show that any positive odd integer is form 6q+1 6q+3 or 6q+5 and q is some integer
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Answered by
62
By Euclid's division lemma:-
a = bq+r where, 0<r<b
Let b = 6
So, a = 6q+r where, 0<r<6
Possible remainders are:- 0,1,2,3,4 and 5
Case(i):-
When r=0
a=6q (it is even)
Case(ii):-
When r=1
a=6q+1 (it is odd)
Similarly, For, r= 3 and 5, a is odd and for r= 2 and 4, a is even.
Thus, any positive odd integer is of the form:- 6q+1, 6q+3, 6q+5
Hence, proved
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Hope it helped you!!
a = bq+r where, 0<r<b
Let b = 6
So, a = 6q+r where, 0<r<6
Possible remainders are:- 0,1,2,3,4 and 5
Case(i):-
When r=0
a=6q (it is even)
Case(ii):-
When r=1
a=6q+1 (it is odd)
Similarly, For, r= 3 and 5, a is odd and for r= 2 and 4, a is even.
Thus, any positive odd integer is of the form:- 6q+1, 6q+3, 6q+5
Hence, proved
______________________________
Hope it helped you!!
Sagittarius29:
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ok
Answered by
15
By Euclid’s division algorithm, for two positive integers a and b, we have
a = bq + r, 0 ≤ r < b
Let b = 6,
r = 0, 1, 2, 3, 4, 5
a = 6q, 6q + 1, 6q + 2, 6q + 3,
6q + 4, 6q + 5
Clearly, a = 6q, 6q + 2, 6q + 4 are even,as they are divisible by 2.
But 6q + 1, 6q + 3, 6q + 5 are odd, as they are not divisible by 2
Hence,
Any positive odd integer is of the form 6q + 1, 6q + 3 or 6q + 5.
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