# New mathematical operation labelled #

1. Apr 25, 2012

### Quark Itself

This is neither homework nor independent study, it was a brain teaser( or so you could call it) I found.

A new mathematical operation has been implemented , call it #
It is used within our knowledge of numbers( it operates with all sets of numbers) and works together with our known operations ( +,-,*,/ etc.)
It also functions in such ways that, for all real values x and y

x # 0 = x (It's # identity is 0, I assumed)

x # y = y # x (it is also commutative)

(x+1) # y = (x # 1) + y + 1

From the information given, find some arbitrary value ; say 12 # 5

Attempt:
Also, I stumbled upon contradiction.
Let x = 5 and y = 0, then using the last rule
(5+1) # 0 = (5 # 1) + 0 + 1
6 # 0 = (5 # 1) +1
but 6#0 should be 6 by one of the rules
so 5#1 = 6-1 = 5

Now, let y = 5 and x = 0, then:
(0+1) # 5 = (0 # 1) + 5 +1
1 # 5 = (0 # 1) +6
Since they commute, one can look at it this way:
5#1 = 1#0 +6
RHS = 5 from test 1
LHS = 1 + 6 = 7 from the rules and addition

Any suggestion would be well appreciated !

2. Apr 25, 2012

### HallsofIvy

Staff Emeritus
More generally, for any x, (x+1)#0= (x#1)+ 1 yet we know that (x+1)#0= x+ 1. That can only be true if x#1= x for all x. But then (x+1)#1= (x#1)+ 1+ 1= x#1+2 which, using "x#1= x for all x" reduces to x+1= x+ 2!! Yes, you are correct- that definition is inconsistent (assuming that "+" means ordinary addition).

3. Apr 26, 2012

### Quark Itself

basically, this arbitrary value can be multiple values and therefore there is no logical solution?
How would one solve it , despite the discrepancy?

4. Apr 26, 2012

### HallsofIvy

Staff Emeritus
Solve what?? This is NOT a valid operation and it makes no sense to ask questions as if it were.