Is this function surjective
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Is this function surjective

[From: ] [author: ] [Date: 12-07-09] [Hit: ]
For instance, there is no (x, y) in QxQ such that T(x, y) = πx + y = π^2.(This follows from π being a transcendental number; that is π does not satisfy any polynomial equation with rational coefficients. More specifically,......
Hello guys :) I have this problem:

Let (x,y) in QxQ,

T: QxQ------------> IR

T(x, y)= Pi x+ y

Is this function surjective? I think it's not.... but i don' t know how to prove it :(

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T is not surjective.

For instance, there is no (x, y) in QxQ such that T(x, y) = πx + y = π^2.
(This follows from π being a transcendental number; that is π does not satisfy any polynomial equation with rational coefficients. More specifically, if x, y in Q did exist, then π is a zero of t^2 - xt - y, which contradicts the fact that π is transcendental.)

I hope this helps!

P.S.: I like the cardinality proof as well.

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QxQ is countable, so T(QxQ) is also countable. IR is not countable, so T is not surjective.
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