Any good jokes ... ?

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I do worry about the fact that each thing has 3 9s - except for the number 1

which is a problem
 

srw

It's a bit more complicated than that...
OK, for us thickoes, can someone please explain 5 and 7?
5 = (√9)! - 9/9. √9 = 3, and 3! = 3 factorial = 3x2x1 = 6. What's actually on the clock is √(9!), which is the square root of 9x8x7x...x3x2x1.

7 = 9 - √9 + 0.9 recurring. "0.9 recurring" means 0.999999.... and so on ad infinitum which is mathematically identical to 1.
 
5 = (√9)! - 9/9. √9 = 3, and 3! = 3 factorial = 3x2x1 = 6. What's actually on the clock is √(9!), which is the square root of 9x8x7x...x3x2x1.

7 = 9 - √9 + 0.9 recurring. "0.9 recurring" means 0.999999.... and so on ad infinitum which is mathematically identical to 1.
Soooo
the position for number one could be 9 - 9 + .9 recurring??

My problem is solved

Thanks
 

cyberknight

As long as I breathe, I attack.
1609176355813.png
 

Profpointy

Legendary Member
I read that as "point nine recurring'. Not the most obvious notation, but it works. If you add in another .9 you break the rule that you can only use three digits in each expression.

(Folks, this is what passes for humour among mathematical pedants.)

I think it's been done before, but if we're doing maths gags ...

"what's Benoit B Mandlebrot's middle name?"

"Benoit B Mandlebrot of course!"
 
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