5

What do we mean when we say "Let $x$ be an element of the set $\mathbb{R}$"?

Does $x$ represents only a single element of set $\mathbb{R}$? or does $x$ represent all the element of set $\mathbb{R}$ simultaneously at the same time?

People say that if $x\in\mathbb{R}$ then $x$ is any real number; that means $x$ represents all real numbers. But if $x$ is any real number then let's say $x=1$; so $x$ is one, then how it can represent all real numbers?

Please help me I am very confused.

Blue
  • 75,673
Shashwat singh
  • 123
  • 5
    If $x \in \Bbb R$, then $x$ is a single real number. It could be $0$, or $-\sqrt2$, or $\pi$, etc. Think of a magic trick where the magician asks you to pick a number from $1$ to $10$. You have a single number in your head: which one? Let's call it $x$. – Théophile Dec 20 '20 at 04:57
  • I don't think this is a duplicate, but you may find this question useful. – Carl Schildkraut Dec 20 '20 at 08:36
  • Everybody gangsta until I write „Let $x$ be an element of the set $\emptyset$“ :) – Hagen von Eitzen Dec 20 '20 at 17:12
  • You should read this post. – user21820 Dec 30 '20 at 12:22
  • @littleO: An if-subcontext like what you said (using "suppose") is not the same as a ∀-subcontext. Suppose you have some complex number $c$ and you want to prove that it is not real. You might say "Suppose $c$ is real. Then ... Hence contradiction. Therefore $c$ is not real." Here you can see clearly that "suppose $c$ ..." did not specify all you know about $c$. Unsurprisingly, this distinction is made completely clear in user-friendly formal systems such as Fitch-style systems. – user21820 Dec 30 '20 at 12:27
  • @user21820 Hmm, if I introduce $x$ by saying "Suppose $x$ is a real number", and then make certain arguments and reach a conclusion about $x$, then we have shown that this conclusion is true for any real number. (In other words, we have shown that "if $x$ is a real number, then such and such conclusion is true.) Regardless, I'm going to delete my previous comment while I ponder this further. – littleO Dec 30 '20 at 18:16
  • @littleO: What you are saying is correct if there is no other governing context regarding $x$. In other words, that "suppose $x$" phrase can be used to make a universally quantifying subcontext if $x$ is currently an unused variable and your formal system allows you to use unused variables without declaring them first. My above example should have been sufficient, but here is another: "Let $r = \sqrt{2}$. Suppose $r$ is rational. Then ... Hence contradiction. Therefore $r$ is not rational." We certainly have no reason to conclude "every rational $r$ leads to contradiction"... – user21820 Dec 30 '20 at 19:02
  • For reference, you can see this Fitch-style system and compare the ⇒intro and ∀intro rules. Not all systems force declaration of all variables, but one should grasp quantification in this clear way. – user21820 Dec 30 '20 at 19:05

4 Answers4

2

The set $\mathbb{R}$ is the familiar real number line, including all "decimal expansions." When we say a number is an element of $\mathbb{R}$, we mean that it's a part of the number line. $1 \in \mathbb{R}$, $\sqrt{2} \in \mathbb{R}$, negative fractions that look weird such as $\frac{-1}{\pi}$ are in the set $\mathbb{R}$, just cause they have a decimal expansion.

When we say $x \in \mathbb{R}$, we just mean that there's a number, just like the ones I mentioned above, that is a real number. We just don't know what it is yet? What is $x$? We don't know, but we can give it a name because for whatever reason it's of interest to us. We just know it lies somewhere on the number line. We don't know what $x$ is, but noting that it's in $\mathbb{R}$ for one reason or another has importance. But "variable" in the title of your question is just giving a letter that we don't know its value.

Moni145
  • 2,142
  • What does decimal expansion mean? Does 0.99999+1.23645i count as a decimal expansion? If not, why? – Divide1918 Dec 20 '20 at 07:56
  • 1
    I said this informally, and no, obviously not. They may not contain an imaginary component. I stuck to the level of rigor that the question was at. It is a number that can be represented by an infinite decimal expansion. They're on the number line, which is enough for the level we're at. Complex numbers are not on the number line. Are you expecting me to construct the reals from an analysis standpoint? – Moni145 Dec 20 '20 at 08:02
2

  • Suppose I say : if x // y and y // z then x // z.

