You are not logged in.
Pages: 1
Ok...discrete math is after me again.
We had the question on a test...
Determine whether the argument below is valid
P^~P
.
. . Q
I said that because P has no relationship to Q, the argument is invalid. I got it wrong. My teacher is not a good enough teacher to explain to me why I am wrong, so I'm hoping some of you can help me. I'm not doubting that I'm wrong, just lacking understanding as to why.
Is the statement above the same as (P^~P)->Q?
Also, on the latest test that hasn't been returned yet, there was the argument
p->q
q->r
.
. . r
I said it was invalid because the truth of p or r has not been established.
I think these out in words
p=I'm in a discrete math class
q=I hate my teacher
r=my head is about to explode
For the first one, I'm in a discrete math class and I'm not in a discrete math class, therefore I hate my teacher? Is that valid?
2nd, if I'm in a discrete math class, then I hate my teacher. If I hate my teacher, then my head is about to explode. Therefore my head is about to explode? Don't we have to establish the validity of either p or q to determine r?
Thanks guys!
Offline
The first one is valid, I would say. Given P^~P, you can conclude that fish wear monocles or Belgium is purple or whatever else you can think of.
You show an implication (P --> Q) is false by finding an example where the predicate P is true, but the consequence Q is false.
eg. x is even --> x is a multiple of 4
This is wrong because 6 is even but not a multiple of 4.
In your case though, there is no situation where the predicate (P^~P) is true, so the implication has to be true even if the consequence is garbage.
I'd agree with you for the second one though. An example of that type of argument would be:
If I'm at Stonehenge, then I'm in England.
If I'm in England, then I'm in Europe.
Therefore, I'm in Europe.
But since I never said I was at Stonehenge, I can't make that last conclusion.
Why did the vector cross the road?
It wanted to be normal.
Offline
Pages: 1