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

Question

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

Nola Robson · Accepted Answer

Proof:
Let m any odd integer.
The definition of even integer gives,

m = 2 p

Here, p is also integer.
According to the question,

3 m + 5 = 3 (2 p) + 5

= 6 p + 5

= 2 (3 p) + 2 (2) + 1

= 2 (3 p + 4) + 1

Let,

(3 p + 4) = a

Here, a is integer.
The above relation implies that,

3 m + 5 = 2 a + 1

The above relation

3 m + 5 = 2 a + 1

implies the definition of odd integer.
Therefore, for all integers m, if m is even then

3 m + 5

is odd.

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

Don Sumner · Accepted Answer

Answer:3m+5 is oddExplanation:Let&#039;s assume that m is an even integer. By definition, this means that m can be expressed as m=2k, where k is an integer.We can substitute this value of m into the expression 3m+5:3m+5=3(2k)+5Simplifying this expression, we get:3m+5=6k+5Now, let&#039;s consider the parity (whether a number is even or odd) of 6k+5.An even number multiplied by an even number is always even. Adding an odd number (5) to an even number will always result in an odd number. Therefore, 6k+5 is odd.Thus, we have shown that if m is even (i.e., m=2k), then 3m+5 is odd (i.e., 3m+5 is of the form 2n+1, where n is an integer).Hence, we have proven that for all integers m, if m is even, then 3m+5 is odd.

Vasquez · Accepted Answer

Let&#039;s assume that m is an even integer. By definition, an even integer can be expressed as m=2k, where k is an integer.Now, we can substitute this value of m into the expression 3m+5 to see if it results in an odd number.3m+5=3(2k)+5=6k+5To show that 6k+5 is odd, we need to prove that it cannot be evenly divided by 2.Suppose, for the sake of contradiction, that 6k+5 is even. Then, by definition, we can write it as 6k+5=2n, where n is an integer.Rearranging the equation, we have:6k=2n−5Dividing both sides by 2:3k=n−52The left-hand side, 3k, is an integer since the product of two integers is always an integer. However, the right-hand side, n−52, is not an integer because subtracting a non-integer (52) from an integer (n) cannot result in an integer.This contradiction arises from assuming that 6k+5 is even. Therefore, our assumption is false, and 6k+5 must be odd.Since 6k+5 is odd for all even integers m=2k, we have proved that for all integers m, if m is even, then 3m+5 is odd.∴ The statement is proved.

RizerMix · Accepted Answer

To prove that for all integers m, if m is even, then 3m+5 is odd, we can use direct proof.Consider that m is an even number. As a result, m can be expressed as m=2k for some integer k.Now, we can substitute this expression into 3m+5:3m+5=3(2k)+5=6k+5Notice that 6k is always even because it can be written as 2(3k), where 3k is an integer.Adding an odd number (5) to an even number (6k) always results in an odd number. This can be proven by noting that an even number plus an odd number is always odd.Therefore, we have shown that if m is an even integer, then 3m+5 is odd.

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

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