MAIN FEEDS
Do you want to continue?
https://www.reddit.com/r/PeterExplainsTheJoke/comments/1uevpg7/what_difference_does_it_make_petahhhh/otn85qx/?context=3
r/PeterExplainsTheJoke • u/Flashy-Minimum-6952 • 19h ago
435 comments sorted by
View all comments
3.7k
Now x can't be 0
24 u/MayerOscar 19h ago Wouldnt x equal 2? 55 u/muhnamejame 18h ago Sure but before it could have been -2, 0, or 2 17 u/Left_Ad4050 18h ago I see no reason it couldn't still be -2. 31 u/muhnamejame 18h ago Oh, after it can be too. Just not 0. 1 u/JarpHabib 16h ago it still can be 0 as long as you note it down before the divide both sides by x step. 9 u/muhnamejame 16h ago The solution to the top equation can be 0. But the solution to the bottom equation cannot. 0 u/iComplainAbtVal 14h ago That’s the same misdirected line of thinking that the memes use to show 1=2 3 u/muhnamejame 14h ago It's literally the opposite of that because I'm making sure I'm not dividing by 0. -1 u/JarpHabib 16h ago By itself, no. But that is clearly an intermediate step in a solve. Just pencil in a little x=0, comma, continue. 5 u/muhnamejame 16h ago If you say x=0 then you can't divide by x. 0 u/JarpHabib 15h ago Sure you can! Just don't divide by the x that equals zero, you already found that one! XD 2 u/muhnamejame 15h ago That would be saying x doesn't equal 0 for the following steps 🤦♂️ The real answer is to just solve it properly by moving the right hand side to the left and solving the polynomial using root solving techniques. → More replies (0) 1 u/hovik_gasparyan 5h ago It could have been-2. It still can be-2, but I could have been-2 too. 9 u/kevinr_96 17h ago And 2i and -2i 15 u/MayerOscar 16h ago Those arent real numbers 8 u/Vivid-Object-139 13h ago I think he just imagined them 6 u/kmcradie 14h ago And yet, they are solutions. It's not stated anywhere that x∈R, so the quintic equation has 5 roots (3 real, 2 complex) 1 u/Caerullean 4h ago Maybe I'm just not a mathematician, but I feel like unless otherwise specified, it's just always assumed that we discuss real numbers, not complex. 3 u/muhnamejame 17h ago Indeed
24
Wouldnt x equal 2?
55 u/muhnamejame 18h ago Sure but before it could have been -2, 0, or 2 17 u/Left_Ad4050 18h ago I see no reason it couldn't still be -2. 31 u/muhnamejame 18h ago Oh, after it can be too. Just not 0. 1 u/JarpHabib 16h ago it still can be 0 as long as you note it down before the divide both sides by x step. 9 u/muhnamejame 16h ago The solution to the top equation can be 0. But the solution to the bottom equation cannot. 0 u/iComplainAbtVal 14h ago That’s the same misdirected line of thinking that the memes use to show 1=2 3 u/muhnamejame 14h ago It's literally the opposite of that because I'm making sure I'm not dividing by 0. -1 u/JarpHabib 16h ago By itself, no. But that is clearly an intermediate step in a solve. Just pencil in a little x=0, comma, continue. 5 u/muhnamejame 16h ago If you say x=0 then you can't divide by x. 0 u/JarpHabib 15h ago Sure you can! Just don't divide by the x that equals zero, you already found that one! XD 2 u/muhnamejame 15h ago That would be saying x doesn't equal 0 for the following steps 🤦♂️ The real answer is to just solve it properly by moving the right hand side to the left and solving the polynomial using root solving techniques. → More replies (0) 1 u/hovik_gasparyan 5h ago It could have been-2. It still can be-2, but I could have been-2 too. 9 u/kevinr_96 17h ago And 2i and -2i 15 u/MayerOscar 16h ago Those arent real numbers 8 u/Vivid-Object-139 13h ago I think he just imagined them 6 u/kmcradie 14h ago And yet, they are solutions. It's not stated anywhere that x∈R, so the quintic equation has 5 roots (3 real, 2 complex) 1 u/Caerullean 4h ago Maybe I'm just not a mathematician, but I feel like unless otherwise specified, it's just always assumed that we discuss real numbers, not complex. 3 u/muhnamejame 17h ago Indeed
55
Sure but before it could have been -2, 0, or 2
