# Implication math problem

1. Sep 11, 2007

### Aeneas

A chapter I am reading says that with "$$\frac{1}{x}$$=$$\frac{1}{2x+1}$$$$\Rightarrow$$2x+1=x", the $$\Rightarrow$$cannot be replaced by $$\Leftrightarrow$$, but if a = b, does not 1/a necessarily = 1/b? Is this a misprint or are they right? If they are right, could you illustrate this with an example please?

2. Sep 11, 2007

### Eighty

a=b implies 1/a=1/b iff $a=b\neq 0$. In this case, x can't be 0, so it won't be the case, but maybe you didn't know that until after that particular step. Once you know that x isn't 0, you can put an equivalence arrow there, but in general you cannot.

3. Sep 11, 2007

### Aeneas

Many thanks Eighty. In this case it seemd to me that the second equation implied that the denominators were not 0 and thus implied the first equation, so is a particular result (that the denominator in not 0) not allowed in a chain of implication? I'm still a bit puzzled here.

4. Sep 11, 2007

### EnumaElish

You are right only with the benefit of hindsight (after having solved for x). Before you solve for it, x can be anything.

5. Sep 12, 2007

### fopc

I don't know what preliminary steps the author took in an attempt to ensure the non-replacement claim.
You haven't provided enough information.

6. Sep 12, 2007

### Aeneas

Implication

There were no preliminary steps; it was just an answer to an exercise. I am aware of the a = b = 0 possibility but 2x+1 = x does not allow this.

I think what I'm really asking is this: can one statement imply a second by virtue of implying a third?

Can 2x+1 = x imply that $$\frac{1}{2x+1}$$ = $$\frac{1}{x}$$ by virtue of implying that neither x, nor 2x + 1 = 0? and is there any way to write such a thing mathematically in one go?

7. Sep 12, 2007

### EnumaElish

I don't know how, short of solving for x or assuming x $\ne$ 0 ex ante.

8. Sep 12, 2007

### Eighty

Yes, when the third implies the second. Ie $$a\Rightarrow b\Rightarrow a$$ gives you $$a\Leftrightarrow b$$. The chain of implications really is:
$$1/x=1/(2x+1) \Rightarrow x=2x+1 \Leftrightarrow x=-1 \Rightarrow 1/x=1/(2x+1)$$
Since $$1/x=1/(2x+1) \Rightarrow x=2x+1$$ and $$x=2x+1\Rightarrow 1/x=1/(2x+1)$$, we have $$1/x=1/(2x+1)\Leftrightarrow x=2x+1$$

But you could just write merely $$1/x=1/(2x+1)\Leftrightarrow x=2x+1$$ if you want. It's true. But if you're doing a proof, you may want to use the implication arrows to indicate your train of thought (if the statement isn't trivial).

9. Sep 13, 2007

### Aeneas

Many thanks Eighty, EnumaElish and fopc. Eighty's answer, then, seems to be saying that the two-way arrow is O.K.here, and that the book is wrong to say that it is not?

Last edited: Sep 13, 2007
10. Sep 15, 2007

### fopc

Apparently no details are given by the author (even in the exercise statement). So you'll have to make some assumptions.
It's a hypothetical formula, so I think it's safe to say it's universally quantified. No problem.
But, the problem domain is another matter. I suspect the author meant for it to be R, because
if he meant say R+ or R-{0}, then clearly <-> holds and the book is wrong.
So assume R. But consider:

Look at the antecedent, and remember that it's universally quantified (for every x in R ...).
Notice there's an element sitting in R that turns it into nonsense.
Specifically, 1/0 = 1/(0+1) is nonsense. It is not true, and it is not false.
This is not the same as saying it has a truth value, but we don't know which one.
It does not have a truth value. So now you have a bound antecedent (a closed subformula)
that does not have a truth value for every x in R.

Last point: Note that the problem says nothing about x = -1.
If this is introduced, then you have a different problem.