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Anonymous No. 16423243

There is an integer between 5 and 6.
Think about it.
It’s obvious.

Anonymous No. 16423251

>>16423243
pi+e

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B00T No. 16423258

OOoOoOoooOOoOH. Boffin Alert

EEEEEEEEhuhahuahauheuehe

Anonymous No. 16423311

1 2 3 4 5 6 7 8 9
[1 2 3 4] ||5|| [6 7 8 9]

0 1 2 3 4 5 6 7 8 9
0 [1 2 3 4] ||5|| [6 7 8 9]
(ontologically, 0 is not a true number as true numbers exist to count a presence of something)

1 2 3 4 5 6 7 8 9 10
[1 2 3 4] ||5|| [6 7 8 9] 10

5 is in the middle.

Anonymous No. 16423580

>>16423243
yeah, it's [math]5.\overline{999}[/math]

Anonymous No. 16423590

>>16423580
I think we can all agree [math] 5 . \overline{ 99 } < 5 . \overline{ 999 } < 5 . \overline{ 9999 } [/math]

Anonymous No. 16423610

That's just an upside down 4.

Anonymous No. 16423631

>>16423243
yeah, sqrt(26)

Anonymous No. 16423929

>>16423243
thrembo - 1

Anonymous No. 16430301

>>16423243
it's true. [math] 4 \leq 5 \leq 6 [/math].

Anonymous No. 16430320

>>16423243
Maybe there aren't ANY integers.

Anonymous No. 16430351

>>16423243
Meds now.

Anonymous No. 16433194

>>16423243
4+1 = 5
5+1 = 6
The difference between any two integers is an integer.
There's an integer between 4 and 5 which would imply that x-4 = 1 and 5-x = 1 or 5-x+x-4 > 1 = 5-4

Anonymous No. 16433263

>>16423243
if its an inclusive range then there will be two integers not just one.