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Definition A real number is irrational if it is not rational.

According to this definition, does "$\sqrt{2}$ is irrational" mean $\lnot$($\sqrt{2}$ is rational)?

This seems right to me, but an example that is why I am not confident is that $\lnot$($i$ is rational) is true, but "$i$ is irrational" is false. However, I feel that complex numbers are irrelevant in this context.

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    $\begingroup$ Your definition says a real number is irrational if it is not rational, so the statement “$x$ is irrational means $\neg$ ($x$ is rational)” only makes sense when $x$ is real. Bringing complex numbers into this would then be no different than saying “a cat is not irrational but $\neg$ (it is rational) is true”. (Of course, that’s according to the definition you wrote down. You could change your definition to include complex numbers, then it would be a different story.) $\endgroup$
    – David Gao
    Mar 2 at 6:56
  • $\begingroup$ The $\sqrt 2$ example is backwards from the $i$ example. In the first case, you start with the correct description of the number and verify that it implies the negation of one thing that the number is not. In the second case, you start with a negation, but it does not give you just the one resulting positive statement about $i$, there are multiple choices. $\endgroup$
    – abiessu
    Mar 2 at 6:56
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    $\begingroup$ A real number is irrational if it is not rational. Need to put in the condition that the thing we are talking about is real. Either $\sqrt{2}\text{ is irrational} \iff (\sqrt{2}\in \mathbb R \land \lnot(\sqrt{2} \text{ is rational}))$ or $\sqrt{2}\in \mathbb R\implies (\sqrt{2}\text{ is irrational} \iff \lnot(\sqrt 2\text{ is rational}))$. The first is the actually definition. The second is a case of us establishing we are only living in the real universe. $\endgroup$
    – fleablood
    Mar 2 at 7:54

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From your definition "A real number is..." means that your definition does not speak to numbers which are not real, so the ir-/rationality of $i$, which is complex but not real, is not resolved by this definition. (Of course, $i$ is irrational, but this definition does not tell you so.)

The given definition allows one to go from $$\sqrt{2} \text{ is irrational}$$ to $$\sqrt{2} \text{ is not rational.}$$ One then has to use the grammatical and syntactical rules of English to move to $$\text{it is not the case that (} \sqrt{2} \text{ is rational),}$$ which is immediately $$ \neg (\sqrt{2} \text{ is rational).}$$

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