# Does n*a ALWAYS mean to a + a + + a (n times) where + is the group operation?

1. Oct 15, 2012

### dumbQuestion

What about in a ring where we have two binary operations defined. I get super confused when I see someone just switch from something like n*a to a + a + ... + a (n times, where * is the binary operation on the "semigroup" part of the ring, and + is the operation on the "group" part of the ring) because I freak out and think, how do I know these two things produce the same result? I mean I get it for familiar number systems like R and C and Z, but does it always work out? I just feel uneasy with generalizing it to everything! Can I? Is there a theorem for that?

2. Oct 15, 2012

### HallsofIvy

Staff Emeritus
If you have a ring, then both "addition" and "multiplication" are defined only for member of the ring so that "n*a" is NOT either of those. "n*x" is defined as "a added to itself n times". They "produce the same result" because "n*x" is defined as shorthand for that sum.

3. Oct 15, 2012

### dumbQuestion

But what I mean is, say you have the group operation and its called + and the other operation (the binary operation on the semigroup part of the ring) say is called &. I get that the shorthand definition n*x is defined as x + x + x +... + x (n times) but is there any relationship between & and *? Or is there any way to tell when such a relationship exists or doesn't exist?

4. Oct 16, 2012

### Stephen Tashi

Are you asking if there can be examples where some special relationship exists ? (I would assume so.) Or are you asking if there is always some relationship? I think the only "always" relationships are things you can prove employing the distributive law of & over +.

Propose examples of some relationships.

5. Oct 18, 2012

### Erland

If the ring contains Z (the integers) as a subring, then the two interpretations of the expression n*a give the same result. Otherwise, the repeated addition interpretation is the only possible one.

6. Oct 21, 2012

### Bacle2

If you're dealing with the ring of polynomials in one variable, then a*b is the

standard polynomial product of a,b .

Erland: do you mean to say that defining multiplication as repeated addition in

the case where the ring contains the integers is the only way of satisfying the

axioms of ring multiplication (together with distributivity props.) ?

Never mind, Erland, sorry.

Last edited: Oct 21, 2012
7. Oct 24, 2012

### Vargo

For every ring R there is a unique ring homomorphism from Z into R. So in that context, n&a only has meaning if you understand n to be the image under that homomorphism. n*a on the other hand, means addition repeated n times. They are equal because:

$n*a = a+ ... + a = 1&a + .... +1&a = (1+....+1)&a = \bar{n}&a$

For example, if the ring is Z mod 8, then 23*a=7&a.

Last edited: Oct 24, 2012