:b\neq 0\:,\:(a,b)=1$

Squaring on both sides

$\sqrt{2} \times \sqrt{2} = \frac{a^2}{b^2}$

$\frac{2}{1} = \frac{a^2}{b^2}$

$a^2 = 2b^2....1$

$a^2\: \bold{is\:divisible\:by\:2}$

$a\: \bold{is\:also\:divisible\:by\:2}$

$a = 2k...2$

2 in 1

$2k.2k = 2b^2$

$2k^2 = b^2$

$b^2 =2k^2$

$b^2\: \bold{is\:divisible\:by\:2}$

$b\: \bold{is\:also\:divisible\:by\:2}$

$b = 2l...3$

HCF of a,b is

$(a,b)=2$

This is impossible/contradiction

∴ our assumption $\sqrt{2}$ is rational is wrong

$\green{\mathfrak{Hence\:\sqrt{2} \:is\:an\:irrational\:number}}$.

HENCE PROVED↑

$\red{\huge{\boxed{\mathfrak{Hope \:its\:help\:You}}}}$