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Ask QuestionPosted by Aditi Singh 5 years, 8 months ago
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Posted by Sachin Kumar 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Let the three consecutive positive integers be n, n + 1 and n + 2, where n is any integer.
By Euclid’s division lemma, we have
a = bq + r; 0 ≤ r < b
For a = n and b = 3, we have
n = 3q + r ...(i)
Where q is an integer and 0 ≤ r < 3, i.e. r = 0, 1, 2.
Putting r = 0 in (i), we get
{tex}n = 3q{/tex}
∴ n is divisible by 3.
{tex}n + 1 = 3q + 1{/tex}
∴ n + 1 is not divisible by 3.
{tex}n + 2 = 3q + 2{/tex}
∴ n + 2 is not divisible by 3.
Putting r = 1 in (i), we get
{tex}n = 3q + 1{/tex}
∴ n is not divisible by 3.
{tex}n + 1 = 3q + 2{/tex}
∴ n + 1 is not divisible by 3.
{tex}n + 2 = 3q + 3 = 3(q + 1){/tex}
∴ n + 2 is divisible by 3.
Putting r = 2 in (i), we get
{tex}n = 3q + 2{/tex}
∴ n is not divisible by 3.
{tex}n + 1 = 3q + 3 = 3(q + 1){/tex}
∴ n + 1 is divisible by 3.
{tex}n + 2 = 3q + 4{/tex}
∴ n + 2 is not divisible by 3.
Thus for each value of r such that 0 ≤ r < 3 only one out of n, n + 1 and n + 2 is divisible by 3.
Posted by Gayathri J S 7 years, 8 months ago
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Posted by S K 7 years, 8 months ago
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Posted by Yogita Baliyan 7 years, 8 months ago
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Posted by Harshit Pandey Pandey 7 years, 8 months ago
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Radhika :) Mishra:} 7 years, 8 months ago
Prakash Mishra 7 years, 8 months ago
Posted by Sushant Lokhande 7 years, 8 months ago
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Posted by Naman Singhal 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Let R and I are rational number and irrational number respectively.
Assume that sum of R and I is a rational number and equal to P<u1:p></u1:p>
So R + I =P<u1:p></u1:p>
or I =P - R......., (1)<u1:p></u1:p>
As P and R both are rational number so P - R is also a rational number.<u1:p></u1:p>
Hence from (1) I is a rational number<u1:p></u1:p>
But this contradict that I is an irrational number.<u1:p></u1:p>
This contradiction has come because we assumed that R+ I is a rational number.<u1:p></u1:p>
Therefore the sum of irrational number and rational number is always an irrational number.<u1:p></u1:p>
Posted by Bragadeeshwaran I A S 7 years, 8 months ago
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Posted by Nanthni Saran 7 years, 8 months ago
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Posted by Payal Yadav 7 years, 8 months ago
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Posted by Anirudh Bhardwaj 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Since, any odd positive integer n is of the form 4m + 1 or 4m + 3.
if n = 4m + 1
n2 = (4m + 1)2
= 16m2 + 8m + 1
= 8(2m2 + m) + 1
So n2 = 8q + 1 ........... (i)) (where q = 2m2 + m is a positive integer)
If n = (4m + 3)
n2 = (4m + 3)2
= 16m2 + 24m + 9
= 8(2m2 + 3m + 1) + 1
So n2 = 8q + 1 ...... (ii) (where q = 2m2 + 3m + 1 is a positive integer)
From (i) and (ii) we conclude that the square of an odd positive integer is of the form 8q + 1, for some integer q.
Posted by P P 7 years, 8 months ago
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Posted by Deepanshu Sahu 7 years, 8 months ago
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Posted by Shweta Kumari 7 years, 8 months ago
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Posted by Harsh Yadav 7 years, 8 months ago
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Posted by Pavan Pavu 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
LCM(p,q) = a3b3
and HCF(p,q) = a2b
LCM(p,q) {tex}\times{/tex} HCF(p,q) = a3b3 {tex}\times{/tex} a2b
LCM (p,q) {tex}\times{/tex} HCF (p,q) = a5b4 .... (1)
and pq = a2b3 {tex}\times{/tex} a3b= a5b4 .... (2)
from eqn (1) and (2)
LCM (p, q) {tex}\times{/tex} HCF (p, q) = pq
Posted by Shivam Jaybhaye 7 years, 8 months ago
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Posted by Rahul Kumar 7 years, 8 months ago
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Aditi Singh 7 years, 7 months ago
1Thank You