Question

cube root of 7 is not a rational number proove that

Answer 1

let us assume that √7 be rational.

then it must in the form of p / q [q ≠ 0] [p and q are co-prime]

√7 = p / q

=> √7 x q = p

squaring on both sides

=> 7q2= p2 ------> (1)

p2 is divisible by 7

p is divisible by 7

p = 7c [c is a positive integer] [squaring on both sides ]

p2 = 49 c2 --------- > (2)

substitute p2 in equ (1) we get

7q2 = 49 c2

q2 = 7c2

=> q is divisible by 7

thus q and p have a common factor 7.

there is a contradiction

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

so that √7 is an irrational.

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Answer 2

Rational number is a number which can be written in the form of p / q where p and q are integers and q ≠ 0.

here cube root of 7 is in place of p.
cube root of 7 is not integer.
as we can see, from the above definition there should be integers p and q and q does not equals to zero.

cube root of 7 is not an integer so it is irrational number not rational.

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