Show that any positive odd integer …
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Sia ? 6 years, 6 months ago
Let a be any positive integer and b = 6
∴ by Euclid’s division lemma
a = bq + r, 0≤ r and q be any integer q ≥ 0
∴ a = 6q + r,
where, r = 0, 1, 2, 3, 4, 5
If a is even then then remainder by division of 6 is 0,2 or 4
Hence r=0,2,or 4
or A is of form 6q,6q+2,6q+4
As, a = 6q = 2(3q), or
a = 6q + 2 = 2(3q + 1), or
a = 6q + 4 = 2(3q + 2).
If these 3 cases a is an even integer.
but if the remainder is 1,3 or 5 then r=1,3 or 5
or A is of form 6q+1,,6q+3 or,6q+5
Case 1:a = 6q + 1 = 2(3q) + 1 = 2n + 1,
Case 2: a = 6q + 3 = 6q + 2 + 1,
= 2(3q + 1) + 1 = 2n + 1,
Case 3: a = 6q + 5 = 6q + 4 + 1
= 2(3q + 2) + 1 = 2n + 1
This shows that odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
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