Question

Prove that root 5 is irrational ​

Answer 1

let us assume to the contrary that √5 is rational. so, we can find coprimes a and b such that√5=a/b,where b≠0

√5b=a

squaring both sides,

5a^2=b^2---(1)

a^2=b^2/5

5 divides b^2, 5 divides b.

so, b=5c for some integer c.s substituting in (1),

5a^2=(5c)^2

5a^2=25c^2 dividing by5,

a^2=5c^2

a^2/5=c^2

5 divides a^2,5 divides a.

so, a and be have atleast 5 as a factor. but this contradicts the fact that a and b are coprimes. this contradiction has arisen because of our incorrect assumption that√5 is rational. so,√5 is irrational

Answer 2

Answer: √5 is an irrational

Step-by-step explanation:

let root 5 be rational

then it must in the form of p/q [q is not equal to 0][p and q are co-prime]

root 5=p/q

=> root 5 * q = p

squaring on both sides

=> 5*q*q = p*p  ------> 1

p*p is divisible by 5

p is divisible by 5

p = 5c  [c is a positive integer] [squaring on both sides ]

p*p = 25c*c  --------- > 2

sub p*p in 1

5*q*q = 25*c*c

q*q = 5*c*c

=> q is divisble by 5

thus q and p have a common factor 5

there is a contradiction

as our assumsion p &q are co prime but it has a common factor

so √5 is an irrational