  • Is this sentence meaningful? In order this sentence to make sense, I should first say : Let x, y and z be straight lines in a given plane D.

  • So, saying " let x belong to R" is a way to state the domain in which a sentence will have meaning, and, therefore, will have a truth value . ( Maybe, allowing x to be some non-real number would yield a meaningless sentence.)

  • Second reason: it might happen that a sentence has meaning in some large domain D, but is false for some values of this domain. In case your goal is to establish a universally true sentence, you restrict the possible values of x to a subset D* of D.

  • Third reason: sometimes , you use " Let x belong to some domain D" as an hypothesis in a conditional proof.

Let x belong to R $\space \space $ (Hypothesis for conditional proof).

x² = 4

$\sqrt{x²} = \sqrt 4$

$|x| = 2$

$x = 2$ OR $x = -2$

$x² =4 \rightarrow (x= 2$ OR $x = -2)$

x belongs to R $\rightarrow [x² =4 \rightarrow (x= 2$ OR $x = -2) ] $

For all $x_{\in R} \space , \space x² =4 \rightarrow (x= 2$ OR $x = -2)$

Note : the conclusion ( for all x belonging to R ... ) is allowed on the ground that, in the hypothesis, x was arbitrary. If the concluson holds for any number in R , it also holds for all numbers in R.

Floridus Floridi
  • 718
1

There are already good answers, but I really struggled with this when first encountering logic and proofs, so I hope my own perspective is useful to someone.

Often we want to to prove something about all elements of a set S. That is, we want to prove that every element in $S$ satisfies some property, call it $P$. For example, say $S = \{1,2,...\}$. Now say $P(x)$ stands for the statement "$x$ is greater than $0$". Clearly $P(1)$ is true, and so is $P(2)$ is true, and indeed it should be obvious that no matter what number we choose from $S$, $P(\text{ 'that number' )}$ will be true.

We want a way to express the truth of the statement in the above example. We know that no matter what object I choose from $S$, the statement $P$ will be true for that object. We do this by reasoning about an arbitrary object from that set. We say $x \in S$, and specify nothing more. Now the key bit here is that $x$ is not simultaneously "all objects in $S$". $x$ represents a specific number from the set $S$, we just don't say which one.

Going back to our example, say I tell you $x \in S$. Now $x$ represents some specific number in $S$, but we don't know which. Suppose $x$ actually represented the number $20$. Well of course $P(20)$ is true, because $20>0$. Similarly, if $x$ actually represented the number $1727361$, then $P(1727361)$ would be true too. The pattern here is clear: no matter what number $x$ actually represents, $P(x)$ is true. It is in this sense that $x$ can represent any/all elements of $S$.

Ultimately, our example can be written as $\forall x \big(x \in S \rightarrow P(x)\big)$. This statement is really saying, "for every object $x$, if this object is a specific a number in the set $S$, then the statement $P(x)$ is true." Note the key point here is that I have said $x$ is specific, yet I have not told you what it is. The meaning here is that in the statement, we treat $x$ as a specific number. After all, we say that $x \in S$ and $P(x)$, which only make sense if $x$ is just one number. But $x$ could represent any specific number in $S$, because every number I use makes $P(x)$ true.

The above is an intuitive and informal approach; but fundamentally this question can be answered formally. This question revolves around notions of variables, quantifiers, and logic. Answers to my own questions and others' here on stack exchange do a very good job of explaining this more formal approach. For example, see this. The answer and Noah Schweber's comment are incredibly useful.

Response to comments by OP

When we say $x>-2$, yes, it is correct to say that $x$ can be any real number. $x$ could also be a function, a group, or any kind of mathematical object. $x$ is simply a symbol. There is no reason to say that $x$ must be a real number larger than $-2$. If I were to then tell you $x$ was the number $-3$, that's perfectly fine, it just means the statement is false for that 'value' of $x$.

Now, if you were to say that $x$ is an arbitrary real number such that $x>-2$, then yes, now $x$ represents a specific (yet unspecified) real number larger than $-2$. In essence we are saying $x$ is an arbitrary element of the set $\{y \in \mathbb{R}: y > -2\}$. Now it makes sense to say that $x$ is an arbitrary element greater than $-2$, because I have told you that is the case.

