Olinguito

If [latex]a[/latex] is a repeated root of a polynomial [latex]p(x)[/latex] then [latex]p(x)=(x-a)^kq(x)[/latex] for some polynomial [latex]q(x)[/latex] and integer [latex]k\geqslant2[/latex]. Thus [latex]p(x)=(x-a)r(x)[/latex] where [latex]r(x)=(x-a)^{k-1}q(x)[/latex] and both [latex]x-a[/latex] and [latex]r(x)[/latex] are not units.

It depends of what your set of axioms is for. For example, your axioms might be for the real numbers as an ordered field. An field [latex]\langle F,+,\cdot\rangle[/latex] is said to be (totally) ordered by [latex]\leq[/latex] iff the following holds: [latex]\forall\,a,b,c\in F[/latex]: [latex]\text{either}\ a\leq b\ \text{or}\ b\leq a[/latex] [latex]a\leq b\ \text{and}\ b\leq a\ \Rightarrow\ a=b[/latex] [latex]a\leq b\ \text{and}\ b\leq c\ \Rightarrow\ a\leq c[/latex] [latex]a\leq b\ \Rightarrow\ a+c\leq b+c[/latex] [latex]0\leq a\ \text{and}\ 0\leq b\ \Rightarrow\ 0\leq ab[/latex] We first show that [latex]0\leq 1[/latex]. By axiom (1) either [latex]0\leq 1[/latex] or [latex]1\leq 0[/latex]. If [latex]1\leq 0[/latex] then [latex]1+(-1)\leq 0+(-1)[/latex] (axiom (4)) i.e. [latex]0\leq -1[/latex]. Then [latex]0\leq -1[/latex] and [latex]0\leq -1[/latex] imply (axiom (5)) [latex]0\leq (-1)(-1)=1[/latex]. (This is actually a contradiction because [latex]1\leq0[/latex] and [latex]0\leq 1[/latex] imply (axiom (2)) [latex]0=1[/latex].) So we can’t have [latex]1\leq 0[/latex]; hence we must have [latex]0\leq 1[/latex]. And now we are done, for [latex]0\leq 1[/latex] [latex]\implies[/latex] (axiom (4)) [latex]1=0+1\leq 1+1=2[/latex]. http://en.wikipedia.org/wiki/Ordered_field

This is an example of Pell’s equation.

That’s the assertion of Murphy’s law, which is clearly not true.

If something can go wrong, it means that the probability of it going wrong is greater than zero. If something will go wrong, it means that the probability of it going wrong is equal to one. [latex]\mathrm{P(go\ wrong)>0\ \not\Rightarrow\ P(go\ wrong)=1}[/latex].

It’s called aposiopesis.

[latex]\frac{(1+3)!}{1!+3!}=\frac{24}7\notin\mathbb N[/latex].

Suspend them and see which one always points north–south.

IMHO this is better: “I would be eating my lunch if I were not typing this response.” Example of a would have been + -ing construction: “If I had carried on waiting, I would have been waiting for two hours by the time she turned up.”

1 hour. The speed of the boat relative to the hat does not depend on the current.

For each [latex]x\in X[/latex], let [latex]x_1=x[/latex]. Then let [latex]x_2=f(1,x_1)[/latex]. Then let [latex]x_3=f(2,x_2)[/latex], [latex]x_4=f(3,x_3)[/latex], [latex]x_5=f(4,x_4)[/latex], … and so on. This is what’s called a recursively defined sequence. The existence of such a sequence is obvious; the corollary asserts that it is the only one possible. By the way, could you state the theorem or proposition to which this result is a corollary?

Basically you are looking at a triangle with side lengths 3, 5, and 7. The angle between the two shorter sides (call it [latex]\theta[/latex]) can be found by the cosine rule: [latex]c^2\ =\ a^2+b^2-2ab\cos\theta[/latex] Now you want to look at the triangle two of whose sides have lengths 3 and 5, the angle between them being [latex]\pi-\theta[/latex]. The length of the opposite side can again be found by the cosine rule.

It can be in any angular unit you like radians, degrees, grads, anything. You simply set your calculator to the angular unit you want and compute the inverse tangent. Note however that in some cases you have to use only radians. For example, if [latex]f(t)=\arctan t[/latex] and you compute [latex]f'(t)=\frac1{1+t^2}[/latex], then in order for the formula to work, [latex]t[/latex] must be in radians. You have to be aware of cases when you must use radians; otherwise you have a choice of radians or degrees.

[latex]2\left[\sum_{i=0}^{n}{i}\right]+(n+1)\ =\ 2\left[\frac{n(n+1)}2\right]+(n+1)\ =\ (n+1)^2[/latex] Also [latex]\left(\sum_{i=0}^n10^i\right)^2\ =\ \sum_{i=0}^n10^{2i}+2\sum_{0\leqslant i<j\leqslant n}^n10^{i+j}[/latex] using the formula [latex]\left(a_n+a_{n-1}+\cdots+a_1+a_0\right)^2\ =\ a_n^2+\cdots+a_0^2+2\sum_{0\leqslant i<j\leqslant n}^na_ia_j[/latex] Given [latex]1\leqslant r\leqslant n[/latex], how many pairs [latex]i,j[/latex] are there with [latex]0\leqslant i<j\leqslant n[/latex] such that [latex]i+j=r[/latex]? Suppose [latex]r=2k[/latex] is even. Then [latex]r=0+2k=1+(2k-1)=\cdots=(k-1)+(k+1)[/latex] so there are [latex]k[/latex] such [latex]i,j[/latex] pairs. Thus the coefficent of [latex]10^r[/latex] is [latex]2k+1=r+1[/latex]. Suppose [latex]r=2k-1[/latex] is odd. Then [latex]r=0+(2k-1)=1+(2k-2)=\cdots=(k-1)+k[/latex] so again there are [latex]k[/latex] such [latex]i,j[/latex] pairs. Then the coefficent of [latex]10^r[/latex] is [latex]2k=r+1[/latex] again. For [latex]n+1\leqslant r\leqslant2n-1[/latex], we let [latex]s=r-n[/latex]; then [latex]r=s+n=(s+1)+(n-1)=\cdots=(s+k)+(n-k)[/latex] giving [latex]k+1[/latex] pairs. Since [latex]s-k<n-k[/latex] we have [latex]n-s>2k[/latex]; therefore [latex]n-s=2k+1[/latex] or [latex]n-s=2k+2[/latex]. If [latex]r[/latex] is even, then so is [latex]n-s=r-2s[/latex]; then [latex]n-s=2k+2[/latex] and the coefficient of [latex]10^r[/latex] is [latex]2(k+1)+1=n-s+1=2n-r+1[/latex]. If [latex]r[/latex] is odd, then so is [latex]n-s[/latex]; so [latex]n-s=2k+1[/latex] and the coefficient of [latex]10^r[/latex] is [latex]2(k+1)=n-s+1=2n-r+1[/latex] again. Hence [latex]\left(\sum_{i=0}^n10^i\right)^2\ =\ 10^{2n}+1+\sum_{r=1}^n(r+1)10^r+\sum_{r=n+1}^{2n-1}(2n-r+1)10^r\ =\ \sum_{r=0}^n(r+1)10^r+\sum_{r=n+1}^{2n}(2n-r+1)10^r[/latex].

Logarithms and square roots of negative numbers do exist. They are complex rather than real.