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.
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
Sorry for taking your time
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.