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a@b = b@a?

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by anayeri » Thu Dec 04, 2008 7:25 pm
A certain mathematical operation may be represented by the symboal @. For all nonzero a and b, does a@b = b@a?

1) m @ m = 1/2 for all nonzero m
2) a @ b = ab/(a^2 + b^2).

OA:B
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Source: — Data Sufficiency |
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by cramya » Thu Dec 04, 2008 7:44 pm
Started with Stmt II

a@b = ab/a^2+b^2

Replace a with b

b@a = ba/b^2+a^2 which is the same as ab/a^2+b^2

So a@b = b@a

SUFF

Stmt I

m@m = 1/2

So a@a=1/2
b@b = 1/2

We know a@a = b@b but we have no way to find out if

a@b = b@a (it is if a=b (both 1/2) and not if a is not equal to b)

INSUFF

Hope I dint miss something here!

B)
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by anayeri » Thu Dec 04, 2008 7:55 pm
Stmt I

m@m = 1/2

So a@a=1/2
b@b = 1/2

We know a@a = b@b but we have no way to find out if

a@b = b@a (it is if a=b (both 1/2) and not if a is not equal to b)

INSUFF

Hope I dint miss something here!
I get statement 2, but here's how I looked at statement 1:

m@m=1/2

can we prove that @ is + or x (and not - or /)?

+: yes, m+m could = 1/2 (m=1/4)
x: yes, m x m could =1/2 (m=+/- sqrt of 1/2)
-: no, m - m can't = 1/2, because it would be 0
/: no, m/m can't = 1/2, because it would be 1

So, @ is + or x, but not - or /

as a result a + b = b + a (check), and a x b = b x a, therefore statement 1: true.

Can someone comment on why I'm wrong?

The only thing I can think of is that perhaps @ could also be sqrt or ^.
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by cramya » Thu Dec 04, 2008 9:36 pm
m @ m = 1/2 for all nonzero m
+: yes, m+m could = 1/2 (m=1/4)
x: yes, m x m could =1/2 (m=+/- sqrt of 1/2)

Agree!

How can we relate this to a and b though?

IMO u r trying to prove For all nonzero a and b, does a@b = b@a which is only possible wiht m@m = 1/2 when both a and b are equal

We dont know if a and b are equal hence INSUFF

Hope this helps!

Regards,
Cramya
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