Does a=b imply 1/a = 1/b in mathematics?

[Aeneas]

A chapter I am reading says that with "\frac{1}{x}=\frac{1}{2x+1}\Rightarrow2x+1=x", the \Rightarrowcannot 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?

 

[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.

 

[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.

 

[EnumaElish]

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

 

[fopc]

A chapter I am reading says that with "\frac{1}{x}=\frac{1}{2x+1}\Rightarrow2x+1=x", the \Rightarrowcannot 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?

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.

 

[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?

 

[EnumaElish]

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

 

[Eighty]

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?

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).

 

[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?

 

[fopc]

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?

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.