geometry - Intersection point of line segments

I need to find the intersection point of 2 line segments (lines are finite, i.e., they have end points).

e.g. segment 1 from $(x_1, y_1)$ to $(x_2, y_2)$ -- segment 2 from $(x_3, y_3)$ to $(x_4, y_4)$

you can assume $m_1$ and $m_2$ are the gradients of segment 1 and segment 2 respectively

similarly, $c_1$ and $c_2$ being the y-intercepts of segment 1 and segment 2 respectively

Using $y=mx+c$ I can easily find the equations of both lines and then derive the intersection point from those equations and check if that point actually lies on both the line segments (intersection point may not lie between the end points).

My problem is when I calculate the gradients $m_1$ and $m_2$ I do $\frac{y_2-y_1}{x_2-x_1}$ there for if one of the lines is vertical then I am going to have problems. How can I deal with this?

Another problem is that if the two segments intersect at a point which is also the same as one of the end points then I want to assume that they are not intersecting.

e.g. if segment 1 is from $(2, 3)$ to $(3, 7)$ and segment 2 is from $(3, 7)$ to $(7, 3)$ then I want to assume they don't intersect.

Answer

Instead of taking $x,y$ as the variables to solve for, write $(x,y)$ $= (x_1,y_1)+t(x_2-x_1,y_2-y_1) $ $= (x_3,y_3)+u(x_4-x_3,y_4-y_3)$ and solve for $t$ and $u$. Then it doesn't matter a bit whether your line segments are horizontal or vertical or whatever, and you can check for being within the segments just by looking at whether $0 < t < 1$ and $0 < u < 1$.

Note that if the line segments are parallel you'll get zero in your denominator, and if they're almost parallel you'll get something very small there; you may want to take care about that unless something in your setup guarantees that the line segments aren't close to being parallel.

$a_1=k,a_{n}=2a_{n-1}+1(ngeq 2).$ Does there exist $kinmathbb N$ such that $a_n,n=1,2,3,cdots$ are all composite numbers?

Let $a_1=k,a_{n}=2a_{n-1}+1(n\geq 2).$

If $k=1$ then $a_n=1,3,7,15,31,63,\cdots$ here $3,7,31$ are prime numbers. I'm interested in this problem:

Does there exist $k\in\mathbb N$ such that $a_n,n=1,2,3,\cdots$ are all composite numbers?

If $k=147$ then $a_n,n=1,2,\cdots 2551$ are all composite, but $a_{2552}$ is prime. So I doubt the existence of such number.

Answer

The numbers you mention are Riesel numbers http://en.wikipedia.org/wiki/Riesel_number
and there is the similar Sierpinski numbers where it is $2a_{n-1}-1$ instead.

http://en.wikipedia.org/wiki/Sierpinski_number

combinatorics - Binomial coefficients identity: $sum i binom{n-i}{k-1}=binom{n+1}{k+1}$

I am trying to prove

$
\sum_{i=1}^{n-k+1} i \binom{n-i}{k-1}=\binom{n+1}{k+1}
$

Whichever numbers for $k,n$ I try, the terms equal, but when I try to use induction by n, I fail to prove the induction step:

Assume the equation holds for $n$. Now by Pascal's recursion formula,

$
\binom{n+2}{k+1}=\binom{n+1}{k+1} + \binom{n+1}{k}\\
=\sum_{i=1}^{n-k+1} i \binom{n-i}{k-1}+\binom{n+1}{k},
$

by induction assumption. In order to complete the proof, I would need to show

$

(n-k+2) \binom{n-(n-k+2)}{k-1} = \binom{n+1}{k}
$

but the left-hand side is zero. What am I doing wrong?

EDIT:

There were links to similar questions when my question was marked as duplicate. However, these links are now gone, so I add them here as they were useful to me:

(I did search, but did not found these.)

Answer

No, to complete the proof along these lines you need to show that

$$\sum_{i=1}^{n-k+2}i\binom{n+1-i}{k-1}=\sum_{i=1}^{n-k+1}i\binom{n-i}{k-1}+\binom{n+1}k\;;\tag{1}$$

you forgot that the upper number in the binomial coefficient changes when you go from $n$ to $n+1$.

