Post #4043290
2026-07-23 18:00 UTC
Replies (2)
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@FishFace@ioc.exchange 2026-07-23 18:31
@rzeta0@mathstodon.xyz In a simple case like this, given an existence assertion, ("for some x, ...") you should be able to give a witness, i.e. an x satisfying everything. The fact that you can't in the second case should be convincing. Note that this is a *convincing* argument but not a *proof* because there are more complicated situations where something may exist (i.e. you have a proof it does) but it may be impractical or even impossible to give a witness. To give a proof, you should rewrite the statements formally as: "for all x, if 3 < x < 2 then 6 < 2x < 4" which is equivalent to "for all x, if False then False", which is true; and "there exists an x, such that 3 < x < 2 and 6 < 2x < 4" which is "there exists an x such that False and False" which is false.
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@consequently@hcommons.social 2026-07-23 18:15
@rzeta0@mathstodon.xyz The second is false because there is no x such that 3<x<2, since 3 is not less than 2, so, a fortiori, there is no x where 6<2x<4 and 3<x<2, i.e., it's not the case that 6<2x<4 for some x where 3<x<2.