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Is proving both:

  • For all natural numbers n, if n is a perfect square, then the root of n is not irrational.
  • For all natural numbers n, if the root of n is irrational, then n is not a perfect square.

The same as proving:

For all natural numbers n, n is either a perfect square or the square root of n is irrational.

  • "For all natural numbers n, n is either a perfect square or the square root of n is irrational." This allows for the possibility both statements are true. You need to prove the *exclusive or". Which to say one or the other is true but not both. That'd be the same as one if and only if the other is false. Which is what the first to statements are. – fleablood Nov 24 '19 at 23:20
  • .... And the other hand. The definition of perfect square implies $n$ being perfect square and square root irrational are incompatible. With that stated then yes, those would e equivalent. – fleablood Nov 24 '19 at 23:21

2 Answers2

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I think you want to ask whether $$\tag1 \forall x\colon \phi(x)\to\neg \psi(x)\quad\land\quad \forall x\colon \psi(x)\to\neg \phi(x)$$ is logically equivalent to $$\tag2 \forall x\colon \phi(x)\not\leftrightarrow \psi(x).$$

It is not. In fact, rather the two parts of $(1)$ are equivalent. However, note that $(1)$ might still be valid even if there exists some $x_0$ with $\neg \phi(x_0)\land\neg\psi(x_0)$, whereas $(2)$ does not allow that. (In your example, the first two statements wouldn't mind if $\sqrt{42}$ were rational)

Hagen von Eitzen
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Let P(n) = n is a perfect square

I(n) = n is irrational

IS(n) : the square root of n is irrational

(1) P(n) --> ~ I(n)

(2) IS(n) --> ~ P(n)

By contraposition (3) yields

(3) P(n) --> ~ IS(n)

Propositions (1) and (3) taken together mean

(4) P(n) --> ~ I (n) & ~ IS(n)

( Read : if n is a perfect square, then neither n is irrational nor the square root of n is irrational).

The sentence you proposed means :

if P is not a perfect square, then the square root of n is irrational

for in general P OR Q means : ~P --> Q