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cxc001
#3
Apr19-10, 11:23 AM
P: 16
Yeah, my bet! a, b are real numbers

I've constructed a linear function f: (0,1)->(0,2) defined by f(x)=2x
such that f(1/2)=1, when x=1/2 (mid point of domain), y=1 (mid point of range)
This linear function is certainly bijection, therefore |(0,1)|=|(0,2)|

But how to prove |(0,1)|=|(a,b)| where a, b are real numbers and a<b???