Discrete Math Proof Method. 1. Give a direct proof of the fact that a^2-5a+6 is even for any integer a. 2. Suppose a and b are integers and a^2-5b is even. Prove that b^2-5a is even.

enmobladatn 2022-07-17 Answered
Discrete Math Proof Method
1. Give a direct proof of the fact that a 2 5 a + 6 is even for any integer a.
2. Suppose a and b are integers and a 2 5 b is even. Prove that b 2 5 a is even.
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Answers (2)

Monica Dennis
Answered 2022-07-18 Author has 13 answers
Step 1
a) a 2 5 a + 6 = ( a 2 ) ( a 3 )
Exactly one of a 2 or a 3 must be even. So the product must be even. (The product of an even and odd number is even).
- If you must, consider cases: Every integer is even or odd. If a is even, then show that so is a 2, and hence so is ( a 2 ) ( a 3 ). And if a is odd, then show that a 3 is even, and hence so is ( a 2 ) ( a 3 )
Step 2
For (b), consider cases: For any two integers a,b, one of the following must be true:
1. a, b both even,
2. a, b both odd,
3. a even and b odd,
4. a odd and b even.
When case (1) or case (2) holds, show that a 2 5 b is even. Then show that given either case (1) or case (2), b 2 5 a is also even.
(In case (3) and in case (4), a 2 5 b is odd, so those cases are irrelevant, since no claim is being made about b 2 5 b when a 2 5 is odd.)

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Libby Owens
Answered 2022-07-19 Author has 3 answers
Step 1
What is required, is knowledge of the rules of logic. Specifically, I will use the fact that logical equivalence ( ) is associative: P ( Q R ) is equivalent to ( P Q ) R , and therefore we can just write P Q R .
Step 2
And what rules do we know about expressions of the form … is even? Well, we know
(1a) a + b  is even a  is even b  is even (1b) a b  is even a  is even b  is even (2) a b  is even a  is even b  is even
Please take some time to make sure you understand these fully. And note how (1b) immediately follows from (1a) by the additional rule ( a )  is even a  is even
Step 2
To prove (a) we can now simply calculate
a 2 5 a + 6  is even = rules  (1a)  and  (1b) a 2  is even 5 a  is even 6  is even = rule  (2) , twice a  is even a  is even 5  is even a  is even 6  is even = logic: simplify; use facts about 5 and 6 a  is even false a  is even true = logic: simplify a bit more, using  false P P  and  P P true true
For (b), what can we do with the assumption that a 2 5 b  is even ? Let's again calculate:
(*) a 2 5 b  is even = rule  (1b) ; rule  (2) , twice a  is even a  is even 5  is even b  is even = use fact about 5; logic: simplify (**) a  is even b  is even
And since the result (**) is symmetric in a and b --in other words: it remains the same if a and b are exchanged--, this proves that the assumption (*) is as well, in other words, we've proven
a 2 5 b  is even b 2 5 a  is even

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