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Jun
21
comment How exactly do we do Gauss elimination?
After Gauss elimination, you get a matrix in reduced row-echelon form. The way I think of it is: The upper left corner is an identity matrix, the upper right is whatever, and the bottom rows are zeroes: $\left[ \begin{smallmatrix} I & F \\ 0 & 0 \end{smallmatrix} \right]$. From there, you can permute the columns if you want.
Jun
21
comment How exactly do we do Gauss elimination?
See how the bottom row is all zeros? This means we know nothing about $z$ at all. So we'll just set it to a "free parameter", $t$, and allow that to vary. In general, you'll get one parameter per zeroed-out-row (after row-reduction, of course).
Jun
21
comment Is Σ* the same as saying L*
Usually, $\Sigma$ refers to the alphabet that you're working over, and $\ast$ means "all finite strings over". In other words, $\Sigma^\ast$ is the set of all strings in your universe. On the other hand, $L^\ast$ has a very different meaning: all strings that can be made by concatenating words in $L$.
Jun
15
comment What is the limit of the ratio of the sum of all real numbers from 0 to 2 over the sum of all real numbers from 0 to 1.
You'll get the same answer: 4. The integral from $0$ to $2$ is $2$, and the integral from $0$ to $1$ is $\frac{1}{2}$.
Jun
3
comment Is the zero vector in the definition of linear dependence arbritary?
Ah, I think I see what you were aiming at. If there's two different ways to make a vector from linear combinations, then the set is dependent. The proof of this is easy: subtract the two combinations, and you'll get a non trivial combination that gives the zero vector. :)
Jun
3
comment Is the zero vector in the definition of linear dependence arbritary?
No, if you change the definition like that, you don't get anything useful. See my edit. Lemme know if you're still stuck.
Jun
3
revised Is the zero vector in the definition of linear dependence arbritary?
added 551 characters in body
Jun
3
comment Is the zero vector in the definition of linear dependence arbritary?
It's independent because there's no linear combination of $\{ 0 \}$ that sums to $v_0$, because $\lambda \cdot 0 = 0 \ne v_0$. As for the second part, my point was: $a$, $b$, and all their linear combinations are independent, and that's not the kind of result you want. I don't think I understand your second question. Is that not the re-definition of "independent" that you wanted?
Jun
3
comment Show $[K : F] = [K : E][E : F]$.
Since $[K : E] = m$, then $K$ is a vector space over $E$ with dimension $m$. So there's some $E$-basis for $K$ with $m$ elements. Similarly, there's an $F$-basis for $E$, with $n$ elements. Can you construct a $F$-basis for $K$? How many elements should it have?
Jun
3
answered Is the zero vector in the definition of linear dependence arbritary?
May
5
awarded  Notable Question
Apr
18
comment Fast method to pick unique random numbers?
If you're drawing a large number of cards, shuffling isn't any slower than drawing randomly, because once you've shuffled, you just need to iterate through the array. You'll still need to make $n$ random picks, but now they're all at the beginning, instead of spaced out.
Apr
1
answered Are all functions that have an inverse bijective functions?
Mar
25
awarded  Enlightened
Mar
20
answered explaining $|a+b|≤|a|+|b|$ in simple terms
Mar
10
comment Paradox: Is $1 \in (0,1)$?
Nothing is wrong about representing them as decimals. You have the clause "they are not all zero (or else $x = 0$)." There is a similar clause you should also add: "they are not all nine (or else $x = 1$)".
Mar
8
comment Given two potatoes, prove that there is a loop of wire which fits around both
Unfortunately, my geometric proofs are a lot worse than my geometric intuition. How would one prove that? You'd have to somehow avoid the case where two flat sides of the potatoes are flush against each other, which seems hard.
Mar
8
awarded  Nice Answer
Mar
8
answered Given two potatoes, prove that there is a loop of wire which fits around both
Mar
6
awarded  Nice Answer