# New Binary operation

• August 15th 2009, 11:03 PM
§uperman
New Binary operation
Hey
Im new here, and i'm not sure where to put this Q,
I was studying for the AMO, and I came across a problem in the 2004 AIMO which has me completely stuck (Headbang)...

A new binary operation ? defined on all positive x and y, satisfies the following conditions:

i x?x = x+2
ii x?y = y?x
iii x?(x+y) = x+y
----x?y------y

Given the above, find 11?8

please explain how to do this, i have been to other forums, and unless I ask, they just give a number.

Thanks
• August 16th 2009, 12:10 AM
running-gag
Hi

From ii
11 ? 8 = 8 ? 11

From iii
8 ? 11 = 8 ? (8 + 3) = (8 ? 3) 11 / 3

Now we need to find 8 ? 3
From ii
8 ? 3 = 3 ? 8

From iii
3 ? 8 = 3 ? (3 + 5) = (3 ? 5) 8 / 5

Now we need to find 8 ? 3
From iii
3 ? 5 = 3 ? (3 + 2) = (3 ? 2) 5 / 2

Now we need to find 3 ? 2
From ii
3 ? 2 = 2 ? 3

2 ? 3 = 2 ? (2 + 1) = (2 ? 1) 3 / 1

Now we need to find 2 ? 1
From ii
2 ? 1 = 1 ? 2

1 ? 2 = 1 ? (1 + 1) = (1 ? 1) 2 / 1

From i
1 ? 1 = 1 + 2 = 3

Therefore you can go up the chain to get 11 ? 8
• August 16th 2009, 12:24 AM
§uperman
Thanks running-gag
just one Q, where did you get (8?3)11/3 ???
then, (3?5)8/5 etc
all the way down?
thanks
• August 16th 2009, 12:28 AM
running-gag
Rule iii is
$\frac{x?(x+y)}{x?y} = \frac{x+y}{y}$

If you substitute x = 8 and y = 3 you get
$\frac{8?(8+3)}{8?3} = \frac{8+3}{3} = \frac{11}{3}$

Therefore $8?(8+3) = (8?3)\:\frac{11}{3}$
• August 16th 2009, 12:31 AM
§uperman
yep, thanks, i see,
just i was keeping x=8, y=11 from the 8?11=8?(8+3), so i got (8?11)11/3
thanks