# Multiplicative Inverses

1. Nov 8, 2007

### mrroboto

In class today, my teacher said,

"the definition for multiplicative inverses in the R (real numbers) does not give an inverse for
the real number [1/n] element of R."

What does this mean? It seems counterintuitive, and I can't figure out why it's true.

Last edited: Nov 8, 2007
2. Nov 8, 2007

### d_leet

What is the real number [1/n]? The reals with usual multiplication and adddition form a field, so every real number does have a multiplicative inverse except for 0.

3. Nov 8, 2007

### mathman

[x] is often used to mean the greatest integer in x. Using this notation [1/n]=0 for n>1.

4. Nov 8, 2007

### d_leet

You do have to admit, though, that this is a rather odd way of referring to 0.

5. Nov 8, 2007

### mrroboto

Oops

I figured it out actually. It has to do specifically with how we defined multiplicative inverses on the real numbers

i.e.

If (an) not equivalent to (0) is a cauchy sequence then (an)^-1 is equal to (bn) where bn = { 1/an when (an) not equal to 0, 1 when (an)=0.

So, if you take the sequence (1/n), in particular, bn = n which is not a cauchy sequence. And because the real numbers are defined as the equivalence classes of cauchy sequences, this inverse does not hold.

6. Nov 9, 2007

### HallsofIvy

Staff Emeritus
Okay, so your [1/n] mean "the set of all Cauchy sequences equivalent to the sequence {1/n}". You really should have said that- especially since there are many other ways of defining the real numbers! That equivalence class also includes the sequence {0} consisting of all 0s- in fact it includes every Cauchy sequence that converges to 0. Now, my question is, what is the definition, in terms of Cauchy sequences, for "multiplicative inverses in R"?