For the following proofs:• If the proposition is false, then the proposed proof is incorrect. In this case, either find theerror in the proof or provide a counterexample showing that the proposition is false.• If the proposition is true, the proposed proof may still be wrong. In this case, identify theerror and provide a correct proof.• If the proposition is true and the proposed proof is incomplete, then indicate how the proofcould be improved by rewriting it.A) Prove: If m is an even integer, then (5m + 4) is an even integer.We see that 5m + 4 = 10n + 4 = 2(5n + 2). Therefore, (5m + 4) is an even integer.B) The notation a—b indicates that a divides b, that is that the result of b/a is aninteger. Prove: For all integers a, b, and c, if a|(bc), then a|b or a|c.We assume that a, b, and c are integers and that a divides bc. So, there exists an integerk such that bc = ka. We now factor k as k = mn, where m and n are integers. We thensee thatbc = mna.This means that b = ma or c = na and hence, a|b or a|c C) Prove: For all integers m and n, if mn is an even integer, then m is even or n iseven.For either m or n to be even, there exists an integer k such that m = 2k or n = 2k. Soif we multiply m and n, the product will contain a factor of 2 and, hence, mn will beeven 2.For the following proofs:• If the proposition is false, then the proposed proof is incorrect. In this case, either find theerror in the proof or provide a counterexample showing that the proposition is false.• If the proposition is true, the proposed proof may still be wrong. In this case, identify theerror and provide a correct proof.• If the proposition is true and the proposed proof is incomplete, then indicate how the proofcould be improved by rewriting it.(a)Prove: If m is an even integer, then (5m + 4) is an even integer.Proof.We see that 5m + 4 = 10n +4=2(5n+2). Therefore, (5m + 4) is an even integer. D(c)Solution:(b)The notation ab indicates that a divides b, that is that the result ofinteger. Prove: For all integers a, b, and c, if al(bc), then alb or alc.Proof.We assume that a, b, and c are integers and that a divides bc. So, there exists an integerk such that be = ka. We now factor k as k = mn, where m and n are integers. We thensee thatbc=mna.This means that b=ma or c=na and hence, alb or alc.Is anSolution:Prove: For all integers m and n, if mn is an even integer, then m is even or 72 iseven.Proof.For either m or n to be even, there exists an integer k such that m = 2k or n = 2k. Soif we multiply m and n, the product will contain a factor of 2 and, hence, mn will beeven.

As you post multiple question according to the guidelines I only solve the first question please…

Answered: Prove: If m is an even integer, then…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

See similar textbooks

Related questions

Question

For the following proofs:
• If the proposition is false, then the proposed proof is incorrect. In this case, either find the
error in the proof or provide a counterexample showing that the proposition is false.
• If the proposition is true, the proposed proof may still be wrong. In this case, identify the
error and provide a correct proof.
• If the proposition is true and the proposed proof is incomplete, then indicate how the proof
could be improved by rewriting it.
A) Prove: If m is an even integer, then (5m + 4) is an even integer.
We see that 5m + 4 = 10n + 4 = 2(5n + 2). Therefore, (5m + 4) is an even integer.

B) The notation a—b indicates that a divides b, that is that the result of b/a is an
integer. Prove: For all integers a, b, and c, if a|(bc), then a|b or a|c.

We assume that a, b, and c are integers and that a divides bc. So, there exists an integer
k such that bc = ka. We now factor k as k = mn, where m and n are integers. We then
see that
bc = mna.
This means that b = ma or c = na and hence, a|b or a|c

C) Prove: For all integers m and n, if mn is an even integer, then m is even or n is
even.
For either m or n to be even, there exists an integer k such that m = 2k or n = 2k. So
if we multiply m and n, the product will contain a factor of 2 and, hence, mn will be
even