State Euclid Division Lemma, using it show that square of a any positive number is either
of the form 5m or 5m ±1 where m is an integer
Answers
Step-by-step explanation:
Given :-
Square of any positive number
To find :-
State Euclid Division Lemma, using it show that square of a any positive number is either of the form 5m or 5m ±1 where m is an integer.
Solution :-
We know that
Euclid's Division Lemma:-
For any two positive integers a and b there exists q and r satisfying a = bq+r, Where, 0≤r<b
Consider a positive number
a = 5q+r -----------(1)
Where, 0≤r<5
The possible values of r = 0,1,2,3,4.
Case-1:-
If r = 0 in (1) then
a = 5q+0
=> a = 5q
On squaring both sides then
=> a² = (5q)²
=> a² = 25q²
=> a² = 5(5q²)
=> a² = 5m ---------------(2)
Where, m = 5q²
Case -2:-
If r = 1 in (1) then
a = 5q+1
On squaring both sides then
=> a² = (5q+1)²
=> a² = (5q)²+2(5q)(1)+1²
Since (a+b)² = a²+2ab+b²
=> a² = 25q²+10q+1
=> a² = 5(5q²+2q)+1
=> a² = 5m +1---------------(3)
Where, m = 5q²+2q
Case -3:-
If r = 2 in (1) then
a = 5q+2
On squaring both sides then
=> a² = (5q+2)²
=> a² = (5q)²+2(5q)(2)+2²
Since (a+b)² = a²+2ab+b²
=> a² = 25q²+20q+4
=> a² = 25q²+20q+5-1
=> a² = 5(5q²+4q+1)-1
=> a² = 5m -1---------------(4)
Where, m = 5q²+4q+1
Case -4:-
If r = 3 in (1) then
a = 5q+3
On squaring both sides then
=> a² = (5q+3)²
=> a² = (5q)²+2(5q)(3)+3²
Since (a+b)² = a²+2ab+b²
=> a² = 25q²+30q+9
=> a² = 25q²+30q+10-1
=> a² = 5(5q²+6q+2)-1
=> a² = 5m -1---------------(4)
Where, m = 5q²+6q+2
From (2),(3) & (4)
We conclude that
"The square of any positive integer is in the form of either 5m or 5m+1 or 5m-1".
Hence, Proved.
Answer:-
The square of a any positive number is either of the form 5m or 5m ±1 where m is an integer.
Used Formulae:-
Euclid's Division Lemma:-
For any two positive integers a and b there exists q and r satisfying a = bq+r, Where, 0≤r<b
→(a+b)² = a²+2ab+b²