Question

show that root 7 is irrational and hence show that 5-3 root 7 is irrational​

Answer 1

Answer:

First Prove √7 is irrational

Assume√7->rational

√7=P/q[P, q are coprimes and q≠0]

q√7=P

Square both sides

7q^2=P^2 ......(1)

=> P^2 divides 7 thus P divides 7 (Irrationality theorem 1.3).......(2)

Similarly P = 7 a (a is some integer)

Subtituting it to (1)

7q^2=(7a)^2

7q^2=49 a^2

q^2=7a^2

=> q^2 divides 7 thus q divides 7 ( Irrationality theorem 1.3 ).......(3)

According to (2) and (3);

7 divides P and q ie..., 7 is the common factor of P and q

But earlier we said P and q are coprimes (have no common factor other than 1)

=> Assumption contradicts

So, √7 is irrational

Second prove 5-3√7 is irrational

Assume they are rational

5-3√7=P/q

3√7=5-P/q

3√7= (5q-P)/q

√7= (5q-P)/3q......(4)

Earlier we proved √7 is irrational or can't be expressed in fractional form (P/q)

here (4) we proved that √7 could be expressed in P/q form with contradicts

Thus our assumption (of 5-3√7 as rational) contradicts

Hence 5-3√7 is irrational

Hope this answer is helpful