Timeline for True, false and meaningless statements in math.
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Feb 9, 2014 at 0:03 | comment | added | tomasz | Just because a statement does not admit a truth value (or is not, in fact, a statement, like the examples you cited) does not mean it's meaningless. | |
| Feb 8, 2014 at 17:17 | comment | added | Carl Mummert | In the normal setting of the field of real numbers, we can't say "$\sqrt{-1} = 1$" or "$\sqrt{-1} \not = 1$", of course, as we have no square root function. Actually we can't even say this in the complex numbers until we extend the language with a square root function in some way. | |
| Feb 8, 2014 at 17:04 | history | edited | amWhy | CC BY-SA 3.0 |
added 463 characters in body
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| Feb 8, 2014 at 16:51 | comment | added | amWhy | Yes, that is true providing $x$ exists and is real. While $x \in \{-1, -2\}$ exists, $\sqrt{-1}, \sqrt{-2}$ does not exist in the reals, so it cannot be compared to the number $1$. What we can say for certain is that $\sqrt{-1} \neq 1$ and $\sqrt{-2} \neq 1$. Even if we allow the complex roots, complex numbers are not comparable save for $z_1 = z_2$ or $z_1\neq z_2$ (the complex numbers are not an ordered field, like the reals are, "less than" and "larger than" are meaningless in the complex numbers.) | |
| Feb 8, 2014 at 16:42 | comment | added | nkoreli | Thanks for the answer, but I am still confused, since I thought that statements (x != 3) and (x>3 or x<3) are equivalent. It is from definition of relation != . | |
| Feb 8, 2014 at 16:36 | history | answered | amWhy | CC BY-SA 3.0 |