Timeline for How do I write this Theorem with quantifiers?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 23 at 21:23 | comment | added | Dr. J | I see. Thank you for your response and answering all my questions. It is greatly appreciated! | |
| Apr 23 at 21:20 | comment | added | ultralegend5385 | @Dr.J: Now $|a-b|<\varepsilon\implies a=b$ is checked for a fixed $\varepsilon$. That is clearly false; take $a=0$, $b=1$, $\varepsilon=2$ then $|0-1|<2\implies 0=1$ is absurd. | |
| Apr 23 at 21:14 | comment | added | Dr. J | Oh, I see. The negation of "property P holds for at least one x" would be property p does not hold for all x. When I had tried using the quantifiers I initially thought it would be this: $$\forall a,b,\varepsilon:\,\Big(|a-b|<\varepsilon\implies a=b\Big)$$. Why was this incorrect? I'm not sure why. | |
| Apr 23 at 20:59 | comment | added | ultralegend5385 | [...] Once you have done that, now think about what the negation of "property $P$ holds for at least one $x$" will be. | |
| Apr 23 at 20:58 | comment | added | ultralegend5385 | @Dr.J: Think about it: you have the statement "property $P$ holds for all $x$". Is the negation "property $P$ does not hold for all $x$ or "property $P$ does not hold for at least one $x$? For your contrapositive, it is correct other than the minor $\geq\varepsilon$ instead of $>\varepsilon$. | |
| Apr 23 at 20:51 | comment | added | Dr. J | Thank you! This makes sense. I'm struggling to understand contradiction and the contrapositive with quantifiers. For example, you negated the for all quantifier for a, and b. I was thinking it would stay the same. Why do we negate it? Also would the contrapositive for the 3rd formula you provided be this: $$\forall a,b:\,\Big(a \neq b\implies(\exists\varepsilon>0:\,|a-b|>\varepsilon)\Big)$$ | |
| Apr 23 at 20:38 | vote | accept | Dr. J | ||
| Apr 23 at 20:29 | comment | added | ultralegend5385 | @Dr.J I edited my answer. | |
| Apr 23 at 20:29 | history | edited | ultralegend5385 | CC BY-SA 4.0 |
added 96 characters in body
|
| Apr 23 at 20:27 | comment | added | Lee Mosher | Yes, $a$ and $b$ do need quantifiers. The "for all" quantifier would go right in front of them. | |
| Apr 23 at 20:25 | comment | added | Dr. J | This does help a lot! What about a, and b? Wouldn’t they have the quantifier for all. Where would that go and how does that factor into the contradiction proof? Thank you! | |
| Apr 23 at 20:23 | history | answered | ultralegend5385 | CC BY-SA 4.0 |