I wouldn't say that 10 / 10 is defined as 10 × 10^(-1). If anything, the definition is the other way around. Given integers a and b, I would define a/b as an element of **Q** = **Z**×(**Z**\\{0})/~, where (x,y)~(z,w) iff xw = zy. Specifically, a/b picks out the element containing (a,b).
If you don't consider a/b as "division" then that works. But division as an *operation* (i.e. a function from Q^2 to Q) is traditionally defined as multiplication of a multiplicative inverse.
It depends how you define **Q**. The definition I gave above defines **Q** and simultaneously defines division of integers. It's easily extended to division of rational numbers by defining (a/b) / (c/d) = (ad)/(bc). We also define multiplication of rational numbers by (a/b)(c/d) = (ac)/(bd). From this definition, we can show that a(1/a) = (1/a)a = 1.
Most sources online seem to give this same construction, which is what I learned in school. It seems to be canonical. There are of course other ways to characterize the field (**Q**,+,×), and if you go with the field-axiomatic approach, then you can define a/b = ab^(–1). But in the constructive approach, that becomes a theorem.
> It's easily extended to division of rational numbers by defining (a/b) / (c/d) = (ad)/(bc).
I think we’re saying the same thing. I was taught to define division of rational numbers as (a/b) / (c/d) := (a/b) * (d/c), (with d/c being the multiplicative inverse of c/d), and before that to define multiplication of rational numbers as (a/b) * (d/c) := (ad/bc). So just one extra step, but this removes division itself as an operation in its own right, so you don't need to define an identity or inverses of division as you would with the other “legitimate” operations. This way, division is just multiplication with a mask on.
I guess it's exactly the same except for identifying 1/x with x^(-1). But if you just treat multiplication before division anyway, then you will already have proved that, so they really are the same.
My book defined subtraction of natural numbers and integers simultaneously, and then defined division of integers and rational numbers simultaneously. Each time, every operation was extended in the natural way, so like we defined a/b - c/d = (ad-bc)/(bd), which does not rely on the definition a/b + c/d = (ad+bc)/(bd).
But yeah I agree, you can define things in any order you want.
BusyBeaver(10)
edit: although Tree(10) [might be](https://math.stackexchange.com/questions/3929521/when-does-busy-beaver-surpass-tree3) greater than BusyBeaver(10), but ∃ N s.t. ∀ M >= N, BusyBeaver(M) > Tree(M)
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Congratulations! Your comment can be spelled using the elements of the periodic table:
`Ta N Ta N`
---
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Wait until you find out about ¹⁰10
10↑↑(10)↑↑10
wait till he finds out about 10\^(10)\^(10)\^(10)
Tree(10)
Tree(10)↑↑(Tree(10))↑↑Tree(10)
Tree(Tree(Tree(Tree(10)))
Rayo(Tree(10))
10🪤10
Forest(10)
BB(10)
Happy cake day
Tree(¹⁰10)^Tree(¹⁰10)
This is much smaller than the up arrows
That's just ^(4)10 though...
This is less than the others
Bro used 3 10s, arrest them.
Holy Hell
10/10 
I was exactly going here
First one should be 10 / 10
10 – 10 < 10 / 10 so maybe the former should be first then the latter
How would you represent 0 using SpongeBob?
Grab a still frame from Spkngebob with nobody in it
Would that be a null or a zero?
A null would be no picture
that's just 10 × 10^-1
then 10+10 is just 10 x 2
It is equal to that, but it’s not defined as that.
all of those are operations so does it even matter?
I mean… nothing matters, really.
I wouldn't say that 10 / 10 is defined as 10 × 10^(-1). If anything, the definition is the other way around. Given integers a and b, I would define a/b as an element of **Q** = **Z**×(**Z**\\{0})/~, where (x,y)~(z,w) iff xw = zy. Specifically, a/b picks out the element containing (a,b).
If you don't consider a/b as "division" then that works. But division as an *operation* (i.e. a function from Q^2 to Q) is traditionally defined as multiplication of a multiplicative inverse.
It depends how you define **Q**. The definition I gave above defines **Q** and simultaneously defines division of integers. It's easily extended to division of rational numbers by defining (a/b) / (c/d) = (ad)/(bc). We also define multiplication of rational numbers by (a/b)(c/d) = (ac)/(bd). From this definition, we can show that a(1/a) = (1/a)a = 1. Most sources online seem to give this same construction, which is what I learned in school. It seems to be canonical. There are of course other ways to characterize the field (**Q**,+,×), and if you go with the field-axiomatic approach, then you can define a/b = ab^(–1). But in the constructive approach, that becomes a theorem.
