Math, asked by varunguptajii9571, 1 month ago

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.​

Answers

Answered by pawansharma941872794
0

Answer:

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Answered by TrustedAnswerer19
4

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.

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