You can rewrite $(1)$ as

$$(n-k+2)\binom{k-1}{k-1}+\sum_{i=1}^{n-k+1}i\left(\binom{n+1-i}{k-1}-\binom{n-i}{k-1}\right)=\binom{n+1}k\;,$$

whose lefthand side reduces to

$$n-k+2+\sum_{i=1}^{n-k+1}i\binom{n-i}{k-2}$$

by Pascal’s identity. This in turn can be rewritten as

$$n-k+2+\sum_{i=1}^{n-k+2}i\binom{n-i}{k-2}-(n-k+2)\binom{k-2}{k-2}=\sum_{i=1}^{n-k+2}i\binom{n-i}{k-2}\;.$$

If you took the right induction hypothesis — namely, that the equation holds for $n$ and all $k$ — then your induction hypothesis allows you to reduce this last summation to a single binomial coefficient, which proves to be the one that you want.

integration - Triple Euler sum result $sum_{kgeq 1}frac{H_k^{(2)}H_k }{k^2}=zeta(2)zeta(3)+zeta(5)$

In the following thread

I arrived at the following result

$$\sum_{k\geq 1}\frac{H_k^{(2)}H_k }{k^2}=\zeta(2)\zeta(3)+\zeta(5)$$

Defining

$$H_k^{(p)}=\sum_{n=1}^k \frac{1}{n^p},\,\,\, H_k^{(1)}\equiv H_k $$

But, it was after long evaluations and considering many variations of product of polylogarithm integrals.

I think there is an easier approach to get the solution, any ideas ?

Answer

Here's a derivation that, while fairly long, is self-contained and uses only basic series manipulation techniques, like partial fractions decomposition, telescoping, swapping the order of summation, etc. It leans heavily on ideas from Borwein and Girgensohn's paper "Evaluation of Triple Euler Sums" (Electronic Journal of Combinatorics 3(1) 1996).

First, some notation. Define the multiple zeta functions by
\begin{align}
\zeta_N(a) &= \sum_{x=1}^N \frac{1}{x^a}, \:\:\: \zeta_N(a,b) = \sum_{x=1}^N \sum_{y=1}^{x-1} \frac{1}{x^a y^b}, \:\:\: \zeta_N(a,b,c) = \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1}\frac{1}{x^a y^b z^c}, \\

\zeta(a,b) &= \sum_{x=1}^{\infty} \sum_{y=1}^{x-1} \frac{1}{x^a y^b}, \:\:\: \zeta(a,b,c) = \sum_{x=1}^{\infty} \sum_{y=1}^{x-1} \sum_{z=1}^{y-1}\frac{1}{x^a y^b z^c}.
\end{align}

We will need the following symmetry relation, as well as expressions for $\zeta(4,1)$ and $\zeta(2,2,1) + \zeta(2,1,2)$. Proofs for all of these are given at the end of the post.
\begin{align}
\zeta_N(a,b) + \zeta_N(b,a) &= \zeta_N(a) \zeta_N(b) - \zeta_N(a+b) \tag{1}\\
\zeta(4,1) &= \zeta(5) - \zeta(3,2) - \zeta(2,3) \tag{2}\\
\zeta(2,2,1) + \zeta(2,1,2) &= \zeta(2,3) + \zeta(3,2) \tag{3}
\end{align}

Given these, we have

The Main Proof:
\begin{align}
\sum_{k=1}^{\infty} \frac{H^{(2)}_k H_k}{k^2} &= \sum_{k=1}^{\infty} \frac{H^{(2)}_{k-1} H_{k-1}}{k^2} + \sum_{k=1}^{\infty} \frac{H^{(2)}_{k-1}}{k^3} + \sum_{k=1}^{\infty} \frac{H_{k-1}}{k^4} + \sum_{k=1}^{\infty} \frac{1}{k^5} \\
&= \sum_{k=1}^{\infty} \frac{H^{(2)}_{k-1} H_{k-1}}{k^2} + \zeta(3,2) + \zeta(4,1) + \zeta(5).
\end{align}
The most complicated sum is the first, so let's look at that more closely.
\begin{align}
\sum_{k=1}^{\infty} \frac{H^{(2)}_{k-1} H_{k-1}}{k^2} &= \sum_{k=1}^{\infty} \frac{1}{k^2} \zeta_{k-1}(2) \zeta_{k-1}(1) \\

