| What was tried to be proven in this proof? Proof. Suppose m is an even integer and is an odd integer. By definition 2g for some integer q and by definition of odd, n= 2r+1 for some in- 4qr+2q = 2(2qr +q). Now 2qr+q is an integer 2-(some integer). %3D of even, m = teger n. Thus mn= because products and sums of integers are integers. Hence mn= and so, by definition of even, mn is even. (2g)(2r+1)

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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3.
What was tried to be proven in this proof?
Proof. Suppose m is an even integer and n is an odd integer. By definition
of even, m =
teger n. Thus mn = (2g)(2r+1)
because products and sums of integers are integers. Hence mn =
and so, by definition of even, mn is even.
2g for some integer q and by definition of odd, n=
2r+1 for some in-
4gr+2q= 2(2qr +q). Now 2qr+q is an integer
2-(some integer).
Transcribed Image Text:3. What was tried to be proven in this proof? Proof. Suppose m is an even integer and n is an odd integer. By definition of even, m = teger n. Thus mn = (2g)(2r+1) because products and sums of integers are integers. Hence mn = and so, by definition of even, mn is even. 2g for some integer q and by definition of odd, n= 2r+1 for some in- 4gr+2q= 2(2qr +q). Now 2qr+q is an integer 2-(some integer).
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