To summarise, the statement $\forall x \in \{y \in \mathbb{R}: y > -2 \} \big(x > -2 \big)$ is true, but the statement $\forall x \in \mathbb{R} \big(x > -2 \big)$ is not. And furthermore, $x > -2$ by itself is not comparable to the other two. This is a well formed formula, and whether it is true or not depends on what $x$ is. Of course, we may restrict $x$ to only take on those values which make $x>-2$ true. In this case, then $x$ represents some specific yet unspecified value from $\{y \in \mathbb{R}: y > -2 \}$.

masiewpao
  • 2,217
  • @masiewpaoThankyou for your explination,I understood that when we write x∈R, here x represents a single element in a set R we just don't specify which element.That's why we say x can be any real number in set R, but then what's the point to- say x represents all elements of R ? I am not able to understand this line.Please help me. – Shashwat singh Dec 20 '20 at 15:11
  • @Shashwatsingh We need to be careful by what we mean when we say "$x$ represents all elements of $\mathbb{R}$". It is correct that $x$ could be any specific real number. For example, $x$ could be $10$, or $\sqrt{2}$. But $x$ is NOT both $10$ and $\sqrt{2}$ at the same time. – masiewpao Dec 20 '20 at 15:30
  • @Shashwatsingh When we say $x \in \mathbb{R}$, we simply say that the variable $x$ may represent any object from the set $\mathbb{R}$. But when we reason and use logic to deal with the variable $x$, $x$ represents just a SINGLE real number! It's just that the logic and reasoning we use will work with any real number whatsoever. So in the example I gave, $P(10)$ is true, and so is $P(11)$. But in one case, $x=10$, and in the other, $x = 11$. At no point is $x$ 'both' of these numbers. It just means I can substitute any number for the variable $x$, and $P(x)$ will still be true! – masiewpao Dec 20 '20 at 15:34
  • @Shashwatsingh Ah, I see. Well the statement "$x$ represents all elements of $\mathbb{R}$" is simply false then. – masiewpao Dec 20 '20 at 15:36
  • 1
    I got the concept. I am just being confused with the use of word "all" and "any". Saying " x can be any real number"means x represents just a SINGLE real number which can be any real number(e.g. 10,12,5,4,etc).since we have not specified which real number x represents,this means Roughly speaking x represents all real numbers but one at a time. Am I correct now ? – Shashwat singh Dec 20 '20 at 15:56
  • @Shashwatsingh yes exactly! – masiewpao Dec 20 '20 at 15:59
  • Also "If something holds for any number in R , then it also holds for all numbers in R" Right ? – Shashwat singh Dec 20 '20 at 16:01
  • @Shashwatsingh Yes that's right. That's the idea behind why proving statements for an arbitrary element of a set means you have proven it for all elements of the set. (Again, that is an informal argument, and you can find formal justifications with rigorous treatments of logic.) – masiewpao Dec 20 '20 at 16:14
  • How "If something holds for any number in R , then it also holds for all numbers in R" say x>-2 here x represents just a SINGLE real number which can be any real number. so x>-2 is true for arbitary x but how can all real numbers follow this like say how real numbers less than -2 will follow this x>-2 ? – Shashwat singh Dec 20 '20 at 16:18
  • @Shashwatsingh Because the statement $\forall x \in \mathbb{R} (x > -2)$ is false. Also the set of all real numbers $x$, such that $x > -2$, is NOT $\mathbb{R}$ – masiewpao Dec 20 '20 at 16:21
  • oh, ok i got that we have to use this concept here - when we say x>-2 here x represents just a SINGLE real number which can be any real number.since we have not specified which real number x represents,this means Roughly speaking x represents all real numbers greater than -2 but one at a time. So this statment x>-2 is true for the arbitary x,thus all the real numbers greater than -2 will make this statment x>-2 true. Now I got it.... – Shashwat singh Dec 20 '20 at 16:36
  • @Shashwatsingh That's not quite right. Let me update my answer to respond to this, as it is getting a bit long to reply in the comments. – masiewpao Dec 20 '20 at 16:47
  • its fine you can answer here I am ok.... – Shashwat singh Dec 20 '20 at 16:48
  • @Shashwatsingh I have updated the answer. Unfortunately extended discussion in the comments is discouraged. – masiewpao Dec 20 '20 at 17:07
  • when we talk in terms of inequation(inequation) then in this statment x>-2, x represents a specific(yet unspecified) real number greater than -2 right ?since x is not specified roughly we say x represents all real numbers but one at a time. Thus, all real numbers greater than -2 are the solution of the inequality. Right ? – Shashwat singh Dec 20 '20 at 18:17
  • @Shashwatsingh If you say $x$ is an arbitrary real number such that the inequality $x>-2$, then yes, $x$ can be thought of as representing any specific real number greater than $2$. – masiewpao Dec 20 '20 at 19:36
  • @Shashwatsingh and yes, if we say the universe of discourse is the set $\mathbb{R}$, this means that variables, things like $x$ or $y$, are placeholders for real numbers. So writing $x>-2$ means that we can replace $x$ with any real number to get a statement that is true or false. For example, if we replace $x$ with $10$, we get the true statement $10 > -2$. If we replace $x$ with $-4.22$, we get the false statement $-4.22>-2$. – masiewpao Dec 20 '20 at 19:41
  • thankyou, you have really helped me a lot. I have just one question more- for the inequality If x represents just a SINGLE real number which can be any real number greater than -2,then why do we say that inequality has infinite solutions ? do this inequality has infinite solutions.... – Shashwat singh Dec 21 '20 at 02:58
  • for the inequality x>-2 If x represents just a SINGLE real number which can be any real number greater than -2,then why do we say that inequality has infinite solutions ? – Shashwat singh Dec 21 '20 at 14:21
  • @Shashwatsingh it has infinite solutions because the set ${y: y > -2}$ is an infinite set. Note that ${x: x>-2}$=${y: y > -2}$, because here $x$ and $y$ are called dummy variables. That means they don't represent any particular object or number. A 'solution' is simply a real number that when plugged into the formula $x>-2$, turns it into a true statement. – masiewpao Dec 21 '20 at 14:29
  • @Shashwatsingh You keep writing $x>-2$ by itself, and this is fairly meaningless. It isn't true or false, it is just a formula. $x$ can represent any real number, in the sense we can replace the symbol $x$ with a number. For example, $3>-2$, or $-10>-2$. But realise that $10$ is NOT a solution to the equation $x>-2$. – masiewpao Dec 21 '20 at 14:37
  • I am confused with the use of "for all x in Set..". I don't understand what do we mean when we say 'for all x in..." or "for each x in ...." I start to visulaise this as -> {x,x,x,x,x,x} something like this. x represents a single a specific real number(yet unspecified).If x is specific (single) real number which can be any real number,The rest other are elements of the set R. Then what does "for every x in.." mean ? – Shashwat singh Dec 22 '20 at 05:34
  • Let us continue this discussion in chat. – masiewpao Dec 22 '20 at 12:54
  • we can continue it here – Shashwat singh Dec 22 '20 at 13:23
  • I don't like the chat room but as you tell ? – Shashwat singh Dec 22 '20 at 13:24
  • @Shashwatsingh I will not respond further here – masiewpao Dec 22 '20 at 13:24
  • then Let us continue this discussion in chat. – Shashwat singh Dec 22 '20 at 13:27
0

