Rewriting the expression using m=2p we have:
Answer:
[tex]m^{2} +5 -1[/tex] is an odd integer but the converse is not true.
Step-by-step explanation:
Even numbers are written as 2n where n is any integer, while odd numbers are written as 2n-1 where n is any integer.
a) To prove that [tex]m^{2} +5m-1[/tex] is an odd integer, we have to prove that it can be written as 2n-1.
By hypothesis, m is an even integer so we will write it as 2p.
Rewriting the original expression using [tex]m=2p[/tex] we have:
[tex]m^{2} +5m-1 = (2p)^{2} +5(2p)-1[/tex]
Solving the expression and factorizing it we get
[tex]4p^{2} +10p -1 = 2(2p^{2}+5p) -1\\ \\[/tex]
And this last expression is an expression of the form 2n-1, and therefore [tex]m^{2} +5m-1[/tex] is an odd integer.
b) The converse would be: if [tex]m^{2} +5m-1[/tex] is an odd integer, then m is an even integer.
We'll give a counterexample, let's make [tex]m=3[/tex], then
[tex]m^{2} +5m-1[/tex]
[tex]3^{2}+5(3)-1 = 23[/tex] is an odd integer but m is odd.
Therefore, the converse is not true.
