- #1
roam
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Homework Statement
For any integer n, let A(n) be the statement:
"If n=4q+1 or n=4q+3 for some [tex]q \in Z[/tex], then n is odd."
(a) Write down the contrapositive of A(n).
(b) Is the contrapositive of A(n) true for all [tex]n \in N[/tex]? Give brief reasons.
(c) Use proof by contradiction to show that the converse of A(n) is true for all [tex]n \in Z[/tex].
Homework Equations
The Attempt at a Solution
(a)
The contrapositive of [tex]A \Rightarrow B[/tex] is [tex]\neg B \Rightarrow \neg A[/tex], so in this case:
A:= If n=4q+1 or n=4q+3 for some [tex]q \in Z[/tex]
B:= n is odd
The contrapositive would be:
If n is even then [tex]n \neq 4q+1[/tex] and [tex]n \neq 4q+3[/tex] for all [tex]q \in Z[/tex].
Is this right?
(b)
I have no idea how to answer this one.
(c)
The converse of the A(n) is: "If n is odd, then n=4q+1 or n=4q+3 for some [tex]q \in Z[/tex]"
A:= n is odd
B:= n=4q+1 or n=4q+3
Now this is what I have to show for a proof by contradiction:
[tex](A \Rightarrow B) \Leftrightarrow \neg (A \wedge \neg B)[/tex]
[tex]A \wedge \neg B[/tex] = n is odd and [tex]n \neq 4q+1[/tex] or [tex]n \neq 4q+3[/tex]
Of course, n is odd [tex]\Leftrightarrow (\exists m \in Z)(2m+1)[/tex]
So, I'm stuck here & I'm not sure exactly how to prove that [tex]A \wedge \neg B[/tex] is false. Any help is appreciated.