this post was submitted on 20 Oct 2024
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Math Memes

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[–] humblebun@sh.itjust.works 11 points 10 months ago (1 children)

You only needed to choose 2 points and prove that they can't be connected by a continuous line. Half of your obviousness rant

[–] lseif@sopuli.xyz 3 points 10 months ago (2 children)
[–] JeeBaiChow@lemmy.world 7 points 10 months ago (3 children)

It's fucking obvious!

Seriously, I once had to prove that mulplying a value by a number between 0 and 1 decreased it's original value, i.e. effectively defining the unary, which should be an axiom.

[–] Sop@lemmy.blahaj.zone 6 points 10 months ago

Mathematicians like to have as little axioms as possible because any axiom is essentially an assumption that can be wrong.

Also proving elementary results like your example with as little tools as possible is a great exercise to learn mathematical deduction and to understand the relation between certain elementary mathematical properties.

[–] friendlymessage@feddit.org 4 points 10 months ago* (last edited 10 months ago) (1 children)

So you need to proof x•c < x for 0<=c<1?

Isn't that just:

xc < x | ÷x

c < x/x (for x=/=0)

c < 1 q.e.d.

What am I missing?

[–] bleistift2@sopuli.xyz 5 points 10 months ago (2 children)

My math teacher would be angry because you started from the conclusion and derived the premise, rather than the other way around. Note also that you assumed that division is defined. That may not have been the case in the original problem.

[–] friendlymessage@feddit.org 2 points 10 months ago* (last edited 10 months ago)

Your math teacher is weird. But you can just turn it around:

c < 1

c < x/x | •x

xc < x q.e.d.

This also shows, that c≥0 is not actually a requirement, but x>0 is

I guess if your math teacher is completely insufferable, you need to add the definitions of the arithmetic operations but at that point you should also need to introduce Latin letters and Arabic numerals.

[–] lseif@sopuli.xyz 1 points 10 months ago (1 children)

isnt that how methods like proof by contrapositive work ??

[–] bleistift2@sopuli.xyz 3 points 10 months ago (1 children)

Proof by contrapositive would be c<0 ∨ c≥1 ⇒ … ⇒ xc≥x. That is not just starting from the conclusion and deriving the premise.

[–] lseif@sopuli.xyz -3 points 10 months ago (1 children)
[–] bleistift2@sopuli.xyz 2 points 10 months ago

Then don’t get involved in this discussion.

[–] superb@lemmy.blahaj.zone 3 points 10 months ago

It can’t be an axiom if it can be defined by other axioms. An axiom can not be formally proven

[–] humblebun@sh.itjust.works 0 points 10 months ago (2 children)

One point on the line

Take 2 points on normal on the opposite sides

Try to connect it

Wow you can't

[–] erin@lemmy.blahaj.zone 6 points 10 months ago (1 children)

This isn't a rigorous mathematic proof that would prove that it holds true in every case. You aren't wrong, but this is a colloquial definition of proof, not a mathematical proof.

[–] humblebun@sh.itjust.works 1 points 10 months ago (2 children)

Sorry, I've spent too much of my earthly time on reading and writing formal proofs. I'm not gonna write it now, but I will insist that it's easy

[–] erin@lemmy.blahaj.zone 1 points 10 months ago

Oh trust me, I believe you. Especially using modern set theory and not the Principia Mathematica.

[–] lseif@sopuli.xyz 1 points 10 months ago

so... maybe its not worth proving then.

[–] davidagain@lemmy.world 1 points 10 months ago* (last edited 10 months ago) (1 children)

Only works for a smooth curve with a neighbourhood around it. I think you need the transverse regular theorem or something.