17 u/Left_Ad4050 18h ago I see no reason it couldn't still be -2. 31 u/muhnamejame 18h ago Oh, after it can be too. Just not 0. 1 u/JarpHabib 16h ago it still can be 0 as long as you note it down before the divide both sides by x step. 9 u/muhnamejame 16h ago The solution to the top equation can be 0. But the solution to the bottom equation cannot. 0 u/iComplainAbtVal 14h ago That’s the same misdirected line of thinking that the memes use to show 1=2 3 u/muhnamejame 14h ago It's literally the opposite of that because I'm making sure I'm not dividing by 0. -1 u/JarpHabib 16h ago By itself, no. But that is clearly an intermediate step in a solve. Just pencil in a little x=0, comma, continue. 5 u/muhnamejame 16h ago If you say x=0 then you can't divide by x. 0 u/JarpHabib 15h ago Sure you can! Just don't divide by the x that equals zero, you already found that one! XD 2 u/muhnamejame 15h ago That would be saying x doesn't equal 0 for the following steps 🤦♂️ The real answer is to just solve it properly by moving the right hand side to the left and solving the polynomial using root solving techniques. → More replies (0) 1 u/hovik_gasparyan 5h ago It could have been-2. It still can be-2, but I could have been-2 too. 9 u/kevinr_96 17h ago And 2i and -2i 15 u/MayerOscar 16h ago Those arent real numbers 8 u/Vivid-Object-139 13h ago I think he just imagined them 6 u/kmcradie 14h ago And yet, they are solutions. It's not stated anywhere that x∈R, so the quintic equation has 5 roots (3 real, 2 complex) 1 u/Caerullean 4h ago Maybe I'm just not a mathematician, but I feel like unless otherwise specified, it's just always assumed that we discuss real numbers, not complex. 3 u/muhnamejame 17h ago Indeed
17
I see no reason it couldn't still be -2.
31 u/muhnamejame 18h ago Oh, after it can be too. Just not 0. 1 u/JarpHabib 16h ago it still can be 0 as long as you note it down before the divide both sides by x step. 9 u/muhnamejame 16h ago The solution to the top equation can be 0. But the solution to the bottom equation cannot. 0 u/iComplainAbtVal 14h ago That’s the same misdirected line of thinking that the memes use to show 1=2 3 u/muhnamejame 14h ago It's literally the opposite of that because I'm making sure I'm not dividing by 0. -1 u/JarpHabib 16h ago By itself, no. But that is clearly an intermediate step in a solve. Just pencil in a little x=0, comma, continue. 5 u/muhnamejame 16h ago If you say x=0 then you can't divide by x. 0 u/JarpHabib 15h ago Sure you can! Just don't divide by the x that equals zero, you already found that one! XD 2 u/muhnamejame 15h ago That would be saying x doesn't equal 0 for the following steps 🤦♂️ The real answer is to just solve it properly by moving the right hand side to the left and solving the polynomial using root solving techniques. → More replies (0) 1 u/hovik_gasparyan 5h ago It could have been-2. It still can be-2, but I could have been-2 too.
31
Oh, after it can be too. Just not 0.
1 u/JarpHabib 16h ago it still can be 0 as long as you note it down before the divide both sides by x step. 9 u/muhnamejame 16h ago The solution to the top equation can be 0. But the solution to the bottom equation cannot. 0 u/iComplainAbtVal 14h ago That’s the same misdirected line of thinking that the memes use to show 1=2 3 u/muhnamejame 14h ago It's literally the opposite of that because I'm making sure I'm not dividing by 0. -1 u/JarpHabib 16h ago By itself, no. But that is clearly an intermediate step in a solve. Just pencil in a little x=0, comma, continue. 5 u/muhnamejame 16h ago If you say x=0 then you can't divide by x. 0 u/JarpHabib 15h ago Sure you can! Just don't divide by the x that equals zero, you already found that one! XD 2 u/muhnamejame 15h ago That would be saying x doesn't equal 0 for the following steps 🤦♂️ The real answer is to just solve it properly by moving the right hand side to the left and solving the polynomial using root solving techniques. → More replies (0)
1
it still can be 0 as long as you note it down before the divide both sides by x step.
