# Homework Help: Propositional logic help please

1. Oct 15, 2011

### mtayab1994

1. The problem statement, all variables and given/known data

for every a in ℝ+:

for every ε>0 : a<ε

2. Relevant equations

prove that a=0

3. The attempt at a solution

is it possible to use contradiction to solve that problem, if not how can I. Urgently need help.

2. Oct 15, 2011

### HallsofIvy

Re: Propositional logic urgent help please

First of all, what you have doesn't make sense. You cannot write "for every x in R+" and then say "prove that a= 0".

You must mean "If for some non-negative x, for every $\epsilon> 0$, $x< \epsilon$, then a= 0." (I also dropped "in R+" in favor of "non-negative" because that is the set of positive real numbers and does NOT include 0.)

3. Oct 15, 2011

### MindWalk

Re: Propositional logic urgent help please

Yes, you can use a proof by contradiction. Suppose that a > 0 and then consider what happens if epsilon = a/2.

4. Oct 15, 2011

### mtayab1994

Re: Propositional logic urgent help please

how so and at the end i am supposed to come out with something opposite of what i need to find right

5. Oct 15, 2011

### CLSabey

Re: Propositional logic urgent help please

I think your best approach is to use the mathematical philosophy constructive for the series of natural numbers. You can reference B.Russell's book "Introduction to Mathematical Philosophy" mainly the first four chapters (series of natural numbers, definition of number, finitude and mathematical induction, and kinds of relations). I would reference them specifically but I am in a bar watching college football. Here are my thoughts based on the above for your problem:

Say that a+ is the successor of a
Say that a can not be the successor of any a+
Say that E is the set whose members have the inductive property of a+1 (successor), and that a is the predecessor of a + 1
Prove a equals zero by contradicting the relation:
E is greater than 0 : a is less than E
If the relations 'greater than' and 'less than' are equal to each other then a equals 0. The use of 'if' is for your propositional logic, so that if a equals 0 then the relation greater than and less than is equal, a contradiction. By this contradiction, if the above relation exists, then a must equal 0. Hope this steers you in the right direction, and hopefully my thinking here is not misleading, if it is, I am sure I can clarify.

6. Oct 16, 2011

### MindWalk

Re: Propositional logic urgent help please

I don't see the previous post as helpful, since the problem statement is in the real numbers, not the natural numbers, so that the successorship relation doesn't seem to be pertinent.

But you have that a < epsilon for any epsilon in R+. Suppose a does not equal 0. Then, since a is positive, a/2 is positive. Since a must be less than epsilon for *any* epsilon in R+, you can choose epsilon to be any positive number you want to and a must still be less than it. So, pick epsilon = a/2. But a isn't less than a/2. So, if a does not equal 0, there is some epsilon > 0 (namely, epsilon = a/2) such that a isn't less than epsilon. But a is less than epsilon for *any* epsilon in R+. Contradiction. Therefore, a = 0.

I'm not sure why this was listed under propositional logic, though.

7. Oct 16, 2011

### CLSabey

Re: Propositional logic urgent help please

The previous post makes some good points; however the conclusion is a contradiction of its premises in assumption of a can't be 0 only to conclude then a = 0. You say more than once for epsilon = a/2, and this works until you conclude a=0 at which point epsilon = a/2 becomes epsilon=0. If this is the case then epsilon cannot be greater than 0 (rather it is equal to) so e > 0 no longer holds. a<e becomes 0 = 0 because a = 0 and epsilon = a/2 (which would be 0 also). So e > 0 : a<0 becomes 0=0:0=0.

I'll have another post later this afternoon in the real numbers with explanation of my starting point in the natural numbers. I'll show the relevance to propositional logic

8. Oct 16, 2011

### MindWalk

Re: Propositional logic urgent help please

No, no, no. The way the proof works is this: First, assume that a > 0. Demonstrate that under that assumption, one can derive a contradiction (the contradiction, in this case, being that for any epsilon > 0, a < epsilon, but that there exists an epsilon [namely, a/2] such that a > epsilon [and therefore such that a is not less than epsilon]). Then, having demonstrated that the assumption that a > 0 leads to a contradiction, one concludes that the assumption must be false.

At this point, one is done with the assumption that a > 0--he is no longer assuming it. The assumption is discharged.

Once one has concluded that the assumption is false (and has discharged it), one concludes that the assumption's negation must be true. The negation of the assumption that a > 0 is that a is not > 0. Since a is selected from the nonnegative reals, this leaves only a = 0. It's true that a/2 is then 0, too, but a/2 is no longer being identified with epsilon. That part is finished--done with--over. That was done only in the conditional proof, in which it was assumed that a was > 0. Once the conditional proof's assumption was discharged, epsilon's being identified with a/2 was over, too.