Question

32. Prove that root 5 is irrational​

Answer 1

Answer:

Step-by-step explanation:

If √5 is rational, then it can be expressed by some number a/b (in lowest terms). This would mean:  

(a/b)² = 5. Squaring,  

a² / b² = 5. Multiplying by b²,  

a² = 5b².  

If a and b are in lowest terms (as supposed), their squares would each have an even number of prime factors. 5b² has one more prime factor than b², meaning it would have an odd number of prime factors.  

Every composite has a unique prime factorization and can't have both an even and odd number of prime factors. This contradiction forces the supposition wrong, so √5 cannot be rational. It is, therefore, irrational.

Answer 2

TO PROVE :- Root of 5 is irrational

PROOF

We can easily prove this by using the cobtradiction method :-

Let root 5 be x/y where x and y are co-primes.

$\sqrt{5} = \frac{x}{y} \\ \\ \\ \implies 5 = \frac{ {x}^{2} }{ {y}^{2} } \\ \\ \\ \implies 5 {y}^{2} = {x}^{2} \\ \\ \\ \implies {y}^{2} = \frac{ {x}^{2} }{5} \\ \\ \\ {x}^{2} is \: divisible \: by \: 5 \: so \: x \: is \: also \: divisible \: by \: 5 \\ \\ \\ now \: let \: \frac{x}{5} \: be \: s \\ \\ \\ \frac{x}{5} = s \\ \\ \\ \implies 5s = x \\ \\ \\ \implies 25 {s}^{2} = {x}^{2} \\ \\ \\ \implies 25 {s}^{2} = 5 {y}^{2} \\ \\ \\ \implies {s}^{2} = \frac{ {y}^{2} }{5} \\ \\ \\ hence \: {y}^{2} is \: divisible \: by \: 5 \: so \: y \: is \: also \: divisible \: by \: 5$

Hence both x and y are divisible by 5 so x and y aren't co-primes.

Therefore it contradicts our assumption so root 5 is irrational.