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Dr.Pepper관련링크
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helpme님이 2010-04-28 14:25:33에 쓰신글
>when f (x) = x2 + x
>and then a, b is positive integer.
>Please establish proof that 4f (a) = f (b) could not be true.
>
f(a) = a2+a
f(b) = b2+b
LHS=4f(a)=4(a2+a) - Left Hand side of given equation
RHS=f(b)=(b2+b) - Right Hand side
a, b are any integers, f(x)=x2+x should be true even when a=b:
f(a) = f(b)
LHS=4f(a)
RHS=f(b)=f(a)
If LHS and RHS are same,
4f(a)=f(a)
If f(a) is cancelled out,
4=1
which is impossible and could not be true.
Too late for you?
>when f (x) = x2 + x
>and then a, b is positive integer.
>Please establish proof that 4f (a) = f (b) could not be true.
>
f(a) = a2+a
f(b) = b2+b
LHS=4f(a)=4(a2+a) - Left Hand side of given equation
RHS=f(b)=(b2+b) - Right Hand side
a, b are any integers, f(x)=x2+x should be true even when a=b:
f(a) = f(b)
LHS=4f(a)
RHS=f(b)=f(a)
If LHS and RHS are same,
4f(a)=f(a)
If f(a) is cancelled out,
4=1
which is impossible and could not be true.
Too late for you?
작성일2010-05-03 18:51
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