I know an answer has already been accepted, but I'll try to answer the heart of the question anyway. All of the following statements mean the same thing:

Let $ x \in \mathbb{R}$. Then $x^2 \in$ {$u \in \mathbb{R}: u \geq 0$}.

Let $ x \in \mathbb{R}$. Then $x^2 \in \mathbb{R}$ and $x^2 \geq 0$.

$ x \in \mathbb{R} \implies x^2 \in$ {$u \in \mathbb{R}: u \geq 0$}.

$ x \in \mathbb{R} \implies \left(x^2 \in \mathbb{R} \text{ and } x^2 \geq 0 \right)$.

If $ x \in \mathbb{R}$, then $x^2 \in$ {$u \in \mathbb{R}: u \geq 0$}

If $ x \in \mathbb{R}$, then $\left(x^2 \in \mathbb{R} \text{ and } x^2 \geq 0 \right)$

If $x$ is any real number then its square is a non-negative real number.

$$$$

The statements are true. However, note that their truth value is independent to whether or not the statements mean the same thing, In other words, false statements can also be equivalent.

Personally, I avoid using the statement with "any" in it, because it can be confusing and I'd rather just get on with the maths than spend time being confused. If someone else uses the statement with "any" in it, then I translate it into one of the other statements.

J. W. Tanner
  • 60,406
Adam Rubinson
  • 20,052