Question

prove √5 is irrational​

Answer 1

$\huge{\mathfrak{\green{\red{\bf\:S}\pink{o}\purple{l}\green{u}\orange{t}\blue{i}\blue{o}\red{n:-}}}}$

$\rm\:Let\:assume\:that\:\sqrt{5}\:be\:rational$

Then,we can write,

$\rm\:\sqrt{5}=\dfrac{a}{b}$ (where a and b are co-primes and b ≠ 0)

Now,

$\rm\:\sqrt{5}=\dfrac{a}{b}\implies\:{a}^{2}=5{b}^{2}..................(i)$

$\rm\implies\:{a}^{2}\:is\:a\:multiple\:of\:5$

$\rm\implies\:a\:is\:a\:multiple\:of\:5$

Let a = 5c for some positive integers c then,

$\rm\:a=5c\implies\:{a}^{2}=25{c}^{2}$

$\rm\implies\:5{b}^{2}=25{c}^{2}\:\:\:\:[using\:eq\:(i)]$

$\rm\implies\:{b}^{2}=5{c}^{2}$

$\rm\implies\:{b}^{2}\:is\:a\:multiple\:of\:5$

$\rm\implies\:b\:is\:a\:multiple\:of\:5$

Thus,5 is common multiple of a and b

This is a contradiction,since a and b are co-primes.

Since the contradiction arises by assuming that √5 is rational,

so,√5 is irrational.

Answer 2

Step-by-step explanation:

Step-by-step explanation:Solution is given in above pic

Step-by-step explanation:Solution is given in above pic Mark it brainlist answer