If $f: [0, 1] \to \mathbb R$ is defined by $() = ^2$, check $f$ is bijective or not?how? I don't think it is onto because 6 doesn't have any pre image but 6 belongs to R
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Do you mean $f(x) = x^2$ ? – Matti P. Dec 23 '20 at 11:00
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Yess it is f(x)=x² – SHUVAM JASWAL Dec 23 '20 at 11:01
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Surely you have some ideas here. Please edit your post to include them. Do you believe the map is bijective? In particular, do you believe it is surjective? – lulu Dec 23 '20 at 11:01
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I am not getting anything I can solve this for N>N R>R but cant find this – SHUVAM JASWAL Dec 23 '20 at 11:03
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Do you know what surjective means? – lulu Dec 23 '20 at 11:04
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Onto function Which has preimage for all of it's range – SHUVAM JASWAL Dec 23 '20 at 11:05
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Post edit, sure! $6$ is not in the image of $f(x)$, and that's all you need. Even more obviously, $-1$ is not in the image of $f(x)$. – lulu Dec 23 '20 at 11:05
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So it is not onto am i right? – SHUVAM JASWAL Dec 23 '20 at 11:06
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Yes, you are right. – lulu Dec 23 '20 at 11:07
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Also, is it one-one? – SHUVAM JASWAL Dec 23 '20 at 11:08
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Is it one-one? How? – SHUVAM JASWAL Dec 23 '20 at 11:11
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No it's not surjective, to be bijective you need it to be both injective and surjective
surjective $\forall$ y $\exists$ x such that $y=f(x)=x^2$
let $y=-3$ (or y=3) then $x=\sqrt(-3)$ but $\sqrt(-3) \notin [0,1]$
injective $f(x_1)=f(x_2)$ if and only if $x_1=x_2$
since the domain is in [0,1], yes it is
yugikaiba
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