Reputation
1,768
Top tag
Next privilege 2,000 Rep.
Edit questions and answers
Badges
1 10 28
Impact
~90k people reached

May
29
accepted Isolating $x$ from $y = (x-1)^2$
May
29
comment Isolating $x$ from $y = (x-1)^2$
@HenningMakholm: If I had $\int_{0}^{1}(x)dy$, I would need to isolate the $x$ from my $y = (x-1)^2$ - if I did that, $x$ could be either $-\sqrt{y}+1$ or $\sqrt{y}+1$ and I'd have to do both cases?
May
29
comment Isolating $x$ from $y = (x-1)^2$
Thanks, I ask because if I have $\int_{0}^{1}(-5y+1)$, I would have to do it twice (one in which $y$ is positive and another when it is negative), right?
May
29
asked Isolating $x$ from $y = (x-1)^2$
May
28
comment If two planes in $\mathbb{R}^3$ pass by the origin, do they necessarily intersect at multiple points?
@user3001408: Nice, my entire life was a lie.
May
28
accepted If two planes in $\mathbb{R}^3$ pass by the origin, do they necessarily intersect at multiple points?
May
28
asked If two planes in $\mathbb{R}^3$ pass by the origin, do they necessarily intersect at multiple points?
May
28
accepted Calculating a basis in $\mathbb{R}^4$.
May
28
asked Calculating a basis in $\mathbb{R}^4$.
May
28
comment Calculating a basis given some constraints.
@David: When I multiplied by $s$, the third term should've been $1$, not the fourth. Ack....... By the way, for this specific method, my basis should always be identical to the solution in the book, right? Or can they differ?
May
28
asked Calculating a basis given some constraints.
May
28
accepted Does linear dependency have anything to do when determining a span?
May
28
comment Does linear dependency have anything to do when determining a span?
Thanks! Yeah I couldn't find the right word. What would you have said?
May
28
comment Does linear dependency have anything to do when determining a span?
Ok so, the dimension of $\mathbb{R}^2$ is 2, therefore its basis must have $2$ vectors. Since a basis contains the minimum number of vectors needed to span $\mathbb{R}^2$, any set of vectors that supposedly span $\mathbb{R}^2$ must have, as a minimum, 2 vectors. Then $\{(1,1),(2,2)\}$ can't be a span because it has only one significant vector, whereas it needs at least 2, right? So a set of 3 vectors can span $\mathbb{R}^2$, as long as two of its vectors are linearly independent, yes?
May
28
asked Does linear dependency have anything to do when determining a span?
May
27
asked Proving that a subset is a subspace by showing a scalar combination.
May
27
accepted When proving if a subset is a subspace, can I prove closure under addition and multiplication in a single proof?
May
27
comment When proving if a subset is a subspace, can I prove closure under addition and multiplication in a single proof?
Reckon this would apply to any problem of this kind, or only certain ones?
May
27
asked When proving if a subset is a subspace, can I prove closure under addition and multiplication in a single proof?
May
27
accepted Proving $||\vec{a}+\vec{b}|| = ||\vec{a}-\vec{b}|| \iff \vec{a} \perp \vec{b}$