0

In Terence Tao's Analysis I, he says that "the statement '$x=3$ if and only if $x^2=9$'" is false. However, isn't it more accurate to say that this statement is not always true? Isn't this statement true sometimes (i.e. when $x\neq-3)$ and false other times (i.e. when $x=-3$). Perhaps the $\forall x$ quanitifer had been left implicit, and what he really means is that the following statement is false: $$ \forall x:x=3 \leftrightarrow x^2=9 \, . $$ However, at this point in the book he hadn't introduced quantifiers, and so I'm unsure what is intended.

To add to my confusion, in this post, it states that the meaning of "If $A$, then $B$" is "whenever $A$ is true, $B$ is true". According to this definition, the statement "$x=3$ if and only if $x^2=9$" is plainly false. However, the negation of $A \leftrightarrow B$ is $A\leftrightarrow \neg B$. So the negation of "$x=3$ if and only if $x^2=9$" is "$x=3$ if and only if $x^2\neq9$", which is also false according to this definition in the linked post. A statement and its negation can't both be false, so I'm unsure what's gone wrong.

Joe
  • 19,636
  • I agree with your analysis. I would recon that the author was being a little informal. – user2628206 Jul 01 '21 at 14:05
  • Something that is "sometimes false" is false. – Randall Jul 01 '21 at 14:08
  • 1
    The statement '$x=3$ if and only if $x^2=9$' contains the implication $x^2=9 \Rightarrow x=3,$ which (in general) is wrong, as $(-3)^2 = 9.$ If, however, we impose further restrictions on $x$, we can obtain similar statements which are true: $\forall x \in [0,\infty) : x^2 = 9 \Leftrightarrow x = 3.$ – LurchiDerLurch Jul 01 '21 at 14:09
  • 2
    "If and only if" means "whenever" both ways not "some of the time" – Mark Bennet Jul 01 '21 at 14:09
  • randall I think there is subtlety there. It may be true in certain models and untrue in others. I think the OP is running up against the distinction of satisfiability vs tautology vs unsatisfyability – user2628206 Jul 01 '21 at 14:10
  • 5
    See page 312: "If x is an integer, then the statement “If $x = 2$, then $x^2 = 4$” is true, regardless of whether x is actually equal to 2 or not (though this statement is only likely to be useful when x is equal to 2)." Why so? because for every value of $x$, if the antecedent is false it is Ok and if the antecedent is true (i.e., $x=2$) then also the consequent is true. Thus, the author is reading - as common mathematical practice - an open formula as universally quantified. – Mauro ALLEGRANZA Jul 01 '21 at 14:10
  • 1
    If we apply the same approach (as you are doing above) to the bi-conditional we get false, because for $x=-3$ we have that $x^2=9$ is true. – Mauro ALLEGRANZA Jul 01 '21 at 14:11
  • In general, if you have two statements, $P$ and $Q$, where $P \implies Q$ and where it is not the case that $Q \implies P$, then the statement $P \iff Q$ is false, by definition of the $\iff$ operator. – user2661923 Jul 01 '21 at 16:06
  • @TonyK: Please see the discussion below. In order to make sense of the statement $x=3 \leftrightarrow x^2=9$, it has to be clear that there is a quantifier over $x$ (either implicitly or explicitly). Otherwise $x=3 \leftrightarrow x^2=9$ does not have a true value at all because $x$ is a free variable. – Joe Jul 01 '21 at 20:09
  • 1
    it states that the meaning of "If A, then B" is "whenever A is true, B is true" Not quite: yes, their meanings can overlap, however, the former might mean that $A\to B$ is true in the current interpretation, while the latter means either that $A\to B$ is a validity or is universally true. The word "whenever" indicates that we are considering $A$ (and $B$)'s truth as the interpretation varies or as its predicate variable varies. – ryang May 30 '22 at 15:35

2 Answers2

4

Indeed, in the statement "$x = 3$ if and only if $x^2 = 9$", the universal quantifier ("for all $x$ it is true that...") has been left implicit. As a logical statement, $x = 3 \leftrightarrow x^2 = 9$ does not have a truth value, as $x$ is still a free variable.

This explains the problem with negation: although the negation of $A \leftrightarrow B$ is $A \leftrightarrow \neg B$, the negation of $\forall x (A(x) \leftrightarrow B(x))$ is $\exists x (A(x) \leftrightarrow \neg B(x))$. And indeed there is an $x$ such that $x = 3$ is false but $x^2 = 9$ is true.

