213 reputation
110
bio website xcrypt-devblog.blogspot.be
location Belgium
age 22
visits member for 2 years, 9 months
seen Apr 16 at 16:04

CS, math, physics and game development hobbyist


Jul
2
awarded  Curious
Oct
26
accepted How is $\sqrt{\frac{\sqrt{3}+2}{4}} = \frac{\sqrt{2}(\sqrt{3}+1)}{4}$
Oct
26
comment How is $\sqrt{\frac{\sqrt{3}+2}{4}} = \frac{\sqrt{2}(\sqrt{3}+1)}{4}$
How did you get 2xy = 1? I think your approach is interesting but I don't understand.
Oct
26
asked How is $\sqrt{\frac{\sqrt{3}+2}{4}} = \frac{\sqrt{2}(\sqrt{3}+1)}{4}$
Oct
23
comment Can we simplify $\sqrt{a}*\sqrt{a}$ to $a$ when $a \in \mathbb{R}$ and we do not know whether a is positive or negative?
@CarlMummert I understand what you just said. But take a look at my comment on Michael's answer. I understand that in order for the expression to be sane, sqrt(a) has to be a real number. Meaning that a has to be nonnegative. But this also means that you still have to figure out whether a is negative or not, else we could all discard the quadratic formula when working with reals. With sqrt(a)^2, basically the question is can we make the entire expression sane in R by temporarily using the complex roots, because it's the entire expression that matters not the individual parts right?
Oct
23
comment Can we simplify $\sqrt{a}*\sqrt{a}$ to $a$ when $a \in \mathbb{R}$ and we do not know whether a is positive or negative?
And what in case of the quadratic formula? You don't really know if what you are taking the square root from is positive or negative, you only know once it is negative it becomes 'undefined'. So we are not entitled to write the quadratic formula. Am I wrong then? @CarlMummert
Oct
23
comment Can we simplify $\sqrt{a}*\sqrt{a}$ to $a$ when $a \in \mathbb{R}$ and we do not know whether a is positive or negative?
@JpMcCarthy If you're going about it that way then how do you want to make a distinction between all the n n-th roots of a complex number?
Oct
23
comment Can we simplify $\sqrt{a}*\sqrt{a}$ to $a$ when $a \in \mathbb{R}$ and we do not know whether a is positive or negative?
I think the problem here is that the definitions for $\sqrt{a}$ is relative which set you are working in. I believe your definition would be correct in $\mathbb{C}$ but incorrect in $\mathbb{R}$
Oct
23
awarded  Yearling
Oct
23
comment Can we simplify $\sqrt{a}*\sqrt{a}$ to $a$ when $a \in \mathbb{R}$ and we do not know whether a is positive or negative?
This is similar to what I was thinking. We can temporarily use complex numbers to simplify $(\sqrt{a})^2$ to $a$, and the first expression which is partly undefined in R would be simplified to a complete defined real number. I guess I asked the question somewhat wrong and it should be more like "is it allowed to temporarily use another number set when working in a set, to achieve results that will later on simplify to the first set?"
Oct
23
accepted Can we simplify $\sqrt{a}*\sqrt{a}$ to $a$ when $a \in \mathbb{R}$ and we do not know whether a is positive or negative?
Oct
23
comment Can we simplify $\sqrt{a}*\sqrt{a}$ to $a$ when $a \in \mathbb{R}$ and we do not know whether a is positive or negative?
I actually didn't change my question, I just gave extra information.
Oct
23
revised Can we simplify $\sqrt{a}*\sqrt{a}$ to $a$ when $a \in \mathbb{R}$ and we do not know whether a is positive or negative?
added 2 characters in body
Oct
23
asked Can we simplify $\sqrt{a}*\sqrt{a}$ to $a$ when $a \in \mathbb{R}$ and we do not know whether a is positive or negative?
Oct
6
comment What is the meaning of dimensions/units in mathematical equations?
@ZevChonoles So the units always have to be consistent? Let's say a person is connected to some kind of device. The device detects how many meters that person has ran, and for each meter the person has ran the device adds 1 gram of some material to a box. Can we not represent this relation by Mass of box in grams = Distance the person ran in meters?
Oct
6
asked What is the meaning of dimensions/units in mathematical equations?
Feb
5
awarded  Informed
Feb
3
comment insight on lemma on linear independence and spanning set
@atricolf because we take the $a_{ij}$ from the $w_i = \sum_j^n{a_{ij}v_j}$, then we choose the unkowns $c_i$ such that the system of linear equations holds. In the end we don't assume that $\sum_{i=1}^{n+1}c_iw_i = 0$. We just end up with $0$ because of the simplification. This is where I seemed to go wrong for hours.
Feb
3
accepted insight on lemma on linear independence and spanning set
Feb
3
comment insight on lemma on linear independence and spanning set
I think you're assuming $\sum_{i=1}^{n+1}{c_ia_{ij}} = 0$ because we looked at the system of linear equations above. But it never stated $\sum_{i=1}^{n+1}{c_ia_{ij}} = 0$. If we assume that we take exactly those $c_i, a_{ij}$ such that the system holds, it makes a little more sense. Hmm, I think I'm starting to understand now.