# Thread: math question urgent thanks :D

1. ## math question urgent thanks :D

The new infix operator is apparently the ampersign: @ . We have no idea what this operator does, but we do know the following are true for any real numbers x and y :

x@0=x

x@y=y@x

(x+1)@y=(x@y) y+1

What is the value of: 12@5

Regards Mathmans

2. Hello,

12@ 5=(11@ 5)5+1.
11@ 5=(10@ 5)5+1 etc. until you get 0@ 5.
Now, 0@ 5=5@ 0=5.

Bye.

3. Though i still wonder how you got the answer
if possible could you explain it to me?

*beg* xD

4. Hello,

This is a kind of recursive definition, the base case being x@ 0.
You can decrease the first operand using (x+1)@y=(x@y) y+1 and you will eventually get 0.

Bye.

5. Originally Posted by wisterville
Hello,

This is a kind of recursive definition, the base case being x@ 0.
You can decrease the first operand using (x+1)@y=(x@y) y+1 and you will eventually get 0.

Bye.
But if i keep decreasing this (x+1)@y=(x@y) y+1 and it ends up becoming 0 what have i proven??

and how do you end up knowing what @ means?

sorry if i'm a bit slow. Just want to understand it as good as possible

6. Hello,

I don' t know what you want to "prove". I just say that you can "compute" x@ y for any nonnegative pair of integers x, y. You don't have to "know" what @ means (Of course, knowing it or paraphrasing @ into something more familiar would help, but that is unnecessary to compute its value.)

Let me explain what I meant by "decreasing the first operand", that is, "x" in the expression "x@ y". To compute the value of (x+1)@ y, all you have to do is get the value of x@ y, multiply it by y and add 1. In other words, if you know the value of x@ y, you also know the value of (x+1)@ y. Since you know that 0@ y=y, you know 1@ y, 2@ y, 3@ y,...

Bye.