How to Prove a Function is Not Surjective(Onto)

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How to Prove a Function is Not Surjective(Onto)
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Upside-down capital E stands for "there exists"?

pinklady
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it really did, thx for the clarification

ifxzrvz
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There is a problem while it is true that x=-2/3 is not an integer -2/3 is not in the codomain. A function is surjective when range(f)=codomain(f)

markgrace
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I have a question and I hope its not too complicated but I'd really appreciate an answer since I am confused..
If I would guess that this is surjective, and would try to solve this like you taught us to:
For all b in the codomain there exists and a in the domain, so:
3a+2 = b I -2
3a = b-2 I:3
a = (b-2)/3
So I go on and fill it in for f(a) = 3*((b-2)/3)+2. The 3 cancels out so f(a) = b-2+2 = b.
So I get f(a) = b.
But that would proof that there exists an a for every b in the codomain. Where is my error?

sophiabnrm
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Is there also a way to prove it without using brute force (trying actual values)?

siraijn
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could you prove this? given f: R -> R, f(x)= x^3 – 2x^2 + x

hasanuciha
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Sir please tell me f(x)=sinx and f(x)=cosx is these trigonometric functions are one one function

HamadAli-knqr
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A y value such that x is not equal or that x is not a memeber of the set?

allysonwalter
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Good video but shouldn't be showing actual values, please make the videos for solving them ganerally.

ashwin