MCQ : Real Numbers - 1

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

Euclid’s division Lemma:

It tells us about the divisibility of integers. It states that any positive integer ‘a’ can be divided by any other positive integer ‘ b’ in such a way that it leaves a remainder ‘r’.

Euclid's division Lemma states that for any two positive integers ‘a’ and ‘b’ there exist two unique whole numbers ‘q’ and ‘r’ such that , a = bq + r, where 0≤ r < b.

Here, a= Dividend, b= Divisor, q= quotient and r = Remainder.

SOLUTION :

Given : a = 3q+r

In this question ,

b = 3

The values 'r’ can take 0 ≤ r < 3.

The greatest number will be multiple of LCM (8, 9)
LCM of 8 and 9 = 72
On verification we find that 99941 when divided by 72 leaves remainder 5.

As the case is of 2 consecutive integers, one will be odd nad the other one even, so their product will always be even which eventually will be divisible by an evn no., i.e. 2, here.

If n is even then it is not possible for n=2. If n is natural number then it is not possible for n=1,2.If n is integer similarly n²- 1 is not divisible by 8.If n is odd the possible for n is 3,5,7. Thus the n is odd number

LCM (p, q) = a³b² 

A=x³y²=x(x)(x)(y)(y)
b=xy³=x(y)(y)(y)
LCM of a and b=xy²(x²×y)
=x³y³
so option C is the correct answer