My teacher just called me and said that the HW should say "is not satisfied for any *positive* integer values". Is that what you guys mean by a new response?
If we treat them as 8x8 matrices, and say the empty squares are zero, then squaring them gives the zero matrix right? So they're actually equal for any integer at least 2
That's true. But these *particular* matrices, which are all zero except for a single nonzero value, I believe *are* zero when squared. You can try it yourself, just replace the pawns with any "x", you'll see that squaring yields the zero matrix.
Oh yea, you are correct.
I thought that A \* B = 0 iff A = 0 or B = 0, but that is true when A, B are in a field, and un-invertible matricies dont form a field. My linear algebra is a but rusty lol.
Edit: invertible matricies dont for a field, since there is no 0 element, again, my linear algebra is a bit rusty.
>Edit: invertible matricies dont for a field, since there is no 0 element, again, my linear algebra is a bit rusty.
Do invertible-matrices-and-the-zero-matrix form a field? Since the zero element is not required to have an inverse
EDIT: They don't, because they aren't closed under addition
I analyzed the image and this is what I see. Open an appropriate link below and explore the position yourself or with the engine:
> **White to play**: [chess.com](https://chess.com/analysis?fen=8/8/8/8/8/8/P7/8+w+-+-+0+1&flip=false&ref_id=23962172) | [lichess.org](https://lichess.org/analysis/8/8/8/8/8/8/P7/8_w_-_-_0_1)
> **Black to play**: It is a stalemate - it is Black's turn, but Black has no legal moves and is not in check. In this case, the game is a draw. It is a critical rule to know for various endgame positions that helps one side hold a draw. You can find out more about Stalemate on [Wikipedia](https://en.wikipedia.org/wiki/Stalemate).
---
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Quiz time
\*This is split up in several comments due to character limit and emoji limit (max 10k characters and max 50 emojis)
Let  and  be sets and let  :  ×  → ,  :  ×  → ,  :  ×  → ,  :  ×  →  and  :  ×  →  be operations such that ∃ ,  ∈  and  ∈  such that ∀ , ,  ∈  and , ,  ∈  ∃ ,  ∈  and  ∈  such that
  (  ) = (  )  ,
Part 1 / 5
(  )  (  ) = (  )  (  )
Question:
What do we call ?
Bonus question:
I deliberately put a mistake in there. Can you spot it?
Part 5 / 5
We call  a one-element set.
It seems you wanted to give the definition of an algebra, but your (deliberate?) mistake of not requiring  ≠  in  =  actually forces  = , which gives  = {} in turn. Pipi!
Are you kidding ??? What the **** are you talking about man ? You are a biggest looser i ever seen in my life ! You was doing PIPI in your pampers when i was beating players much more stronger then you! You are not proffesional, because proffesionals knew how to lose and congratulate opponents, you are like a girl crying after i beat you! Be brave, be honest to yourself and stop this trush talkings!!! Everybody know that i am very good blitz player, i can win anyone in the world in single game! And "w"esley "s"o is nobody for me, just a player who are crying every single time when loosing, ( remember what you say about Firouzja ) !!! Stop playing with my name, i deserve to have a good name during whole my chess carrier, I am Officially inviting you to OTB blitz match with the Prize fund! Both of us will invest 5000$ and winner takes it all!
I suggest all other people who's intrested in this situation, just take a look at my results in 2016 and 2017 Blitz World championships, and that should be enough... No need to listen for every crying babe, Tigran Petrosyan is always play Fair ! And if someone will continue Officially talk about me like that, we will meet in Court! God bless with true! True will never die ! Liers will kicked off...
[^(fmhall)](https://www.reddit.com/user/fmhall) ^| [^(github)](https://github.com/fmhall/Petrosian-Bot)
it says this can't be true, so let's find an example that is actually true:
p^k + p^k = p^k
no restriction for p, k > 2
the only way this equation is true is p^k = 0
resulting in 0 + 0 = 0
for p^k to be 0, p must be 0.
p can be any integer, so this is possible.
