√7 is
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 contradicting to the fact that a and b have no common factor.This happend because of our wrong assumption.
hence,√7 is irrational
For positive integers a and 3, there exist unique integers q and r such that a = 3q + r, where r must satisfy:
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.
Given : a = 3q+r
In this question ,
b = 3
The values 'r’ can take 0 ≤ r < 3.
For some integer q, every odd integer is of the form
We know that, odd integers are 1, 3, 5, ...
So, it can be written in the form of 2q + 1,
where, q = integer = Z
or q = ..., -1, 0,1,2,3, ...
∴ 2q + 1 = ..., -3, -1, 1, 3,5, ...
Alternate Method
Let 'a' be given positive integer. On dividing 'a' by 2, let q be the quotient and r be the remainder. Then, by Euclid's division algorithm, we have
a = 2q + r, where 0 ≤ r < 2
⇒ a = 2q + r, where r = 0 or r = 1
⇒ a = 2q or 2q + 1
when a = 2q + 1 for some integer q, then clearly a is odd.
Find the greatest number of 5 digits, that will give us remainder of 5, when divided by 8 and 9 respectively.
Greatest 5-Digit number = 99999
LCM of 8 and 9,
8 = 2 × 2 × 2
9 = 3 × 3
LCM = 2 × 2 × 2 × 3 × 3 = 72
Now, dividing 99999 by 72, we get
Quotient = 1388
Remainder = 63
So, the greatest 5-digit number divisible by 8 and 9 = 99999 - 63 = 99936
Required number = 99936 + 5 = 99941
The product of two consecutive integers is divisible by
As the case is of 2 consecutive integers, one will be odd and the other one even, so their product will always be even which eventually will be divisible by an even no., i.e. 2, here.
If two positive integers a and b are written as a = x3y2 and b = xy3, where x, y are prime numbers, then LCM(a, b) is
Here, a = x3y2 and b = xy3.
⇒ a = x * x * x * y * y and b = xy * y * y
∴ LCM(a, b) = x * x * x * y * y * y = x3 * y3 = x3y3
LCM = x3y3
n2 - 1 is divisible by 8 if n is
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.
If two positive integers p and q can be expressed as p = ab2 and q = a3b; where a, b being prime numbers, then LCM (p, q) is equal to
As per question, we have,
p = ab2 = a × b × b
q = a3b = a × a × a × b
So, their Least Common Multiple (LCM) = a3 × b2
The least perfect square number which is divisible by 3, 4, 5, 6 and 8 is
L.C.M. of 3, 4, 5, 6, 8 = 120 ;
Required number =2 x 2 x 2 x 2 x 3 x 3 x 5 x 5 = 3600.
The ratio between the LCM and HCF of 5,15, 20 is:
5, 15 = 5 x 3, 20 = 2 x 2 x 5
LCM(5, 15, 20) = 5 x 3 x 2 x 2 = 60
HCF(5, 15, 20) = 5
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