Logic in the use of variables

Question

I've been questioning some things about the use of variables that I hadn't paid much attention to. I'm sorry if this question is too obvious. I know how to use variables, I've worked with them many times. My question is on some logical concerns I have.

In mathematics we often state an identity using the variable $x$, for example: $(x+1)^2=x^2+2x+1$ $\forall x \in \mathbb{R}$ or we describe a set like $A=\{(\cos(t),\sin(t)) | t\in \mathbb{R} \}$. Why is it logically valid to use the same symbol for different numbers in the same context? The first identity is valid for $x=3$ and for $x=4$ and although both numbers are the same letter $x$, I certainly can't say $3=x=4$ because $x$ can't be 3 and 4 at the same "time".

Similarly in $A$, $t$ takes all the possible real values, so the same symbol is used for infinitely many objects.

When I describe the unit circle I say that it is the set of all the points $(x,y) \in \mathbb{R}^2$ that satisy $x^2+y^2=1$, Why can I use the same symbol "$(x,y)$" to refer to all those points and not imply an equality between them?

Another doubt I have concerning variables is: when I name an object with a letter (stating a definition) Is it the same "type" of equality as $1+1=2$ where the right hand side is the same as the left hand side? For example, in the previous example $A=\{(\cos(t),\sin(t)) | t\in \mathbb{R} \}$. An equality between sets happens when they both have the same elements, but how can A have the same elements as the right hand side when it isn't even defined before the equality. I know some people use $:=$ instead of $=$ to state a definition, so is there a mathematical distinction?

I apologize if this question is too elemental. I'm not sure how to tag this question, please correct me if I'm wrong.

Answer 1

Variables are "syntactical" objcets used to write expressions in mathematics.

When we write an identity, like: $(x+1)^2=x^2+2x+1$ we are assertingbthat the expression holds for any value of $x$ (as you correctly write: $\forall x \in \mathbb R$.

Thus, using a fully formalized expression, we are asserting the senetence:

$\forall x \in \mathbb R \ [(x+1)^2=x^2+2x+1]$.

In this formula we have a quantifier (the universal one) binding the variable $x$, whose occurrences in the formula are called bound.

The identity holds for every possible value of $x$ (in the domain of real numbers) and thus also for $x=3$ and $x=4$.

But this does not mean that we can conclude that $3=x=4$.

From the universally quantified formula, we are licensed to assert every instance of it, i.e. that the instance obtained assigning to $x$ the value $3$: $(3+1)^2=3^2+2 \times 3+1$ holds, as well as the instance obtained assigning to $x$ the value $4$, and so on.

Basically, the previous case can be derived form the "functional notation", typical of mathematics.

We write $f(x)$ to denote a one-argument function, i.e. a rule that assigns to every value in input an output value.

In the example above, $(x+1)^2$ is a one-valued real function, call it $f(x)$ and $x^2+2x+1$ in another one, call it $g(x)$.

The above identity amounts to asserting:

$\forall x \in \mathbb R \ (f(x)=g(x))$.

In the "functional notation", the variable $x$ is a place-holder: we need it to refer to the argument place to be filled with the input values.

We may as well write $f(\quad)$; we use $x,y,\ldots$ according to the tradition, but also to manage functions with many argumets, like $f(x,y)$ or $f(x_1, \ldots, x_n)$.

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The example using the set-builder operator: $A = \{ (cos(t),sin(t)) \mid t \in \mathbb R \}$ in not very different.

In order to understand it, we have to start from the syntax of $A = \{ x \mid \varphi(x) \}$.

It denotes a set: the set of all and only those $x$ such that the "condition" expressed by the formula $\varphi(x)$ holds of them.

$\varphi(x)$ is a formula with a free variables, i.e. a parameter, and is again a "functional" expression.

When we assign to $x$ values (from a specified domain) we may have that the formula holds of them or does not hold.

The set-builder notation "collect" all the values that satisfy the formula into a set.

Having said that, the formula $A = \{ (cos(t),sin(t)) \mid t \in \mathbb R \}$ can be "unwinded" as follows:

$A = \{ x \mid \exists t \ \exists r \ \exists s \ [ t \in \mathbb R \land x=(r,s) \land r=cos(t) \land s=sin(t)] \}$.

