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Discrete Math
Help me with the following.
2. Prove that the law of syllogism is a tautology.
8. If a divides b, then a <= b.
14. Prove or disaprove - The product of two odd integers is odd.
8 and 14 I can do.Originally Posted by j-down
8: If a | b, where a and b are positive whole numbers then b = ak for k is some positive whole number: k = 0, 1, 2, 3.... Therefore, a <= b.
14: Since the form of an odd number is 2n + 1 for n = 0, 1, 2...
Let 2a + 1 and 2b + 1 be two odd integers
(2a + 1)(2b + 1) = 4ab + 2a + 2b + 1 = 2(2ab + a + b) + 1 which is an odd integer.
here's 14:
The statement is true:
Proof
let a and b be two odd integers. then a = 2n + 1 and b = 2m + 1 for some integers m and n.
so a*b = (2n + 1)(2m + 1) = 4mn + 2m + 2n + 1 = 2(2mn + m + n) + 1 = 2c + 1, where c = 2mn + m + n is an integer
so we see a*b is odd
QED
what is the law of syllogism?
Note, if a and b can be positive OR negative integers, then if a | b, it can be shown that |a| <= |b|, however, if b is negative then b <= a. Therefore, I assume this problem is only considering a and b to be positive whole numbers.
Okay, so if my memory serves me right, the law of syllogism goes like this:
for statements P, Q and R
if (P=>Q) and (Q=> R) then (P=>R)
the proof is done using truth tables, see the image below. notice that the last column has all true values. a statement that is true in every instance is called a taughtology (of course i don't remember if this is the exact law of syllogism, but whatever it is, it's a taughtology. so it will at least give you an idea for the solution
Not completely irrelevant. Though it is very likely that we are using the real number system, given that the class is Discrete Math it is not outside of the realm of reason to use modular Mathematics, in which case the theorem doesn't hold.Good of you to make that completely irrelevant point.
Just pointing that out.
-Dan
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