Question

Show that every positive even integer is of the form 2m every positive odd integer is of the form 2m + 1 where m some integer.​

Answer 1

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Answer 2

Answer:

Let a be any positive integer and b = 2. Then, vy Euclid's division lemma there exist integers

m and r such that

$\bf \: a = 2m + r \: \: \: where \: \: \: 0 \leqslant r < 2 \\ \\ \sf \: now \\ \bf \: 0 \leqslant r < 2 \\ \bf \implies \: 0 \leqslant r < 1 \\ \bf \implies \:r = 0 \: \: or \: r = 1 \: \: \: \: \pink{ \{ \because \: r \: is \: an \: integer \}} \\ \\ \bf \therefore \: a = 2m \: \: or \: \: a = 2m + 1 \\ \\ \sf \: if \: a \: = 2m \: then \: \: a \: is \: n \: even \: integer. \\ \\ \sf \: we \: know \: that \: an \: integer \: can \: be \: either \\ \sf \: even \: or \: o dd. \: therefore \: any \: odd \: integer \\ \sf i s\: of \:the \: form\: \: \pink{ \: 2m + 1}$

Hence showed.

Other method :

Let a be any positive integer.

On dividing a by 2,

Assume m be the quotient.

And r be the remainder.

a = 2m + r,

where 0 ≤ r < 2.

Therefore,

a = 2m or (2m + 1), for some integer m.

Case-1:

in this case, a is clearly even.

Case-2:

In this case, a is clearly odd.

Hence, every positive even integer is of the form 2m and every positive odd integer is of the form (2m + 1) for some integer m.