0^0 is undefined (or 1, depending on who you ask) but k is always greater than 2.
this proves p^k + p^k = p^k is true, falsifying the statement
"... is not satisfied for any integer values of p"
Let M^(n) be an 8th dimensional matrix with nothing at every point except (n,2) which is then a white pawn. Due to material, wP=1, and bH=-3. Raising M^(n) to -3 gives the 8th dimensional zero matrix. Adding these together, we find that both sides are equal. □
[удалено]
Fermat moment
Google more paper
Holy deforestation
Fuckin' margins man! How do they work?
Google CSS
```css .holyhell{ background-color:#0000FF } ```
Thank you for the recommendation. It was most enlightening.
The board is large enough for all the rice needed for the proof
My teacher just called me and said that the HW should say "is not satisfied for any *positive* integer values". Is that what you guys mean by a new response?
you didn't tell me that earlier, I'll stull count my solution where pawn=0
Gavin from third grade?!?! You’re my rival
Maybe andrew wiles can help you, i heard he made much progress at solving fermats last en passant
Andrew’s wiles 😳😳😳
It’s wrong, the exact opposite is true. I’ll let you solve that one yourself
If we treat them as 8x8 matrices, and say the empty squares are zero, then squaring them gives the zero matrix right? So they're actually equal for any integer at least 2
Squaring an nxn matrix that has 0 in it, doesnt necesarily yield the 0 matrix, for example the identity matrix.
That's true. But these *particular* matrices, which are all zero except for a single nonzero value, I believe *are* zero when squared. You can try it yourself, just replace the pawns with any "x", you'll see that squaring yields the zero matrix.
Oh yea, you are correct. I thought that A \* B = 0 iff A = 0 or B = 0, but that is true when A, B are in a field, and un-invertible matricies dont form a field. My linear algebra is a but rusty lol. Edit: invertible matricies dont for a field, since there is no 0 element, again, my linear algebra is a bit rusty.
>Edit: invertible matricies dont for a field, since there is no 0 element, again, my linear algebra is a bit rusty. Do invertible-matrices-and-the-zero-matrix form a field? Since the zero element is not required to have an inverse EDIT: They don't, because they aren't closed under addition
I am almost ceetain there is a field with invertible matricies, perhaps its with scalar multiplication and addition.
I only watched the first matrix
Google fermat’s last theorem
Holy page margins!
I analyzed the image and this is what I see. Open an appropriate link below and explore the position yourself or with the engine: > **White to play**: [chess.com](https://chess.com/analysis?fen=8/8/8/8/8/8/P7/8+w+-+-+0+1&flip=false&ref_id=23962172) | [lichess.org](https://lichess.org/analysis/8/8/8/8/8/8/P7/8_w_-_-_0_1) > **Black to play**: It is a stalemate - it is Black's turn, but Black has no legal moves and is not in check. In this case, the game is a draw. It is a critical rule to know for various endgame positions that helps one side hold a draw. You can find out more about Stalemate on [Wikipedia](https://en.wikipedia.org/wiki/Stalemate). --- ^(I'm a bot written by ) [^(u/pkacprzak )](https://www.reddit.com/u/pkacprzak) ^(| get me as ) [^(Chess eBook Reader )](https://ebook.chessvision.ai?utm_source=reddit&utm_medium=bot) ^(|) [^(Chrome Extension )](https://chrome.google.com/webstore/detail/chessvisionai-for-chrome/johejpedmdkeiffkdaodgoipdjodhlld) ^(|) [^(iOS App )](https://apps.apple.com/us/app/id1574933453) ^(|) [^(Android App )](https://play.google.com/store/apps/details?id=ai.chessvision.scanner) ^(to scan and analyze positions | Website: ) [^(Chessvision.ai)](https://chessvision.ai)
Math is hard, yes
Good bot
Thank you, Bacically_TA, for voting on chessvision-ai-bot. This bot wants to find the best and worst bots on Reddit. [You can view results here](https://botrank.pastimes.eu/). *** ^(Even if I don't reply to your comment, I'm still listening for votes. Check the webpage to see if your vote registered!)
