Question

prove that √7 is an irrational number and hence show that 3-2√7 is irrational number.

plz answer fast!​

Answer 1

Answer:

Lets assume that √7 is rational number. ie √7=p/q.

suppose p/q have common factor then

we divide by the common factor to get √7 = a/b were a and b are co-prime number.

that is a and b have no common factor.

√7 =a/b co- prime number

√7= a/b

a=√7b

squaring

a²=7b² .......1

a² is divisible by 7

a=7c

substituting values in 1

(7c)²=7b²

49c²=7b²

7c²=b²

b²=7c²

b² is divisible by 7

that is a and b have atleast one common factor 7. This is contridite to the fact that a and b have no common factor.This is happen because of our wrong assumption.

√7 is irrational

Step-by-step explanation:

Answer 2

Let _/7 is a rational no

_/7 = p/ q where pand q both are co- prime no.s whose H.C.F. is 1

now,

_/7 = p/ q

On squaring both sides

7= p^2/ q^2

p^2 = 7q^2

p= _/7q

•°• _/7 is the factor of p

Now, p= 7m

put the value of p in equation ll

•°• _/7 q = (7m)^2

q = _/7 m

•°• again _/7 is the factor of q

•°• Our assumption is wrong _/7 is an irrational no. becoz it has more than two H.C.F


ll• 3 - 2 _/7


Let 3- 2 _/7 is a rational no

3 - 2_/7 = p/ q , where p and q both are co - prime whose H.C.F is 1

3 - 2 _/7 = p/ q

- 2 _/7 = p - 3q by q

_/7 = -( p - 3q ) by 2q

_/7 = 3q + p by 2q

•°• Our Contradiction is wrong becoz irrational never = rational no.

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