# The definition of multiplication

1. Aug 14, 2012

### GarageDweller

Warning: Semantics battle may ensue, tread lightly
So I was wondering the other day, the repeated-addition definition of multiplication only works for integers, for example you cannot use this to calculate the square of e or pi.
So is there a rigorous definition for multiplication that is viable for all numbers?

2. Aug 14, 2012

### SteveL27

If you believe in multiplication of integers, it extends to the rationals and reals when you construct the rationals and reals out of integers.

The rationals are constructed as the field of quotients of the integers.

http://en.wikipedia.org/wiki/Field_of_fractions

(The article's a little general. In the Construction section, just think "integers" when they say "commutative pseudo-ring" and the construction's the same).

Then the reals are constructed as equivalence classes of Cauchy sequences of rationals, or alternately, Dedekind cuts of rationals.

http://en.wikipedia.org/wiki/Dedekind_cut

In each construction, from the integers to the rationals and from the rationals to the reals, addition and multiplication are defined based on what we already have. In other words multiplication of rationals is defined using multiplication of integers, and multiplication of reals is defined using multiplication of rationals.

I'm leaving out all the details, but this is the outline of the definition of multiplication of real numbers based on multiplication of integers.

We can go down the other way too. Where does multiplication of integers come from? It's ultimately defined in terms of the successor operation in the Peano axioms of the natural numbers.

http://en.wikipedia.org/wiki/Peano_axioms

3. Aug 15, 2012

### coolul007

The definition of multiplication is area. (my never ending mantra)

4. Aug 15, 2012

### Staff: Mentor

Your definition is not very useful (or correct). If my walking speed is 3.5 miles/hr and I walk for 2 hours, I go a distance of 7 miles. That in no way should be considered an "area."

5. Aug 15, 2012

### coolul007

In the land of units, my definition is correct. As soon as you apply measurement of specific units you have changed the system from mathematics to miles/hour. there are no fluid ounces in that system either.

6. Aug 15, 2012

### micromass

Please explain $i^2=-1$ or $(-1)^2=1$ as area...

7. Aug 15, 2012

### Staff: Mentor

"In the land of units" - did you mean in problems where there are no units, such as mi/hr, fl. oz., etc.? Or did you mean dimensionless units?

Your mantra did not specify that you were talking only about the product of two numbers. Certainly, any product of two real numbers could be interpreted as an area, but that's not the only interpretation.

8. Aug 15, 2012

### coolul007

Of course it is not the only definition, however, as a fundamental definition, it is the one that is the most intuitive. as for i, you may spend the next few centuries explaining where that fits on the number line. I use are to explain multiplication when defining multiplication as "repetitive addition" fails. It fails if both numbers are not integers.

9. Aug 15, 2012

### pwsnafu

That is an interpretation not a definition.

Or you could spend 5 minutes explaining the Argand plane.

And your explanation fails in other spaces.

10. Aug 15, 2012

### coolul007

OK, here's a definition, multiplication is an operation.

11. Aug 15, 2012

### coolul007

btw, operations on i are defined, we do not know how to multiply i*i, we define its result. As far as multiplication goes my definition comes from the historical need to define ownership of land.

12. Aug 16, 2012

### HallsofIvy

Your first statement is not true- we define i first, then show that i*i= -1. And, as has been pointed out before, "multiplication is area" is not even a "definition".

13. Aug 16, 2012

### GarageDweller

Calculating areas is one of the applications of multiplication, similarly vector fields historically arose from the concept of fluid velocity fields, but the concept can be applied to electromagnetism

14. Aug 16, 2012

### Studiot

Why just numbers?

You talk of vectors in later posts. Do you wish to exclude vector multiplication?

coolul, if you are willing to learn you are not so far from the mark with this, although it needs tidying up with some conditions.

Multiplication is a binary operation on a set.

Depending upon circumstance we may wish to restrict the output of this operation to other members of the set or we may wish to allow it to map to another set.

We can add further useful restrictions again depending upon circumstance eg

the requirement that ab = ba (or not)

etc.

15. Aug 16, 2012

### coolul007

The reason defining i first does not solve the i*i result. Because if we take i = $\sqrt{-1}$. it fails for i*i because we get 2 results depending on order of operations. Because of rules of exponents, $\sqrt{-1}$ * $\sqrt{-1}$ can result in $\sqrt{(-1)(-1)}$ or $\sqrt{(-1)^2}$, now if we square first the answer is +1, if we don't it's -1, that is why i*i must be defined, because complex operations must be restricted.

16. Aug 16, 2012

### micromass

The rules of exponents do not hold for negative numbers.

17. Aug 16, 2012

### Bacle2

You can define i as the complex number with r=1 and arg =Pi/2 .

Then, multiplication z1z2 is given by the product of the

moduli and the sum of the arg. Yes, in some cases you have to mod out the sum

of arguments by 2π , but that is not a problem here. I don't see your point.

18. Aug 16, 2012

### jgens

This is probably circular. The theorems showing that you can represent complex numbers this way usually involve i already. Even if you could eliminate this issue, you are going to be doing acrobatics to define the complex numbers and prove these theorems without mentioning i.

19. Aug 16, 2012

### HallsofIvy

But we don't. The standard definition of the complex numbers is as pairs of real numbers, (x, y) with addition defined by (a, b)+ (c, d)= (a+ c, b+ d) and multiplication by (a, b)*(c, d)= (ac- bd, ad+ bc). We can map the real numbers to that by the identification of x with (x, 0). We define "i" to be (0, 1) and then show that i*i=(0, 1)*(0, 1)=(0*0- 1*1, 0*1+ 1*0)= (-1, 0), the complex number identified with the real number -1.

20. Aug 16, 2012

### coolul007

You said the magic word "define", that has been my assertion all along, calculations involving i require the result to be defined. We have rules that govern the results, but no operation takes place, like no operation takes place with 3x, until we know the value of x. 3x is defined as 3x. Last comment by me as all of you are much more knowledgeable about these matters. Do me one last favor, calculate the area of a triangle without multiplication.

21. Aug 16, 2012

### micromass

What will that show?? If you want to say that multiplication is somehow useful in calculating areas, then you are right.
The thing is that you can't define multiplication as just "area". The problem is that you never really defined area. So the definition is quite shady.

22. Aug 16, 2012

### pwsnafu

We define the area of a rectangle to be it's base times height.
We do not define multiplication to be the area the rectangle with said base and height.
Do you understand the difference?

23. Aug 16, 2012

A simple definition of multiplication is 'do to the left-hand-side what the right-hand-side does to unity'.

On the reals unity is 1, so for 2 * 3, what the right-hand-side does to unity is to physically stretch three-fold.

On the complex numbers unity is 1+0i. so for z * (2+2i) the right hand side stretches unity by sqrt(8) and rotates it by 45 degrees.

On matrices 'unity' means the identity matrix, and the right-hand-side represents a scale, rotation, stretch, sheer upon identity.

24. Aug 16, 2012

### Number Nine

That really only works if multiplication is already defined, since 3 only "stretches" unity by being multiplied with it.

25. Aug 17, 2012