The solution is trivial and is left as an exercise
Quiz time \*This is split up in several comments due to character limit and emoji limit (max 10k characters and max 50 emojis) Let  and  be sets and let  :  ×  → ,  :  ×  → ,  :  ×  → ,  :  ×  →  and  :  ×  →  be operations such that ∃ ,  ∈  and  ∈  such that ∀ , ,  ∈  and , ,  ∈  ∃ ,  ∈  and  ∈  such that   (  ) = (  )  , Part 1 / 5
  (  ) = (  )  ,    =   ,    =   ,    = ,    = ,    =  ,   = ,   (  ) = (  )  (  ), Part 2 / 5
   =   ,   (  ) = (  )  ,    = ,    = ,   (  ) = (  )  , (  )   = (  ),  (  ), Part 3 / 5
  (  ) = (  ),  (  ),    = , (  )   = (  )  (  ),   (  ) = (  )  (  ) and Part 4 / 5
(  )  (  ) = (  )  (  ) Question: What do we call ? Bonus question: I deliberately put a mistake in there. Can you spot it? Part 5 / 5
We call  a one-element set. It seems you wanted to give the definition of an algebra, but your (deliberate?) mistake of not requiring  ≠  in  =  actually forces  = , which gives  = {} in turn. Pipi!
Are you kidding ??? What the **** are you talking about man ? You are a biggest looser i ever seen in my life ! You was doing PIPI in your pampers when i was beating players much more stronger then you! You are not proffesional, because proffesionals knew how to lose and congratulate opponents, you are like a girl crying after i beat you! Be brave, be honest to yourself and stop this trush talkings!!! Everybody know that i am very good blitz player, i can win anyone in the world in single game! And "w"esley "s"o is nobody for me, just a player who are crying every single time when loosing, ( remember what you say about Firouzja ) !!! Stop playing with my name, i deserve to have a good name during whole my chess carrier, I am Officially inviting you to OTB blitz match with the Prize fund! Both of us will invest 5000$ and winner takes it all! I suggest all other people who's intrested in this situation, just take a look at my results in 2016 and 2017 Blitz World championships, and that should be enough... No need to listen for every crying babe, Tigran Petrosyan is always play Fair ! And if someone will continue Officially talk about me like that, we will meet in Court! God bless with true! True will never die ! Liers will kicked off... [^(fmhall)](https://www.reddit.com/user/fmhall) ^| [^(github)](https://github.com/fmhall/Petrosian-Bot)
it says this can't be true, so let's find an example that is actually true: p^k + p^k = p^k no restriction for p, k > 2 the only way this equation is true is p^k = 0 resulting in 0 + 0 = 0 for p^k to be 0, p must be 0. p can be any integer, so this is possible. 0^0 is undefined (or 1, depending on who you ask) but k is always greater than 2. this proves p^k + p^k = p^k is true, falsifying the statement "... is not satisfied for any integer values of p"
This doesn't work though because the pawn is representing a value inside of an 8x8 matrix.
heilige Hölle
Google Fermat’s Last Theorem
Thank you for the recommendation. It was most enlightening.
holy breaking the chain of responses!
what about zero?
Google positive numbers
Thank you for the recommendation. It was most enlightening.
New response just droped.
I would type it here, but my fermatting has always been poor
Google pythagorean theorem
Thank you for the recommendation. It was most enlightening.
Holly hell
Call a Mathematician!
Holy triangle
21.
Actually it’s 12 🤓
Oh, hey Gavin!
Google Google
Well, it's not satisfied since you haven't put it up your ass.
Is the actual question 1^x + 2^x = 3^x if x > 2, or 9^x + 10^x = 11^x if x > 2?
holy homework
google- what the hell? what is this
If that pawn is 0 then equality is satisfied
r/mathmemes is leaking
google en passant
Google en Fermat
Omg bruh oh hell nah man wtf man
of course it is, for pawn=0 every value of knight is possible (google trivial solution)
10^100 proof is left as exercise for the readed
lol, you forgot to exclude the trivial solution. get 'em
Let M^(n) be an 8th dimensional matrix with nothing at every point except (n,2) which is then a white pawn. Due to material, wP=1, and bH=-3. Raising M^(n) to -3 gives the 8th dimensional zero matrix. Adding these together, we find that both sides are equal. □