Timeline for True or false or not-defined statements
Current License: CC BY-SA 3.0
24 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 20, 2023 at 16:50 | comment | added | ryang | My answers are here and here. | |
| Sep 23, 2012 at 12:12 | comment | added | hmakholm left over Monica | My answer here describes various ways of dealing formally with formulas like this. | |
| Sep 23, 2012 at 11:34 | comment | added | Tobias Kienzler | related: Does “This is a lie” prove the insufficiency of binary logic? | |
| Sep 23, 2012 at 5:46 | answer | added | Hannes R. | timeline score: 0 | |
| Sep 23, 2012 at 3:01 | history | tweeted | twitter.com/#!/StackMath/status/249704914232676352 | ||
| Sep 23, 2012 at 2:12 | comment | added | André Nicolas | Things get complicated. For example, in school mathematics, it is said that the "identity" $\frac{\sin(2x)}{\sin x}=2\cos x$ is called true even though the left side is undefined when $x$ is a multiple of $\pi$, while the right side is always defined. | |
| Sep 23, 2012 at 0:47 | vote | accept | Thomas | ||
| Sep 23, 2012 at 0:44 | answer | added | Michael Hardy | timeline score: 4 | |
| Sep 23, 2012 at 0:08 | answer | added | Ben Millwood | timeline score: 4 | |
| Sep 22, 2012 at 23:51 | comment | added | Lemon | How can a meaningless expression have truth or falsity? | |
| Sep 22, 2012 at 23:37 | comment | added | André Nicolas | The division "operation" in unpleasant to try to accommodate in a formal system. For when we are defining term, we cannot say that if $a$ and $b$ are terms, then $a\div b$ is a term. There are workarounds, but nothing direct. | |
| Sep 22, 2012 at 23:30 | comment | added | Trevor Wilson | How come all the debate is focusing on the first question? I want to know if sdfjinrivodinvr is true! | |
| Sep 22, 2012 at 23:25 | answer | added | Qiaochu Yuan | timeline score: 9 | |
| Sep 22, 2012 at 23:25 | comment | added | Clive Newstead | @Thomas: Perhaps! It all depends on whether $\frac{1}{0}$ is a constant in your language ;) | |
| Sep 22, 2012 at 23:25 | answer | added | Makoto Kato | timeline score: 7 | |
| Sep 22, 2012 at 23:25 | comment | added | Thomas | @CliveN.: That was what I was trying to ask. So I guess in my mind the $\frac{1}{0} = 1$ isn't true or false because it is not a valid formula since dividing by $0$ "isn't allowed". | |
| Sep 22, 2012 at 23:24 | comment | added | Clive Newstead | @TrevorWilson: As above, I guess. | |
| Sep 22, 2012 at 23:20 | comment | added | Trevor Wilson | @CliveN. Would "$1/0 \ne 1$" also be false for the same reason? Or is the equality relation special? | |
| Sep 22, 2012 at 23:20 | comment | added | Clive Newstead | @Thomas: I guess that would depend what you meant by each side of the equation. On a more formal level, if you have an interpretation of a logical system then you can only declare some formula $\phi$ in that system to be true or false if $\phi$ is actually a valid formula. So if $\phi$ is not then the truth or falsity (or otherwise) of $\phi$ is meaningless. | |
| Sep 22, 2012 at 23:18 | comment | added | Thomas | @CliveN.: So would then also the statement $\frac{1}{0} = \frac{2}{0}$ be false? (here both sides not being well-defined) | |
| Sep 22, 2012 at 23:18 | comment | added | Trevor Wilson | The question reminds me of Pauli's remark: en.wikipedia.org/wiki/Not_even_wrong | |
| Sep 22, 2012 at 23:16 | comment | added | Michael Dyrud | Your first example is false. Something you are comparing something well defined with something undefined by way of the equivalence relation =. The relation holds if and only if 1 and 1/0 are equivalent, which they are not. | |
| Sep 22, 2012 at 23:16 | comment | added | Clive Newstead | Your first expression is false, because the right-hand side is defined, and so if it were true then the left-hand side would also be defined. Your second statement is not a mathematical problem. | |
| Sep 22, 2012 at 23:11 | history | asked | Thomas | CC BY-SA 3.0 |