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Suppose I want to prove that the function $f: (0, \infty) \to (0, \infty)$ defined by $f(x) = x^2$ is bijective. Let $a, b \in (0, \infty)$ and $f(a) = f(b)$. Then $a^2 = b^2 \implies a = b$ since everything is non-negative we can simply take square roots. Therefore $f$ is injective. To prove...

Find ${5^{2009}}^{1492}\mod{503}.$ How do you calculate a beast like this?

Wonderful explanations, thanks!

Could someone please explain how they're getting the answers in the table, for example $g = (123)$.

Thanks. I wonder whether there's a systematic way of working this out if one has to find $g^i$ for all $2 \le i \le 27$?

Say an element $g$ in a group has order $28$. How do I find the order of say $g^8$?

You've to put it in the augmented matrix form then row reduce it (not to echelon form necessarily). $\begin{aligned} \begin{pmatrix}\begin{array}{rrr|r} 5&6&7&6 \\ -7 & -4 & 1 & 30 \\ -4 & 4 &16 &h \end{array}\end{pmatrix} & \xrightarrow{R_1 \to R_1+R_2}\begin{pmatrix}\begin{array}{rrr|r}...

For $(v)$ we have $ (\mathbf{a}-\mathbf{b}) \cdot \mathbf{w}(t) = 0 \implies <6,-6,-6><2, t, 1> = 0\implies 6t = 6 \implies t = 1$, so the answer to (v) is $(C)$. For $(iv)$ let $P = (a, b, c)$ now $|PA| = 2|PB| \iff [(4,1,2)-(a, b, c)] = 2[(-2,7,4) -(a, b, c)] \iff (a, b, c) = (-8, 13, 10).$...

For (iii) we have $\mathbf{a} \times \mathbf{b} = (18, -12, 30) \implies ||\mathbf{a} \times \mathbf{b}|| = \sqrt{1368} \implies \frac{1}{2}||\mathbf{a} \times \mathbf{b}|| = \sqrt{342} \implies \triangle ^2 = 342$. The reason being the area of the triangle is half the area of the parallelogram...

Thank you. The midpoint between the two points will be a point on the plane. This point is $\left(\frac{4-2}{2}, \frac{1+7}{2}, \frac{4-2}{2} \right) = (1,4,1)$. The direction vector between the two points, which is also a normal to the plane, is $(-2-4, 7-1, 4+6) = (-6, 6, 6)$ The general...

I would appreciate any help with this questions because I truly horrid at geometry questions. I've only done (i) to which I've found the answer to be (E). I can't do from from part (ii) on.

I'm trying to answer the question below in the attachment. Could someone please check my answer to part (c), as I'm not sure. Is it correct? Is that that the right geometric explanation for the planes? (a) The equation is $\begin{pmatrix}1 & -1& 2& 1 \\ 2 & 1 & 1 & -1 \\ 1 & -2 & \lambda & 3...

Woah, I've solved it I think (Happy) Since $v_1 = (1,0,-1)$ and $v_2 = (2,1,3)$ span the plane, we can write any point in the plane as: $\begin{align*} \begin{pmatrix}x\\y\\z\end{pmatrix}=tv_1+rv_2=t\begin{pmatrix}1\\0\\-1\end{pmatrix}+r\begin{pmatrix}2\\1\\3\end{pmatrix}...

So $(1,2,9) = (a, 0,-a)+(2b, b, 3b) = (a+2b, b, 3b-a) \implies b = 2 \implies a = -3. $

I was actually trying to get the system of equations in the cases, i.e. the one I was supposed solve (I completely misread the problem)! (Rofl) Thank you. I'll study your post.