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(1) ∀x ∀y (P(x, x, y)∧ ∼≈ (x, 0) → L(0, x)),

(2) ∀x ∀y (P(x, x, y)∧ ∼≈ (x, 0) → L(0, y)),

where P(x, y, z)is meant to be x · y = z; L(x, y) is interpreted as x < y; and ≈ (x, y) is interpreted as x = y

I'm trying to figure out whether the statements above are true or false under the domains: all integers as well as all natural numbers.

Do the two x's in P(x, x, y) have to be the same integer? If so, are they both false under the pretense that if y is a negative number, no integer for x can multiply to y therefore it's false under the domain of all integers.

And then, if y is 2 or anything number >2, no x integer can multiply by itself to produce 2.

But if the two x's in the statement can be different, there will be a huge change. Again, do the two x's in P(x, x, y) have to be the same integer?

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1 Answer 1

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Hint

(1) means :

for all $x$ and $y$, if $x \times x = y$ and $x \ne 0$, then $0 < x$, i.e. $x$ is positive.

Thus, YES, the two occurrences of $x$ in $P(x,x,y)$ must "name" the same number.

Thus, to check e.g. in $\mathbb N$ the truth-value of the statement, we can proceed this way :

(i) consider a couple of numbers $n, m \in \mathbb N$;

(ii) if they do not satisfy $n \times n = m$, then the antecedent of the conditional is false, and thus the conditional is true;

(iii) if they do, check if $n = 0$; if so, again conclude with true;

(iv) if $n \ne 0$, clearly (we are in $\mathbb N$) $n$ must be positive and thus the conditional is again true ($T \to T$ is $T$).

Conclusion : the statement is true in $\mathbb N$.

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  • $\begingroup$ Why is it true? If y is 5, there isn't a single positive x integer where when multiplied by itself gives a 5. $\endgroup$
    – Helpsun
    Sep 30, 2015 at 7:31
  • $\begingroup$ @Helpsun - exactly ! thus $n \times n = 5$ is false for every $n \in \mathbb N$; thus $n \times n = 5 \land n \ne 0$ is false for every $n \in \mathbb N$; thus $(n \times n = 5 \land n \ne 0) \to 0 < n$ is true for every $n \in \mathbb N$, because a conditional with false antecedent is ALWAYS true. $\endgroup$ Sep 30, 2015 at 7:36
  • $\begingroup$ I finally see! I had the order or operations wrong and was focused on seeing is as (function ∧ ( function → function)) I redid my work and I was hoping if you believe it's done well. For (1) and (2) it is true in N since y in L(0, y)), must still be a positive integer. For (1) and (2) it is true in Z since in L(0, y)). when given a negative integer, the statement because false and a false is a vacuous true! $\endgroup$
    – Helpsun
    Sep 30, 2015 at 7:52
  • $\begingroup$ @Helpsun - I see... Without parentheses, the usual "convention" is : (i) $\lnot$ applies as little as possible; (ii) given (i), $\lor$ and $\land$ apply to as little as possible. But you have to check it with your instructor ... $\endgroup$ Sep 30, 2015 at 8:56

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