# Read a book of analysis

1. Feb 2, 2005

### C0nfused

Hi everybody,
I have recently read a book of analysis, which starts with some stuff about Set Theory before moving on to functions. By reading the book, I realised (it was written in it) that the simple operations that we do (addition,multiplication) are actually functions.

Let's speak for the real numbers only:according to the book addition and multiplication are two functions from RxR to R, and actually functions are sets . So, addition of a pair (x,y) with x,y real numbers is a number that is the value of the function "addition" at the "point" (x,y) and is written x+y . So x+y=S((x,y)) if we name the function of addition S. The same applies to multiplication , so the value of the function M(=multiplication) is written xy and xy=M((x,y)).

So here comes the question: When we write x+y (example 5+3 etc) or xy , x+y or xy is actually one number and not an expression? To make myself more clear, x+y is not an expression that when evaluated gives the sum of x and y but a number that we, in order to find it have to add x and y? (I don't know if anyone else understands what i am asking!) So actually x+y ( or x+y+2z etc) is always one number, which is the "image" of a function(addition or multiplication) and we can write it in many ways because addition ( or multiplication) is not a "1-1" function? And we have just figured out some algorithms in order to find that image through calculations using the (x,y) pair , from which the image is "produced" , or generally transformate these pairs with others that have the same image , aslo making sure that the axioms referring to addition and multiplication are true?

To sum up the expression x+y (or a more complicated one) represents/is equal to one unique real number , so when we write x+y ,even if we don't calculate the sum, this still is equal to this unique number z with
z=S((x,y), and is not just an expression whose value is equal to z?
Are similar functions defined for operations in a vector space generally?

That's all(for now). Sorry for the weird language-English is not my mother-tongue. I hope that you have understood what i am trying to ask!(of course it may be just nonsense)
Thanks

Last edited: Feb 3, 2005
2. Feb 2, 2005

### JasonRox

I didn't read everything you wrote, but be careful when you said a function is a set.

f is a set (or map) of pairs.

f(x) is not a set of pairs.

Try breaking up your post. I have a hard time reading large paragraphs. I understand that you don't know English too well, but if you just press Enter a couple times, it would make it so MUCH better.

Thanks.

3. Feb 2, 2005

### jcsd

Yes though some people like to call the set of ordered pairs (a,f(a)) for an arbitary function f:A-->B the graph of the function (though it's cera that knowing the set of ordered pairs means you know everything there is to know about the function in question, so that's why I guess many prefer to identify the set of orderd pairs with the function).

But yes the expression 3 + 1 (it's still an expression) is equal to 4 and is just another way of writing 4.

In genral any binary operation is a function f:AxA -->A

Last edited: Feb 2, 2005
4. Feb 2, 2005

### mathwonk

the reason it is interesting, or at least efficient, to think of operations like multiplication as functions, is that then in calculus for example you can take the derivative of a product by the chain rule, composing with the derivative of the product function!

i.e. multiplication is a function from RxR to R, and at the point (a,b) the gradient of multiplication is (b,a). i.e. if we change x and y a little bit, we get the product as

(a+dx)(b+dy) = ab + bdx + ady + dxdy, so the linear part is (b,a).(dx,dy).

then if f,g are any two functions from R to R, together they give a function from R to RxR, and fg is the composition with multiplication. Then the derivative of fg at t, is

the dot product of (f'(t),g'(t)) with the gradient (g(t),f(t)),

i.e. we get f'(t)g(t) + g'(t)f(t).

now this is a lot easier and more natural than proving the product rule by some trick of adding and subtracting the same terms.

this sort of approach is in the great book "foundations of modern analysis" by jean dieudonne, somewhat hard to find nowadays, and pricey.

but almost the only time people actually think of functions as sets of ordered pairs, i.e. as graphs , is when they are doing geometry on the graphs, like taking tangent planes etc... or doing some clever proof where it is useful to factor a function as an injection of the domain onto the graph, followed by a projection from the graph to the range.

Last edited: Feb 2, 2005