Dottie Parra

2021-01-08

Prove that for all integers m, if even then $3m\text{}+\text{}5$ is odd.

Nola Robson

Skilled2021-01-09Added 94 answers

Proof:

Let m any odd integer.

The definition of even integer gives,

$m=2p$

Here, p is also integer.

According to the question,

$3m\text{}+\text{}5=3(2p)\text{}+\text{}5$

$=6p\text{}+\text{}5$

$=2(3p)\text{}+\text{}2(2)\text{}+\text{}1$

$=2(3p\text{}+\text{}4)\text{}+\text{}1$

Let,$(3p\text{}+\text{}4)=a$

Here, a is integer.

The above relation implies that,

$3m\text{}+\text{}5=2a\text{}+\text{}1$

The above relation$3m\text{}+\text{}5=2a\text{}+\text{}1$ implies the definition of odd integer.

Therefore, for all integers m, if m is even then$3m\text{}+\text{}5$ is odd.

Let m any odd integer.

The definition of even integer gives,

Here, p is also integer.

According to the question,

Let,

Here, a is integer.

The above relation implies that,

The above relation

Therefore, for all integers m, if m is even then

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Expert2021-11-03Added 2605 answers

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