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submitted 1 month ago by silverchase@sh.itjust.works to c/mathmemes
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[-] friendlymessage@feddit.org 3 points 1 month ago* (last edited 1 month ago)

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 1 month ago

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.

[-] lseif@sopuli.xyz 2 points 1 month ago

isnt that how methods like proof by contrapositive work ??

[-] bleistift2@sopuli.xyz 3 points 1 month ago

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 1 points 1 month ago
[-] bleistift2@sopuli.xyz 2 points 1 month ago

Then don’t get involved in this discussion.

[-] friendlymessage@feddit.org 1 points 1 month ago* (last edited 1 month 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.

this post was submitted on 20 Oct 2024
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