I just began the topic of functions in my Mathematics textbook and in the first paragraph itself, two conditions about a relation being a function were mentioned.
They were :
1) For every a∈A, there exists b∈B such that (a,b)∈f I think that is another way of saying that both the input and output must lie in the given sets
2) If (a,b)∈f and (a,c)∈f, then b = c And I think that this is another way of saying that an input has a unique outputLet me know if I have misunderstood these statements.
Now, let's think of the square root function...
The square root of 4 is ±2, which means that there are two outputs 2 and -2 for one input 4, so there are two images of the same input, so it defies the second condition that states that a function must have a unique output/image of some input
Let R represent the set of real numbers, then if square root is a function, it means that both the input and output must lie in the set R. In the case of square root of -1, the output will be i (iota) which is not a member of the set R but is a member of C (the set of complex numbers), this defies the first condition that makes a relation a function.
Please let me know if I have misunderstood some concept related to functions and if square root is really a function or not.
Thank You...
$\endgroup$ 14 Answers
$\begingroup$Not quite. There are two numbers that square to $4$. Those numbers are $2$ and $-2$. The (usual) square root function takes the input $4$ to the output $2$.
In principle, the square root function must choose which of two outputs to provide for all but one input. (The one input is $0$, for which there is only one possible output, also $0$.) You can make many non-standard square root functions by making different choices for particular inputs.
Now regarding the square root of real numbers... You recite
For every $a \in A$, there exists $b \in B$ such that $(a,b) \in f$.
There is no requirement that $A = B$, so the set of inputs and the set of outputs need not be the same. You already know that the square of any real number is either zero or positive. Consequently, if the output set of a square root function is going to be the real numbers, the input set can only be the set of zero or positive real numbers.
$\endgroup$ 1 $\begingroup$To amplify the other comment and answer: we are used to the algebraic step of seeing an equation like $x^2=4$ and realizing that it is equivalent to $x=\pm2$. We often say "take the square root of both sides" to describe this algebraic step; but that's not exactly accurate. The square root function, by definition, is a function whose values are all nonnegative. So the algebraic step is related to taking square roots, but it's not the same as taking square roots.
Indeed, the pedentically correct formulation of the "square root" algebraic step is this: $x^2=y^2$ if and only if $|x|=|y|$. So if we see $x^2=4$, we realize that this is the same as $x^2=2^2$ and thus deduce that equivalently $|x|=|2|$, which implies that $x=\pm2$. (Or we can see $x^2=4$ as $x^2=(-2)^2$ if we like, which is similarly equivalent to $|x|=|-2|$, leading to the same outcome.)
$\endgroup$ $\begingroup$A historical note. As others have said, it is now customary to build in to the notion of a function the idea that a a function is (at most) single-valued. But mathematicians used to be more accommodating to talk of multi-value functions. For example in Hardy's great A Course of Pure Mathematics, after giving some examples of single valued total functions, he goes on to remark that the general notion of a function allows for functions which are partial and/or multi-valued. He cheerfully treats the squareroot function as a two-valued function for positive numbers.
$\endgroup$ 1 $\begingroup$The terminology is subtle.
$4$ has two square roots, namely $2$ and $-2$, because $2^2=(-2)^2=4$.
But the square root function, denoted with the symbol $\sqrt\ $ is obviously a function, always defined to be a positive value. Hence
$$\sqrt4=2\text{ and }\sqrt4\ne-2.$$
But you can very well say that the square roots of $4$ are $\pm\sqrt 4$.
$\endgroup$
