Determine whether the rule describes a function with the given domain and codomain

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Determine whether the rule describes a function with the given domain and codomain

f:N ---> N where f(n) = (squared root of n)

How can I solve this question? I tried putting many different numbers on the function and seeing if it will return just one value but I am not sure if this is the way I am supposed to solve this, can someone explain it to me?

Thanks
 

stapel

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Determine whether the rule describes a function with the given domain and codomain
Is "codomain" the "range of the function", or the set from which the range is taken, or something else? Please provide your book's specific definition.

f:N ---> N where f(n) = (squared root of n)
Does "N" stand for "the natural numbers", being the set {1, 2, 3, 4, ...} (or possibly {0, 1, 2, 3, 4, ...})? Does "squared root" mean "square root", or something else?

How can I solve this question? I tried putting many different numbers on the function and seeing if it will return just one value but I am not sure if this is the way I am supposed to solve this
Is there any way for this relation (or rule) to return more than just one value? So can this relation possibly not be a function?

So you've answered the "is it a function" part. Now you're left with the "given domain and codomain" part. So look at your list of outputs. Is the value of f(3) included within the set of natural numbers? Is the value of f(5) included? And so forth. What can you conclude? ;)
 

pka

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Determine whether the rule describes a function with the given domain and codomain
f:N ---> N where f(n) = (squared root of n)
There is no universal agreement on this definition. As anyone can see the two major sources disagree on the definitions. This website agrees with reply #2, while this website is the way that I have always used.
The Texas school of analysis calls for an initial set(domain) and a final set(range), which agrees with MathWorld.

But in any case, \(\displaystyle f:n\mapsto\sqrt n\) is NOT a function \(\displaystyle f:\mathbb{N}\to\mathbb{N}\). You can see that \(\displaystyle \sqrt 2\notin\mathbb{N}\)
 

Dale10101

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Interestng

There is no universal agreement on this definition. As anyone can see the two major sources disagree on the definitions. This website agrees with reply #2, while this website is the way that I have always used.
The Texas school of analysis calls for an initial set(domain) and a final set(range), which agrees with MathWorld.

But in any case, \(\displaystyle f:n\mapsto\sqrt n\) is NOT a function \(\displaystyle f:\mathbb{N}\to\mathbb{N}\). You can see that \(\displaystyle \sqrt 2\notin\mathbb{N}\)
I would have interpreted "squared root of n" as f(n) = ((n)1/2)2 . At which point one might be tempted to reduce this to:

f(n) = n, but I think that would be wrong. I believe the rules of precedence (working from within the parentheses first) would require taking the root first in which case the domain would be limited to only those values of N (natrual numbers) which have an natural number root. (?)
 
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ksdhart

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Let's say we define "squared root of n" as \(\displaystyle f\left(n\right)=\left(\sqrt{n}\right)^2\). Anything squared is always positive, so the function evaluates to \(\displaystyle f\left(n\right)=\left|n\right|\:for\:n\in \left[0, \infty \right)\). Further, because we're taking a square root and thus limited to only positive numbers, it does, in fact, simplify down to \(\displaystyle f(n)=n\). The task is to determine if \(\displaystyle f:n \mapsto n\) is a function \(\displaystyle f:\mathbb{N}\mapsto \mathbb{N}\). And, a function which returns the original input satisfies the required mapping.
 

Dale10101

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OK, still

Let's say we define "squared root of n" as \(\displaystyle f\left(n\right)=\left(\sqrt{n}\right)^2\). Anything squared is always positive, so the function evaluates to \(\displaystyle f\left(n\right)=\left|n\right|\:for\:n\in \left[0, \infty \right)\). Further, because we're taking a square root and thus limited to only positive numbers, it does, in fact, simplify down to \(\displaystyle f(n)=n\). The task is to determine if \(\displaystyle f:n \mapsto n\) is a function \(\displaystyle f:\mathbb{N}\mapsto \mathbb{N}\). And, a function which returns the original input satisfies the required mapping.
First, I hope this is not confusing to the original poster of the question but rather complementary to his/her understanding. Second, I did add a question mark (?) because I too am still learning and so am not trying to speak beyond my level of understanding ... so thank you for your response ksdhart.

I am still at a loss. I am supposing that f(n) = (n1/2)2 defines a procedure for the evaluation of f(n). My understanding is that the order of precedence requires beginning the evaluation within the parentheses. Going back to the offered example f(2) =(21/2)2 one must evaluate 21/2 first, but there is no such number if one is confined to the set of natural numbers.

I see what you are saying but it runs contrary to this other line of thought. Is this a case where the rule of precedence fails? Again, this is inquiry not debate.
 
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