Bad Math Jokes

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Nothing in the tax code requires that we operate in base 10.

So I calculate my income in Hexadecimal.
 
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Ivan Seeking said:
Nothing in the tax code requires that we operate in base 10.

So I calculate my income in Hexadecimal.
They may send you a response in binary Esperanto.
 
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BillTre said:
They may send you a response in binary Esperanto.
I will need Bill Shatner!
 
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If I ever get caught I will just tell them I've done a lot of programming. It's an easy mistake to make.
 
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1736350583807.png
 
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472735196_1047322564078612_5196476120040524529_n.jpg
 
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Given a matrix: " And the Eigenvalues are Real?".
" Yes, the Eigenvalues are Real, but their names were changed to protect the innocent."
 
berkeman said:
A little Medic humor...
Is that a gold-plated deformed rebar I see, held before me?
 
Topologist, responding to accusations: " Yes, I slept with your wife. But up to Homotopy, she's my wife too."
 
WWGD said:
Topologist, responding to accusations: " Yes, I slept with your wife. But up to Homotopy, she's my wife too."
Bold. It means he knows they have the exact same number of piercings.
 
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Orodruin said:
Bold. It means he knows they have the exact same number of piercings.
I doubt your average Math prof's wife will have piercings that can't be smoothed out.
 
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Orodruin said:
Bold. It means he knows they have the exact same number of piercings.
So probably not Jane Pierce.
 
jbriggs444 said:
So probably not Jane Pierce.
Didn't see any references on her having piercings, certainly not non-trivial ones.
 
New Math Logic book:
" Ultraproducts, America's New Supermodel".
 
etotheipi said:
If 2005 humour is still funny... the English cat 'one two three' and the French cat 'un deux trois' had a swimming race to decide after which country the Channel should be named. The un deux trois cat sank.
Ahh, un deux trois quartre cinq
 
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DrGreg said:
All numbers are equal! Let ##a## and ##b## be any two numbers and define$$
c = a + b.
$$Multiply both sides by ##a-b##:$$
(a - b)c = (a - b)(a+b).
$$Expand:$$
ac - bc = a^2 - b^2.
$$Rearrange:$$
b^2 - bc = a^2 - ac.
$$Add ##ab## to both sides:$$
ab + b^2 - bc = a^2 + ab - ac.
$$Factorise:$$
(a+b-c)b = (a+b-c)a.
$$Cancel:$$
b=a
$$QED.
Nope, because C = A + B, so
A+B-C = 0,
Let's see.

c = a + b.
$$Multiply both sides by ##a-b##:$$
(a - b)c = (a - b)(a+b).
$$Expand:$$
ac - bc = a^2 - b^2.
$$Rearrange:$$
b^2 - bc = a^2 - ac.
$$Add ##ab## to both sides:$$
ab + b^2 - bc = a^2 + ab - ac.
$$Factorise:$$
(a+b-c)b = (a+b-c)a. = 0xB = 0xA
$$Cancel:$$
b=a
$$QED.
 
Re Pde's, the song " Looking for love in all the wrong places, looking for love in Sobolev spaces.
 
Screenshot 2025-03-01 at 10.41.38 AM.png
 
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An old car chuggs up a hill of 1 mile for an average speed of 15 mph.
The car then travels downhill for 1 mile.
What must the downhill speed be, so that for the whole trip the average speed is 30 mph.

Ludicrous Speed
 
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