Does a real number - say, r - such that [eqn]\forall n \in N:rx<1[/eqn]

>>8977065
Does a real number exist*

sorry for my incompentence

No, real numbers are an Archimedean field.
You can do this with surreal numbers though.

What is [math]x[/math]? What is [math]N[/math]? If [math]x=n[/math] and [math]N = \mathbb{N}[/math], then yes, for instance [math] r = -1 [/math].

>>8977065
0

>>8977065
rn < 1
r < 1/n > 0
r < 0

>>8977088
>r < 1/n > 0
>r < 0
What the fuck are you doing, that's not how inequalities work

>>8977068
This is the correct answer.

>>8977079
>>8977081
quit being autists

>>8977065
Look here
https://en.wikipedia.org/wiki/Infinitesimal

>>8977065
No. It's possible if r is an infinitesimal in the hyperreals though.

>>8978582
Lmao, you mad because they saw something you did not? It's true that if r is 0 or any negative number, then rx < 0 for any natural number x.

>>8978591
rx < 1 *

yes if x=0.999 and r=0.999
>>8977079
n=rx you brainlet