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I asked something similar a while ago, but there were no responses. Suppose we are working over the real numbers.

Consider the term $0x_2 + x_1 + 1$. By itself, this is a function of two variables. However, after simplification, the term becomes $x_1 +1$, which represents a function of only one variable.

So, what is going on? Is there a solution to this puzzle? I would appreciate some clarification of this.

Blue
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user107952
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    Who says $x_1+1$ has to represent a function of only one variable? – Eric Wofsey May 08 '21 at 02:56
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    Ummm... what's the "puzzle"? – David G. Stork May 08 '21 at 03:03
  • What you can do is define $f(x_1,x_2) = x_1 +1$. Since you have that the coefficient of $x_2$ is 0, it doesn't change the fact that this is still a function of both $x_1, x_2$. You can conclude that in the $x_2$ direction, graphically, it is a constant function at $0$. – Joey May 08 '21 at 03:04
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    I don't understand this question. What exactly is the "puzzle" here? What is problematic about having a two-variable function which treats one variable trivially? I don't really know what you're looking for in an answer. – Noah Schweber May 08 '21 at 03:13
  • @NoahSchweber This is related to my question of associating functions to terms, which I asked a while ago and you answered. It shows that simplifying terms can change the arity of the function. – user107952 May 08 '21 at 19:03
  • @user107952 Well, sort of. As I said in my answer to said question, a given term doesn't really have a single function associated to it. What you can do by simplifying is change the "minimal arity." But I don't see what's problematic about any of this. – Noah Schweber May 08 '21 at 19:05
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    @NoahSchweber So, I guess that when simplifying terms, you have to provide some data that shows what the arity of the function is. – user107952 May 08 '21 at 20:28
  • Simplification commonly does remove data (which must be supplemented to retain identical two-way implication). In the case of rational equations of polynomials with a vanishing factor that was unique to the non-simplified denominiator, that equates to ∀-quantification. In the case of a[n erroneously] vanishing variable, that equates to ∃-quantification (which correlates to arity- or dimensionality- retention). – user946772 Jan 01 '22 at 02:45

2 Answers2

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The answer is that you actually have a function of two variables, just that one of them happens to be canceled out by its 0 coefficient. I'm not sure how far into calculus you are, but what you really have a is a function $f:\mathbb{R}^2\to\mathbb{R}$, that is, a function that takes in a pair of real numbers and gives you back a single real number. Your example function is $f(x_1,x_2)=0x_2+x_1+1$, so if you input $(x_1=-1,x_2=2)$, you get $f=2$. You can make the $x_2$ coordinate whatever you want, but it doesn't change the value of $f$. That's what's going on, it's a sneaky function that looks like it's one variable, but is actually two.

Jake Brown
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You have two functions $f$ (with 2 variables) and $g$ (with 1 variable) given by:

$~~~~~~\forall a,b \in R:f(a,b) = 0a+b+1$

$~~~~~~\forall a\in R:g(a)=a+1$

It turns out that:

$~~~~~~~\forall a,b \in R:f(a,b)=g(b)$

Dan Christensen
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