1. ## 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.

2. Originally Posted by j-down
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.

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.

3. Originally Posted by j-down
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.
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?

4. Originally Posted by j-down
8. If a divides b, then a <= b.
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.

5. Originally Posted by j-down
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.
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

6. Number 8 is not true.

2 divides -2 but yet,

2<=-2

You should have used absolute values.

7. Originally Posted by ThePerfectHacker
Number 8 is not true.

2 divides -2 but yet,

2<=-2

You should have used absolute values.

Originally Posted by ecMathGeek
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.
Good of you to catch that.

8. Originally Posted by ecMathGeek
Good of you to catch that.
Hmmmm....

And 4 divides 2 a modulo 6....

-Dan

9. Originally Posted by topsquark
Hmmmm....

And 4 divides 2 a modulo 6....

-Dan
Good of you to make that completely irrelevant point.

10. Originally Posted by ecMathGeek
Good of you to make that completely irrelevant point.
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.

Just pointing that out.

-Dan