Question

Prove that for all integers m, if even then 3m + 5 is odd.

Transformation properties
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asked 2021-01-08
Prove that for all integers m, if even then \(3m\ +\ 5\) is odd.

Answers (1)

2021-01-09
Proof:
Let m any odd integer.
The definition of even integer gives,
\(m = 2p\)
Here, p is also integer.
According to the question,
\(3m\ +\ 5 =3(2p)\ +\ 5\)
\(= 6p\ +\ 5\)
\(= 2(3p)\ +\ 2(2)\ +\ 1\)
\(= 2(3p\ +\ 4)\ +\ 1\)
Let, \((3p\ +\ 4) = a\)
Here, a is integer.
The above relation implies that,
\(3m\ +\ 5 = 2a\ +\ 1\)
The above relation \(3m\ +\ 5 = 2a\ +\ 1\) implies the definition of odd integer.
Therefore, for all integers m, if m is even then \(3m\ +\ 5\) is odd.
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