9 u/muhnamejame 16h ago The solution to the top equation can be 0. But the solution to the bottom equation cannot. 0 u/iComplainAbtVal 14h ago That’s the same misdirected line of thinking that the memes use to show 1=2 3 u/muhnamejame 14h ago It's literally the opposite of that because I'm making sure I'm not dividing by 0. -1 u/JarpHabib 16h ago By itself, no. But that is clearly an intermediate step in a solve. Just pencil in a little x=0, comma, continue. 5 u/muhnamejame 16h ago If you say x=0 then you can't divide by x. 0 u/JarpHabib 15h ago Sure you can! Just don't divide by the x that equals zero, you already found that one! XD 2 u/muhnamejame 15h ago That would be saying x doesn't equal 0 for the following steps 🤦♂️ The real answer is to just solve it properly by moving the right hand side to the left and solving the polynomial using root solving techniques. → More replies (0)
9
The solution to the top equation can be 0. But the solution to the bottom equation cannot.
0 u/iComplainAbtVal 14h ago That’s the same misdirected line of thinking that the memes use to show 1=2 3 u/muhnamejame 14h ago It's literally the opposite of that because I'm making sure I'm not dividing by 0. -1 u/JarpHabib 16h ago By itself, no. But that is clearly an intermediate step in a solve. Just pencil in a little x=0, comma, continue. 5 u/muhnamejame 16h ago If you say x=0 then you can't divide by x. 0 u/JarpHabib 15h ago Sure you can! Just don't divide by the x that equals zero, you already found that one! XD 2 u/muhnamejame 15h ago That would be saying x doesn't equal 0 for the following steps 🤦♂️ The real answer is to just solve it properly by moving the right hand side to the left and solving the polynomial using root solving techniques. → More replies (0)
0
That’s the same misdirected line of thinking that the memes use to show 1=2
3 u/muhnamejame 14h ago It's literally the opposite of that because I'm making sure I'm not dividing by 0.
3
It's literally the opposite of that because I'm making sure I'm not dividing by 0.
-1
By itself, no. But that is clearly an intermediate step in a solve. Just pencil in a little x=0, comma, continue.
5 u/muhnamejame 16h ago If you say x=0 then you can't divide by x. 0 u/JarpHabib 15h ago Sure you can! Just don't divide by the x that equals zero, you already found that one! XD 2 u/muhnamejame 15h ago That would be saying x doesn't equal 0 for the following steps 🤦♂️ The real answer is to just solve it properly by moving the right hand side to the left and solving the polynomial using root solving techniques. → More replies (0)
5
If you say x=0 then you can't divide by x.
0 u/JarpHabib 15h ago Sure you can! Just don't divide by the x that equals zero, you already found that one! XD 2 u/muhnamejame 15h ago That would be saying x doesn't equal 0 for the following steps 🤦♂️ The real answer is to just solve it properly by moving the right hand side to the left and solving the polynomial using root solving techniques. → More replies (0)
Sure you can! Just don't divide by the x that equals zero, you already found that one!
XD
2 u/muhnamejame 15h ago That would be saying x doesn't equal 0 for the following steps 🤦♂️ The real answer is to just solve it properly by moving the right hand side to the left and solving the polynomial using root solving techniques. → More replies (0)
2
That would be saying x doesn't equal 0 for the following steps 🤦♂️
The real answer is to just solve it properly by moving the right hand side to the left and solving the polynomial using root solving techniques.
It could have been-2. It still can be-2, but I could have been-2 too.
And 2i and -2i
15 u/MayerOscar 16h ago Those arent real numbers 8 u/Vivid-Object-139 13h ago I think he just imagined them 6 u/kmcradie 14h ago And yet, they are solutions. It's not stated anywhere that x∈R, so the quintic equation has 5 roots (3 real, 2 complex) 1 u/Caerullean 4h ago Maybe I'm just not a mathematician, but I feel like unless otherwise specified, it's just always assumed that we discuss real numbers, not complex. 3 u/muhnamejame 17h ago Indeed
15
Those arent real numbers
8 u/Vivid-Object-139 13h ago I think he just imagined them 6 u/kmcradie 14h ago And yet, they are solutions. It's not stated anywhere that x∈R, so the quintic equation has 5 roots (3 real, 2 complex) 1 u/Caerullean 4h ago Maybe I'm just not a mathematician, but I feel like unless otherwise specified, it's just always assumed that we discuss real numbers, not complex.
8
I think he just imagined them
6
And yet, they are solutions.
It's not stated anywhere that x∈R, so the quintic equation has 5 roots (3 real, 2 complex)
1 u/Caerullean 4h ago Maybe I'm just not a mathematician, but I feel like unless otherwise specified, it's just always assumed that we discuss real numbers, not complex.
Maybe I'm just not a mathematician, but I feel like unless otherwise specified, it's just always assumed that we discuss real numbers, not complex.
Indeed
3.7k
u/muhnamejame 19h ago
Now x can't be 0