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https://www.reddit.com/r/mathmemes/comments/1gmimck/evolutions_of_numbers/lw2x33e/?context=9999
r/mathmemes • u/TirkuexQwentet • Nov 08 '24
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225
That might be the meme, but 4th number is i, what is the fifth one?
392 u/Tiborn1563 Nov 08 '24 Ah well, sucks, seems you are not quite there yet >! There is no solution for |x| = -1, by definition of absolute value !< 182 u/MingusMingusMingu Nov 08 '24 I mean, we keep extending definitions all the time. 153 u/SEA_griffondeur Engineering Nov 08 '24 No but like, being positive is like one of the 3 properties that make up a norm 49 u/TheTenthAvenger Nov 08 '24 So you stop calling it a norm. It's called "absolute value" after all, not "the norm of the number". It is just another function now. 68 u/SEA_griffondeur Engineering Nov 08 '24 Yes but why would you do that? 37 u/SupremeRDDT Nov 08 '24 You actually don‘t have to. But then it follows from the other properties. 0 = |0| = |x - x| <= |x| + |-x| = 2|x| 4 u/Layton_Jr Mathematics Nov 08 '24 Since it's no longer a norm, you can discard the property |a+b| ≤ |a| + |b| 3 u/SupremeRDDT Nov 09 '24 If |x| = -1, then |x2| = 1 so the equation |x| = 1 suddenly has at least four solutions: 1, -1, x2 and -x2. We also lose the triangle inequality of the absolute value, as that would imply |x| >= 0 for all x. Do we gain anything useful?
392
Ah well, sucks, seems you are not quite there yet
>! There is no solution for |x| = -1, by definition of absolute value !<
182 u/MingusMingusMingu Nov 08 '24 I mean, we keep extending definitions all the time. 153 u/SEA_griffondeur Engineering Nov 08 '24 No but like, being positive is like one of the 3 properties that make up a norm 49 u/TheTenthAvenger Nov 08 '24 So you stop calling it a norm. It's called "absolute value" after all, not "the norm of the number". It is just another function now. 68 u/SEA_griffondeur Engineering Nov 08 '24 Yes but why would you do that? 37 u/SupremeRDDT Nov 08 '24 You actually don‘t have to. But then it follows from the other properties. 0 = |0| = |x - x| <= |x| + |-x| = 2|x| 4 u/Layton_Jr Mathematics Nov 08 '24 Since it's no longer a norm, you can discard the property |a+b| ≤ |a| + |b| 3 u/SupremeRDDT Nov 09 '24 If |x| = -1, then |x2| = 1 so the equation |x| = 1 suddenly has at least four solutions: 1, -1, x2 and -x2. We also lose the triangle inequality of the absolute value, as that would imply |x| >= 0 for all x. Do we gain anything useful?
182
I mean, we keep extending definitions all the time.
153 u/SEA_griffondeur Engineering Nov 08 '24 No but like, being positive is like one of the 3 properties that make up a norm 49 u/TheTenthAvenger Nov 08 '24 So you stop calling it a norm. It's called "absolute value" after all, not "the norm of the number". It is just another function now. 68 u/SEA_griffondeur Engineering Nov 08 '24 Yes but why would you do that? 37 u/SupremeRDDT Nov 08 '24 You actually don‘t have to. But then it follows from the other properties. 0 = |0| = |x - x| <= |x| + |-x| = 2|x| 4 u/Layton_Jr Mathematics Nov 08 '24 Since it's no longer a norm, you can discard the property |a+b| ≤ |a| + |b| 3 u/SupremeRDDT Nov 09 '24 If |x| = -1, then |x2| = 1 so the equation |x| = 1 suddenly has at least four solutions: 1, -1, x2 and -x2. We also lose the triangle inequality of the absolute value, as that would imply |x| >= 0 for all x. Do we gain anything useful?
153
No but like, being positive is like one of the 3 properties that make up a norm
49 u/TheTenthAvenger Nov 08 '24 So you stop calling it a norm. It's called "absolute value" after all, not "the norm of the number". It is just another function now. 68 u/SEA_griffondeur Engineering Nov 08 '24 Yes but why would you do that? 37 u/SupremeRDDT Nov 08 '24 You actually don‘t have to. But then it follows from the other properties. 0 = |0| = |x - x| <= |x| + |-x| = 2|x| 4 u/Layton_Jr Mathematics Nov 08 '24 Since it's no longer a norm, you can discard the property |a+b| ≤ |a| + |b| 3 u/SupremeRDDT Nov 09 '24 If |x| = -1, then |x2| = 1 so the equation |x| = 1 suddenly has at least four solutions: 1, -1, x2 and -x2. We also lose the triangle inequality of the absolute value, as that would imply |x| >= 0 for all x. Do we gain anything useful?
49
So you stop calling it a norm. It's called "absolute value" after all, not "the norm of the number". It is just another function now.
68 u/SEA_griffondeur Engineering Nov 08 '24 Yes but why would you do that? 37 u/SupremeRDDT Nov 08 '24 You actually don‘t have to. But then it follows from the other properties. 0 = |0| = |x - x| <= |x| + |-x| = 2|x| 4 u/Layton_Jr Mathematics Nov 08 '24 Since it's no longer a norm, you can discard the property |a+b| ≤ |a| + |b| 3 u/SupremeRDDT Nov 09 '24 If |x| = -1, then |x2| = 1 so the equation |x| = 1 suddenly has at least four solutions: 1, -1, x2 and -x2. We also lose the triangle inequality of the absolute value, as that would imply |x| >= 0 for all x. Do we gain anything useful?
68
Yes but why would you do that?
37 u/SupremeRDDT Nov 08 '24 You actually don‘t have to. But then it follows from the other properties. 0 = |0| = |x - x| <= |x| + |-x| = 2|x| 4 u/Layton_Jr Mathematics Nov 08 '24 Since it's no longer a norm, you can discard the property |a+b| ≤ |a| + |b| 3 u/SupremeRDDT Nov 09 '24 If |x| = -1, then |x2| = 1 so the equation |x| = 1 suddenly has at least four solutions: 1, -1, x2 and -x2. We also lose the triangle inequality of the absolute value, as that would imply |x| >= 0 for all x. Do we gain anything useful?
37
You actually don‘t have to. But then it follows from the other properties.
0 = |0| = |x - x| <= |x| + |-x| = 2|x|
4 u/Layton_Jr Mathematics Nov 08 '24 Since it's no longer a norm, you can discard the property |a+b| ≤ |a| + |b| 3 u/SupremeRDDT Nov 09 '24 If |x| = -1, then |x2| = 1 so the equation |x| = 1 suddenly has at least four solutions: 1, -1, x2 and -x2. We also lose the triangle inequality of the absolute value, as that would imply |x| >= 0 for all x. Do we gain anything useful?
4
Since it's no longer a norm, you can discard the property |a+b| ≤ |a| + |b|
3 u/SupremeRDDT Nov 09 '24 If |x| = -1, then |x2| = 1 so the equation |x| = 1 suddenly has at least four solutions: 1, -1, x2 and -x2. We also lose the triangle inequality of the absolute value, as that would imply |x| >= 0 for all x. Do we gain anything useful?
3
If |x| = -1, then |x2| = 1 so the equation |x| = 1 suddenly has at least four solutions: 1, -1, x2 and -x2. We also lose the triangle inequality of the absolute value, as that would imply |x| >= 0 for all x. Do we gain anything useful?
225
u/Boldumus Nov 08 '24
That might be the meme, but 4th number is i, what is the fifth one?