ล่ 2. 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. The notation a b indicates that a divides b, that is that the result of is an integer. Prove: For all integers a, b, and c, if al(be), then alb or alc. Proof. We assume that a, b, and c are integers and that a divides bc. So, there exists an integer k such that be = ka. We now factor k as k=mn, where m and n are integers. We then see that be=mna. This means that b=ma or e=na and hence, alb or alc. Solution: even. Prove: For all integers m and n, if mn is an even integer, then m is even or 1 is Proof 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 0 even.
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) 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
B) 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
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