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I try to understand, why domain of $x^2$ is the set of all real numbers.

My doubts: The domain of square root is not defined for negative numbers. Reason to that (If I am not wrong) is that function is supposed to have output with only one input leading to it. Therefore 9 cannot be square-rooted to -3 and 3, only to the positive number (in this case it's 3). Is not the situation with $x^2$ the same? Should not a domain for it be limited to non-negative numbers? Otherwise, we are having an ambiguity of 3 and -3 leading to 9...

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  • $\begingroup$ What is the definition of the domain of a function? Once you have that, the answer is immediate. $\endgroup$ – quasi Sep 30 '17 at 17:10
  • $\begingroup$ the domain of $x^2$ is the set of all real numbers, this is right $\endgroup$ – Dr. Sonnhard Graubner Sep 30 '17 at 17:11
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    $\begingroup$ The set of all possible inputs. Is there any number $x$ that you can't plug in to get a result? The domain is not concerned with the nature of the results. As long as there is a result, the input is legal. $\endgroup$ – quasi Sep 30 '17 at 17:11
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    $\begingroup$ Show me a real number that you can't square. $\endgroup$ – quasi Sep 30 '17 at 17:18
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    $\begingroup$ Yes, a function needs one output for every legal input. What you said: "for each output, only one input" is not the correct concept of a function. In other words, using arrows, for each legal input, an arrow has to go from that input to some output. It can't split and go to two different outputs. However, two or more inputs can share the same output. As an extreme example, consider a constant function, say $f(x) = 4$. Then every input goes to the output $4$. $\endgroup$ – quasi Sep 30 '17 at 17:25
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You're right about one thing: the square root function is the inverse function to the squaring function, after the squaring function is restricted to a domain on which it's one-to-one. That domain is nonnegative numbers.

However, I think you're overthinking this problem. When a problem asks for the the domain of a function defined by an algebraic expression, the task is to calculate the entire subset of real numbers which can be substituted into the expression. For instance, if the expression is $\frac{1}{1-x^2}$, you're supposed to notice that the denominator cannot be zero for this to make sense. So the domain must carve out any numbers which do that, namely $\pm 1$. Therefore the domain is $\mathbb{R}\setminus\{-1,1\}$.

But the expression you're given is just $x^2$. This is defined for all real numbers. So the domain is $\mathbb{R}$.

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The domain is $\Bbb R$, the set of all real numbers. The two inputs $\pm3$ both have output $9$, yes, but this doesn't stop either from being an input: $x^2$ still a function.

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A function $f$ on a set $X$ is a subset of the cartesian product of $X$ with itself such that if $(x,y) \in f$ and $(x,z) \in f$ then $y=z$. The domain of $f$ is the set $d$ of left coordinates of ordered pairs of $f$. In short, the domain is something you can have the distinct privilege of specifying.

I.e. if you have a function $f$ and a domain $d$ such that $d$ has more than, (for example, $10$ real numbers). Then you can create a new function $g$ by creating a new domain $d'$ by removing one element from $d$ and setting $g$ equal to the restriction of values of $f$ defined over $d'$.

In short, one domain for the function $f(x)=x^{2}$ is $\mathbb{R}$ because for each real number $x$ in the domain you get a unique real number, (namely $x^2$.)

You might opine that this is a "natural" domain for the function. But then again, we can specify the same equation defining the function $f$ and just restrict our domain set $d$ to be a new domain, $d = \{274848638463926284\} $ instead of all of $\mathbb{R}$ and the result is still a domain for $f(x)=x^{2}$. Although, the graph of this function is a single point in the plane-- which is rather boring compared to a parabola.

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I try to understand, why domain of $x^2$ is the set of all real numbers.

The key idea is to distinguish these two things:

  • the image of the function, i.e. the "output" of the function, and here you're right: it's not $\mathbb{R}$, but $\mathbb{R}^+$

  • the domain of definition, i.e. the set of all authorized numbers as "input" of the function: here all real numbers can be squared. Can $2.317$ be squared? Yes. Can $-3$ be squared? Yes. Indeed: $(-3)^2 = -3 \times (-3) = 9$. Thus the domain of definition of the function $x \mapsto x^2$ is $\mathbb{R}$.

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