Question

please solve this question​

Answer 1

Hello,

Here we go,

Let n be some arbitrary positive integer.

When when divided by positive integer to then let its quotient be q and remainder be r.

According to equality division lemma,

n = 2q + r

Where r = 0<=r<2

So in this case are can be 0 or 1

n = 2q when r is 0

so when r is zero then the given equation is even where Q is some positive integer

n = 2q + 1

here reminder is one so the given equation is odd positive integer for some integer q.

I hope my answer will help you.

Answer 2

Answer:

Show that every positive even integer is of the form 2q and every positive odd integer is of the form 2q+1, where q is a whole number.

December 26, 2019avatar

Sanjeev Garg

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ANSWER

(i) Let 'a' be an even positive integer.

Apply division algorithm with a and b, where b=2

a=(2×q)+r where 0≤r<2

a=2q+r where r=0 or r=1

since 'a' is an even positive integer, 2 divides 'a'.

∴r=0⇒a=2q+0=2q

Hence, a=2q when 'a' is an even positive integer.

(ii) Let 'a' be an odd positive integer.

apply division algorithm with a and b, where b=2

a=(2×q)+r where 0≤r<2

a=2q+r where r=0 or 1

Here r



=0 (∵a is not even) ⇒r=1

∴a=2q+1

Hence, a=2q+1 when 'a' is an odd positive integer.

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