Mees de Vries
  • 26,947
  • Wow, there are so many explainations here. –  Jul 01 '21 at 14:11
  • 2
    Thank you for this. You and Paul Frost's answers seem to be the only ones that have understood and addressed my confusion. – Joe Jul 01 '21 at 14:41
  • 1
    Joe I second that. The other answers all contain misunderstandings. – user2628206 Jul 01 '21 at 14:46
  • @user2628206: Thank you. By the way, if you want to notify someone when you comment, then you should write @name. See here. – Joe Jul 01 '21 at 14:52
  • Ah thanks! @Joe – user2628206 Jul 01 '21 at 14:55
  • 1
    I have a question: I've done some research and it turns out that formulae that contain a free variable are not assigned truth values. Only closed formulae are assigned truth values. So to give a formula we have to bind all the variables within it, either by putting a quantifier over them, or assigning a particular value to one of the variables. Here is one edge case I need to clarify: the words "let $x=5$" binds the variable $x$. However, do the words "let $y$ be a solution to $y^2=36$" bind the variable $y$, given that this equation has two solutions? – Joe Jul 01 '21 at 16:46
  • Good question, but I think you are making a category error. Variable binding usually refers to connecting a variable to a quantifier within a given formula. The variable is said to be 'within the scope' of the quantifier. The Let statement is made when writing proofs in a non formal (as in formal system) way when one wants to introduce variable that satisfies a certain condition to reason upon. How this is done formally depends on the proof system that one is using, but there is a clear precedent for its usage in prose. These are, to my understanding, fundamentally different processes – user2628206 Jul 01 '21 at 18:11
4

In fact the quantifier $\forall x$ is implicit here. Although Tao has not formally introduced quantifiers when he considers the statetement in question on p. 358, he has made it clear verbally what is going on. Quotation:

Thus for instance, if $x$ is a real number, then the statement "$x = 3$ if and only if $2x = 6$" is true: this means that whenever $x = 3$ is true, then $2x = 6$ is true, and whenever $2x = 6$ is true, then $x = 3$ is true. On the other hand, the statement "$x = 3$ if and only if $x^2 = 9$" is false; while it is true that whenever $x = 3$ is true, $x^2 = 9$ is also true, it is not the case that whenever $x^2 = 9$ is true, that $x = 3$ is also automatically true (think of what happens when $x = -3$).

Thus the claim is

If $x$ is a real number, then the statement "$x = 3$ if and only if $x^2 = 9$" is false.

This is the verbal formulation of

$\forall x \in \mathbb R$ : $x = 3 \leftrightarrow x^2 = 9$

which is false.

Joe
  • 19,636
Paul Frost
  • 76,394
  • 12
  • 43
  • 125
  • Thanks for this answer, Paul. It does help to put things in context. I asked Mees de Vries this question, but I'd also like to hear your opinion: I've done some research and it turns out that formulae that contain a free variable are not assigned truth values. Only closed formulae are assigned truth values. So to give a formula we have to bind all the variables within it, either by putting a quantifier over them, or assigning a particular value to one of the variables. – Joe Jul 01 '21 at 16:48
  • Here is one edge case I need to clarify: the words "let $x=5$" binds the variable $x$. However, do the words "let $y$ be a solution to $y^2=36$" bind the variable $y$, given that this solution has two solutions? – Joe Jul 01 '21 at 16:48
  • 1
    @Joe Yes. On p. 373 Tao writes The symbol $\forall$ can be used instead of "For all", thus for instance "$\forall x \in X : P(x)$ is true" or "$P(x)$ is true $\forall x \in X$" is synonymous with "$P(x)$ is true for all $x\in X$". You see that $X$ specifies the possible values for the free variable $x$ occurring in $P(x)$. The phrase "Let $y$ be a solution to $y^2 = 36$" itself is not a statement, but it specifies the values $y$ may take in some statement $P(y)$ depending on the free variable $y$. – Paul Frost Jul 01 '21 at 21:42
  • Formally you can define $Y = { y \in \mathbb R\mid y^2 = 36}$ and then you get "$\forall y \in Y : P(y)$" which may be true or false. – Paul Frost Jul 01 '21 at 21:42
  • Okay, but when we write $\forall y \in Y: P(y)$, $y$ is a bound variable. Is that correct? – Joe Jul 01 '21 at 22:14
  • 1
    @Joe Yes, correct. – Paul Frost Jul 01 '21 at 22:15