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#1 Help Me ! » Arguments...validity...humbug! » 2008-10-14 10:19:19

nsherman2006
Replies: 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!

#2 Re: Help Me ! » Principle of Mathematical Induction » 2008-09-24 14:48:08

Thanks!

I actually realized during my study group today as I was writing it out that I missed that part...details.

So how would I go about proving it for an odd case?

I'll chip away at it when I have a chance but any input is appreciated!

Neal

#3 Help Me ! » Principle of Mathematical Induction » 2008-09-23 13:54:16

nsherman2006
Replies: 3

Hello!

Quick background on the newbie...I'm a junior at eastern Connecticut State University studying to be a secondary mathematics educator...so I'm working on my undergrad degree in math. Always thought I was pretty good at math, has always been my strongest subject and I have an eye, memory, and understanding of numbers. However, I'm in a Discrete Structures course with a terrible instructor and I'm stuck. We're going over proofs using the Principle of Mathematical Induction (PMI) and I understand the concept. I had a problem getting through a proof (the 2nd one below), but in typing it I figured it out. However, it seems like there has to be an easier way to do it...any chance someone can guide me where I need to go?

The first example we had to prove was 2+4+6...+2n=n(n+1)

Simple enough...basis step yields 2=2(2)=2(2+1) => 6=6 => true

Inductive step assumes 2+4+6...+2k=k(k+1) is true and then finally we arrive at 2+4+6...+2k+(2k+1)=(k+1)(k+2)

Substituting k(k+1) for {2+4+6...+k} nets k(k+1)+(2k+1)=(k+1)(k+2)...both sides simplify to k²+3k+2, so the proof is complete

Now...next problem that we run into is more involved...

We have a 2 part formula:

c sub 1 = 0
c sub n = c sub (floor [n/2]) + n² for all x>1

so c sub 2 = c sub (floor 2/2) + 2² = 0+4 = 4

and c sub 3 = c sub (floor 3/2) + 3² = 0+9 = 9

and c sub 4 = c sub (floor 4/2) + 4² = 4+16 = 20

and c sub 5 = c sub (floor 5/2) + 5² = 4+25 = 29

and c sub 6 = c sub (floor 6/2) + 6² = 9+36 = 45

I am attempting to prove c sub n < 4n² for all n≥1

So...basis step:

c sub 1 = 0
4(1)² = 4
0<4, hence, true

So we assume:

c sub k < 4k²
so
4k²>c sub (floor k/2) + k², which simplifies to
3k²>c sub (floor k/2)

Now for even k values, (floor k/2) = (floor [k+1]/2)
for odd k values, (floor [k+1]/2) = (floor k/2) + 1

So let's do k+1 for even k
4(k+1)²> c sub (floor [k+1]/2) + (k+1)², so
3(k+1)²> c sub (floor [k+1]/2)

As stated above (floor k/2) = (floor [k+1]/2), and 3k²>c sub (floor k/2)

3(k+1)² > 3k² true for all k≥1

Let's do an odd k
4(k+1)²> c sub (floor [k+1]/2) + (k+1)², so
3(k+1)²> c sub (floor [k+1]/2)

(floor [k+1]/2) = (floor k/2) + 1, so adding 1 to the left side of 3k²>c sub (floor k/2) yields 3k² + 1 > c sub (floor k/2) + 1

3(k+1)² > 3k² + 1 true for all k≥1

So it can be proven in both cases...but there has to be a more straightforward way of doing this...any ideas?

Thanks!

Neal

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