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I encountered the following question in an exam:

If $\mathbb{R}$ is the set of real numbers and $f: \mathbb{R} →\mathbb{R}$ is defined by $f(\sqrt{x})=x$, the '$f$' is:

a) Injective but not surjective

b) Surjective but not injective

c) Bijective

d) Not a function

I chose option (a), but the answer, as I found out later, was (d). I cannot explain how it is so, can anyone help?

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darkv0id
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  • @Babak You mean even if we consider the given relation to be a function, it clearly cannot exist for $x<0$. But since the domain has been defined as $\mathbb{R}$, $f$ is not a function (via contradiction) – darkv0id Jan 12 '13 at 08:57
  • This is a dreadful question! It's like saying: "snargle" is a word, defined by "bloop snargle fnork". Is "snargle" (a) a noun, (b) an adjective, (c) not a word? In other words: if the definition is incomprehensible, how can we tell? darkvoid, are you sure you have it right? – TonyK Jan 12 '13 at 09:04
  • As it is defined like that and you put $x$ under the square root; you can't have a function. $\sqrt{x},x\in\mathbb R, x<0$ is a contradiction. – Mikasa Jan 12 '13 at 09:04
  • @TonyK, perhaps the question would have been more lucid if it described $f$ as a relation and not a function? – darkv0id Jan 12 '13 at 09:09
  • @BabakSorouh so if we consider $f$ to be a relation,we can say via contradiction that it is not a function, i.e. option (d) is corrrect. – darkv0id Jan 12 '13 at 09:11
  • Yes. What was given below is practical when you'd like to make your relation a function. But if you didn't want $f$ to be a function and just to know the correct option, $d$ looks fine. – Mikasa Jan 12 '13 at 09:15

1 Answers1

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First of all, we cannot take all $\mathbb R$ as whole domain, because $\sqrt{x}$ is just defined for positive real numbers $x$, so the relation you are working on is not a function. Now set $$\sqrt{x}=t$$ then $$x=t^2$$ and so $f(t)=t^2$ or $f(x)=x^2$. If $x_1=x_2$ then clearly $f(x_1)=f(x_2)$. And if $x_1^2=x_2^2$ then $x_1=x_2$ (We already restrict the domain to be $\mathbb R\cup\{0\}$) so the map is well-defined injective function. The function is not surjective. Just imagine what would be $x\in\mathbb R^-\subset \text{range}(f)$ if we wanted to solve $t^2=x$.

Mikasa
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