this post was submitted on 20 Oct 2024
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Math Memes

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Memes related to mathematics.

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1: Memes must be related to mathematics in some way.
2: No bigotry of any kind.

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[–] [email protected] 3 points 4 weeks ago (2 children)
[–] [email protected] 7 points 4 weeks 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.

[–] [email protected] 5 points 3 weeks 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.

[–] [email protected] 3 points 4 weeks ago

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

[–] [email protected] 3 points 3 weeks ago* (last edited 3 weeks 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?

[–] [email protected] 5 points 3 weeks 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.

[–] [email protected] 2 points 3 weeks ago (1 children)

isnt that how methods like proof by contrapositive work ??

[–] [email protected] 3 points 3 weeks 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.

[–] [email protected] -2 points 3 weeks ago (1 children)
[–] [email protected] 2 points 3 weeks ago

Then don’t get involved in this discussion.

[–] [email protected] 1 points 3 weeks ago* (last edited 3 weeks 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.

[–] [email protected] 0 points 3 weeks 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

[–] [email protected] 6 points 3 weeks 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.

[–] [email protected] 1 points 3 weeks 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

[–] [email protected] 1 points 3 weeks ago

so... maybe its not worth proving then.

[–] [email protected] 1 points 3 weeks ago

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

[–] [email protected] 1 points 3 weeks ago* (last edited 3 weeks ago) (1 children)

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

[–] [email protected] 1 points 3 weeks ago