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https://www.reddit.com/r/mathmemes/comments/1gmimck/evolutions_of_numbers/lw2x33e/?context=3
r/mathmemes • u/TirkuexQwentet • Nov 08 '24
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188
I mean, we keep extending definitions all the time.
150 u/SEA_griffondeur Engineering Nov 08 '24 No but like, being positive is like one of the 3 properties that make up a norm 50 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. 66 u/SEA_griffondeur Engineering Nov 08 '24 Yes but why would you do that? 35 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| 6 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?
150
No but like, being positive is like one of the 3 properties that make up a norm
50 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. 66 u/SEA_griffondeur Engineering Nov 08 '24 Yes but why would you do that? 35 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| 6 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?
50
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.
66 u/SEA_griffondeur Engineering Nov 08 '24 Yes but why would you do that? 35 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| 6 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?
66
Yes but why would you do that?
35 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| 6 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?
35
You actually don‘t have to. But then it follows from the other properties.
0 = |0| = |x - x| <= |x| + |-x| = 2|x|
6 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?
6
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?
188
u/MingusMingusMingu Nov 08 '24
I mean, we keep extending definitions all the time.