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
a@b = b@a?
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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)
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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I get statement 2, but here's how I looked at statement 1: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!
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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+: yes, m+m could = 1/2 (m=1/4)m @ m = 1/2 for all nonzero m
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