use Euclid division lemma to show that the square of any positive integer is either of the form of 3m or 3m+1
Answers
Step-by-step explanation:
Given Question:-
use Euclid division lemma to show that the square of any positive integer is either of the form of 3m or 3m+1
Solution:-
Let "a" be the square of an integer .
Given positive integers a and b there exists unique pair of integers q and r satisfying a = bq+r where 0≤r<b------(1)
On applying Euclid's Division Algorithm with a and b = 3
Since 0≤r<3 then the possible remainders are 0,1,2
Case -1:-
If r = 0 then (1) becomes
=>a= 3q+0
=>a= 3q
On squaring both sides
=>a^2 = (3q)^2
a^2 = 9q^2
=>a^2 = 3(3q^2)
=>a^2 = 3m, where m =3q^2-----(2)
Case -2:-
If r = 1 then (1) becomes
=>a=3q+1
On squaring both sides then
=>a^2 = (3q+1)^2
=>a^2=(3q)^2+2(3q)(1)+(1)^2
=>a^2=9q^2+6q+1
=>a^2=3(3q^2+2q)+1
=>a^2=3m+1 , where m =3q^2+2q -----(3)
Case -3:-
If r=2 then (1) becomes
=>a=3q+2
on squaring both sides then
=>a^2=(3q+2)^2
=>a^2=(3q)^2+2(3q)(2)+(2)^2
=>a^2=9q^2+12q+4
=>a^2= 9q^2+12q+3+1
=>a^2=3(3q^2+4q+1)+1
=>a^2=3m+1,where m=3q^2+4q+1-----(4)
From (2) ;(3) and (4)
It is clear that the square of any positive integer is of the form 3m or 3m+1
Hence ,proved .
Euclid's Division Lemma :-
Given positive integers a and b there exists unique pair of integers q and r satisfying a = bq+r where 0≤r<b
Answer:
It is the correct answer.
Step-by-step explanation:
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