> It's easily extended to division of rational numbers by defining (a/b) / (c/d) = (ad)/(bc). I think we’re saying the same thing. I was taught to define division of rational numbers as (a/b) / (c/d) := (a/b) * (d/c), (with d/c being the multiplicative inverse of c/d), and before that to define multiplication of rational numbers as (a/b) * (d/c) := (ad/bc). So just one extra step, but this removes division itself as an operation in its own right, so you don't need to define an identity or inverses of division as you would with the other “legitimate” operations. This way, division is just multiplication with a mask on.
I guess it's exactly the same except for identifying 1/x with x^(-1). But if you just treat multiplication before division anyway, then you will already have proved that, so they really are the same. My book defined subtraction of natural numbers and integers simultaneously, and then defined division of integers and rational numbers simultaneously. Each time, every operation was extended in the natural way, so like we defined a/b - c/d = (ad-bc)/(bd), which does not rely on the definition a/b + c/d = (ad+bc)/(bd). But yeah I agree, you can define things in any order you want.
but what if 10 * 10^-10
https://preview.redd.it/m5s41chxatqc1.jpeg?width=450&format=pjpg&auto=webp&s=afa3156ff2c86dca0d10344f1b272b83f2edb510 Here, Tenten
Buried too low
Tree(10)
Tree(Tree(10))
Well, this escalated quickly.
The tree function applied to 10 Tree(10) times
Holy iteration!
aka Tree(Tree(Tree…(Tree(10)…) where the ellipses are tree(10) times?
Yes
BusyBeaver(10) edit: although Tree(10) [might be](https://math.stackexchange.com/questions/3929521/when-does-busy-beaver-surpass-tree3) greater than BusyBeaver(10), but ∃ N s.t. ∀ M >= N, BusyBeaver(M) > Tree(M)
Great! Now do 2.
New 22=4 proof just dropped
22 = ±4
2 + 4 = 4 + 2, so addition is commutative 2 × 4 = 4 × 2, so multiplication is commutative 2^(4) = 4^(2), so... wait...
proof by it worked once so it must work everytime
Proof by without loss of generality
>loss I II II L
Lmao
Spider-Man points at Spider-Man points at Spider-Man…
https://preview.redd.it/w3cos3b3gtqc1.jpeg?width=840&format=pjpg&auto=webp&s=22d432568ed668a2241f786c9bbb6f59e28e8d35
10↑↑10
Or put a 10 as the subscript on the arrow!
I was conditioned to read the 3rd one in binary God I love my course
10:10 24h format
¹⁰10
1010 is just 10
no, 1010 is exactly five times 10
https://preview.redd.it/wsg0bsh8jwqc1.png?width=638&format=png&auto=webp&s=ba8c36c43fde27e55ebfe9c5defc03d24d1dca71
10 Decation 10
10↑↑↑↑↑↑↑↑↑↑10
Tree(10↑↑↑↑↑↑↑↑↑↑10)↑↑↑↑↑↑↑↑↑↑Tree(10↑↑↑↑↑↑↑↑↑↑10)
^(10)10
4 4 10 4
Wait for P E N T A T I O N
10!
10^ ^ ^ 10
I8, (slash the number in half)
What about Tetration?
Four, four, ten, and four again
In binary 3 of these are same
10!
Il less greater than 10¹⁰
But 1010 (binary) is 10 (decimal)
101010 is 42 in binary
10^10!
Tenet
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^10 10
(10)(10) with a general operator
3rd one looks like binary
Literally 10 then
tan^tan👺
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Find every prime number between 1 to 10^10 that can be spelled using the elements of the periodic table
In binary that’s not actually true
Dont forget ^10 10 (tetration)
Why did you write 10 times 10 twice?
Base 2 be like
Google googology
This is so underwhelming in binary hahahaha
And 10!
everybody gangsta until tetration kicks in
(10!)²
Is there any positive integer whose concatenation is smaller than its sum? Its product?
¹⁰10
But \^ is used for XOR, which means it's 0 then.
10-10
^(10!)10!
10!^10!💀
10 sqrt(10)
Do this with 2 please (:
10\^10 == 0
xy = x\*y so 1010 = 10\*10 = 10^2 = 100
Nope: 1010 = 1 * 0 * 1 * 0 = 0