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I just learned the definition of implication and have been considering a few problems.

Given $x$ is real, prove $x = 1 \Rightarrow x = 2$. Either $x = 1$ or $x \neq 1$. If $x=1$, then the implication is false, and if $x \neq 1$ is true, then the implication is true. Since we don't know the value of $x$, then this statement cannot be proved.

Also consider given $x$ is real, prove it is false that $x \ge 0 \implies \text{abs}(x) > 0$. Textbook answers usually go: when x is $0$, $\text{abs}(x)$ is $0$ and therefore the statement is wrong.

But isn't this the proof of the NEGATION of the UNIVERSAL statement (for all real $x$, if $x\le 0$, then $\text{abs}(x) > 0$) by proving there exists a value of $x$ (i.e. 0) such that $\text{abs}(x)$ is not greater than $0$?

To prove this statement false, we need to prove $x \ge 0$ AND $\text{abs}(x) \le 0$. But we don't know the value of $x$, so rigorously speaking, we cannot prove this, right?

Maybe there is some sort of convention regarding this question?

However,

suppose now the question is instead if $x > 0$, prove $\text{abs}(x) >0$. In this scenario, to me, it is clear that it is the question about this SPECIFIC value of x that we do not know that is being asked (it just sounds right, and you often see problems start with an unknow e.g. the possibility of something is p...), even though the universal statement about all $x$ implies this and is probably most people would use to deduce this statement.

So it seems like there is some convention regarding proving the falsehood of such a statement, but not for the truth?

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    $\begingroup$ When we say a statement such as "Given $x$ is real, $P(x)$" we mean to say that for all $x$ in the universe of discourse we are interested in (in this case all real numbers) the statement $P(x)$ holds true. The first example you give, that $x=1\implies x=2$ is false so long as you are considering a universe in which $1$ is an element and $1\neq 2$, which you imply you are when you said "given $x$ is real." $\endgroup$
    – JMoravitz
    Jan 3, 2022 at 13:30
  • $\begingroup$ The implication $A\implies B$ is false if and only if $A$ is true and $B$ is false. $x>0\implies |x|>0$ can be proven since we can show that for every $x>0$ we also have $|x|>0$ (in fact , $x>0$ implies $|x|=x$). So, the value of $x$ need not be known in general to determine the truth of such an implication. $\endgroup$
    – Peter
    Jan 3, 2022 at 13:34
  • $\begingroup$ @JMoravitz If asked to prove something is false, should I prove that statement is false for all values of variables (x, y, z, etc.), or should I prove the falsehood of the universal statement involving all the variables(i.e. to prove there exists some values that the statement is false)? $\endgroup$
    – TFR
    Jan 3, 2022 at 14:18
  • $\begingroup$ Make sure you understand the difference between "$P(x)$" and "$(\forall x) P(x)$". When you talk about "that statement" it is unclear whether you mean $P(x)$ or whether you mean $(\forall x)P(x)$. To prove "$(\forall x)P(x)$" is false (in other words proving $\neg((\forall x)P(x))$ is true) you need only find one example of an $x$ in the universe of discourse such that $\neg P(x)$ (though there may be multiple $x$ that you can do this for). This is different than trying to prove $(\forall x)\neg P(x)$. $\endgroup$
    – JMoravitz
    Jan 3, 2022 at 15:11
  • $\begingroup$ @JMoravitz You mentioned "When we say a statement such as "Given x is real, P(x)" we mean to say that for all x in the universe of discourse we are interested in (in this case all real numbers) the statement P(x) holds true." So if asked to prove it is false that Given x is real, P(x), you prove ¬((∀x)P(x)) rather than (∀x)¬P(x), right? It is confusing because one could have tried to prove Given x is real, it is false that P(x). $\endgroup$
    – TFR
    Jan 4, 2022 at 1:55

2 Answers 2

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Given that $x$ is real, prove that $$x = 1 \implies x = 2.\tag1$$ Since we don't know the value of x, then this statement cannot be proved.

given that $x$ is real, prove that it is false that $$x\geq0\implies\textrm{abs}(x) >0.\tag2$$ To prove that this statement false, we need to prove that $x\geq0$ AND $\textrm{abs}(x) \leq 0.$ But we don't know the value of $x,$ so, rigorously speaking, we cannot prove this, right?

$(1)$ and $(2)$ are open formulae and thus not statements, let alone provable statements. However, if we treat them as being implicitly universally quantified, i.e., $$\forall x\,\big(x = 1 \implies x = 2\big)\tag{1a}$$ and $$\forall x\,\big(x\geq0\implies\textrm{abs}(x) >0\big)\tag{2a},$$ then the counterexamples $x=1$ and $x=0,$ respectively, show that they are false statements.

If asked to prove something is false, should I prove that statement is false for all values of variables (x, y, z, etc.), or should I prove the falsehood of the universal statement involving all the variables(i.e. to prove there exists some values that the statement is false)?

The latter. Providing the counterexamples above is essentially disproving $(1\mathrm a)$ and $(2\mathrm a)$ by proving their negations $$\exists x\,\big(x = 1 \;\text{and}\; x \ne 2\big)\tag{1n}$$ $$\exists x\,\big(x\geq0 \;\text{and}\; \textrm{abs}(x) \leq0\big)\tag{2n}.$$

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Hint 1: Proving $\neg \forall x:[x\in R \to [x=1 \to x=2]]$

Suppose to the contrary. Assuming $~1\in R~$ and $~1\neq 2$, obtain a contradiction.


Hint 2: Proving $\neg \forall x:[x\in R \to [x\leq 0 \to |x|\gt0]]$

Suppose to the contrary. Assuming $~0\in R, ~|0|=0,~0\leq 0$ and $~0 \ngtr 0$, obtain a contradiction.

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