The sat $A$ is a set of pairs; the complex formula $\exists t \ \exists r \ \exists s \ [ t \in \mathbb R \land x=(r,s) \land r=cos(t) \land s=sin(t)]$ has three bound variables and a free one: the parameter $x$.

Thus, it has the "form": $\varphi(x)$, expressing a "condition" on $x$.

It amounts to:

$(r,s) \in A \text { iff } r= cos(t) \text { and } s=sin(t)$, for some value $t \in \mathbb R$.

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Additional topis

The formula $1+1=2$ means that we have two different expressions naming the same thing: the number two.

The formula $A = \{ (cos(t),sin(t)) \mid t \in \mathbb R \}$ is a definition: we made the stipulation of calling $A$ the set defined by the expression on the right-hand side.

Answer 2

These are expressions: $$ (x+1)^2\\ x^2+2x+1\\ \cos(t) $$ These are variables $$ x\\ t $$ These are values $$ 3\\ \pi $$

In an expression, you can evaluate any variable to any value you want. The symbol $|$ is used for this operation: $$ \left.(x+1)^2\right|_{x=3}=(3+1)^2=16\\ \left.x^2+2x+1\right|_{x=3}=3^2+2\cdot3+1=16\\ \left.\cos(t)\right|_{t=\pi}=\cos(\pi)=0 $$

An equality, inequality or any other proposition, can be true or false according the evaluation you do.

• $(x+1)^2=x^2+2x+1$ will be true for every real $x$. This is an identity, i.e. an equality which holds for every value of the variable,
• $\cos(t)^2+\sin(t)^2=1$ is another identity, holds for any real $t$,
• $-1<\cos(t)<1$ for every real $t$. This is an inequality identity,
• $(x-1)^2=4$ is true only for some values of $x$. This is an equality between variables, an equation,
• $y=x^2$ is true only for some values of $x$ and $y$. This is an equality between those variables. As you see not with a single solution,
• $x^2=-1$ is true for no real number. Again another equality, but with no real solution.

From here, several notations and objects can arise, which you must have to interprete one by one, as they appear in your route:

• $(x,y)$ is a point in the 2D space, $(x,y,z)$ in the 3D space.
• If $x,y,z$ are free, these expression can represent any point in the 2D|3D space,
• $(x,y),x=y$ is a point, and a equality. This means both holds. $x,y$ can take any value, but also must satisfy the given equality. In this case after an analysis, this describe a line in the 2D space. Only one line can satisfy $y=x$ in the 2D space,
• $(x,y),x^2+y^2=1$ is a curve and one equality. Like the previous example, this constraint the point, to be only the points in a curve in the 2D space, a circle,
• $(x,y),x=\cos(t), y=\sin(t)$ is a point, and two equalities. Because $t$ is free and $x,y$ given by equalities, the point will depend on how $t$ varies. Every value of $t$ evaluates two values of $x$ and $y$ smoothly, and because of that, the point $(x,y)$ defines a curve on the 2D space. Because of algebra manipulation, you can reduce these two last equalities $x=\cos(t),y=\sin(t)$ plus the identity $\cos(t)^2+\sin(t)^2=1$ to the previously given equality $x^2+y^2=1$.
• $(x,y),y=x,x=\cos(t),y=\sin(t)$ is a point and three equalities. Now the case starts to be difficult. The first equality defines a curve, and the next two equalities again define another curve. In this case, because all expressions must hold, this defines an intersection, i.e. the intersection of all curves, which is finally just two points.

Answer 3

I don't understand your problem. If one says

• $(x+1)^2=x^2+2x+1\; \forall x \in \mathbb{R}$

one does not say that $x$ is $3$ and $4$ at the same time. It means, that

• if $x$ is a real number then $(x+1)^2=x^2+2x+1.$

And from this follows that

• if $x=3$ then $(x+1)^2=x^2+2x+1$ holds, so $(3+1)^2=3^2+2\cdot 3+1$

and

• if $x=4$ then $(x+1)^2=x^2+2x+1$ holds, so $(4+1)^2=4^4+2\cdot 4+1$

because 3 and 4 are reale numbers.

But there is no conflict. Because the variable name is valid only in the sentence where it is used. It is a bounded variable. This range of validity is often called the scope of the variable. One can replace a variable in its scope by another variable without changing the meaning of anything. So if we replace the last sentence by

• if $y=4$ then $(y+1)^2=y^2+2y+1$

all the meaning and therefore the validity of no statement changes.