&= \sum_{k=1}^{\infty} \frac{1}{k^2} (\zeta_{k-1}(2,1) + \zeta_{k-1}(1,2) + \zeta_{k-1}(3)), \text{ by (1)} \\
&= \zeta(2,2,1) + \zeta(2,1,2) + \zeta(2,3), \text{ by definition of the multiple zeta functions} \\
&= 2\zeta(2,3) + \zeta(3,2), \text{ by (3)}.
\end{align}
Thus
\begin{align}
\sum_{k=1}^{\infty} \frac{H^{(2)}_k H_k}{k^2} &= 2 \zeta(2,3) + \zeta(3,2) + \zeta(3,2) + \zeta(5) - \zeta(3,2) - \zeta(2,3) + \zeta(5), \text{ by (2)} \\
&= \zeta(2,3) + \zeta(3,2) + 2 \zeta(5) \\
&= \zeta(2) \zeta(3) - \zeta(5) + 2 \zeta(5), \text{ by (1)} \\
&= \zeta(2) \zeta(3) + \zeta(5).

\end{align}

---

---

Proof of (1):
\begin{align}
\zeta_N(a,b) + \zeta_N(b,a) &= \sum_{x=1}^N \sum_{y=1}^{x-1} \frac{1}{x^a y^b} + \sum_{x=1}^N \sum_{y=1}^{x-1} \frac{1}{x^b y^a} \\
&= \sum_{y=1}^N \sum_{x=y+1}^N \frac{1}{x^a y^b} + \sum_{x=1}^N \sum_{y=1}^{x-1} \frac{1}{x^b y^a}, \\
& \:\:\:\:\:\: \text{ swapping the order of summation on the first sum} \\
&= \sum_{x=1}^N \sum_{y=x+1}^N \frac{1}{y^a x^b} + \sum_{x=1}^N \sum_{y=1}^{x-1} \frac{1}{x^b y^a}, \text{ relabeling variables on the first sum} \\
&= \sum_{x=1}^N \sum_{y=1}^N \frac{1}{y^a x^b} - \sum_{x=1}^N \frac{1}{x^{a+b}}, \text{ combining sums} \\
&= \zeta_N(a) \zeta_N(b) - \zeta_N(a+b). \square

\end{align}

Proof of (2):
\begin{align}
\zeta(4,1) &= \sum_{x=1}^{\infty} \sum_{y=1}^{x-1} \frac{1}{x^4 y} \\
&= \sum_{x=1}^{\infty} \sum_{y=1}^{x-1} \frac{1}{x^4 (x-y)}, \text{ reindexing the second sum} \\
&= \sum_{x=1}^{\infty} \sum_{y=1}^{x-1} \left(-\frac{1}{x^4 y} - \frac{1}{x^3 y^2} - \frac{1}{x^2y^3} - \frac{1}{x y^4} + \frac{1}{(x-y)y^4}\right), \\
&\:\:\:\:\: \text{ by partial fractions decomposition}\\
&= - \zeta(4,1) - \zeta(3,2) - \zeta(2,3) + \sum_{x=1}^{\infty} \sum_{y=1}^{x-1} \left(\frac{1}{(x-y)y^4} - \frac{1}{x y^4} \right) \\
&= - \zeta(4,1) - \zeta(3,2) - \zeta(2,3) + \sum_{x=1}^{\infty} \sum_{y=1}^{x-1} \frac{1}{y^4} \left(\frac{1}{x-y} - \frac{1}{x} \right) \\

&= - \zeta(4,1) - \zeta(3,2) - \zeta(2,3) + \sum_{y=1}^{\infty} \frac{1}{y^4} \sum_{x=y+1}^{\infty} \left(\frac{1}{x-y} - \frac{1}{x} \right), \\
& \:\:\:\:\: \text{ swapping the order of summation} \\
&= - \zeta(4,1) - \zeta(3,2) - \zeta(2,3) + \sum_{y=1}^{\infty} \frac{1}{y^4} \sum_{x=1}^y \frac{1}{x}, \text{ as the sum telescopes} \\
&= - \zeta(4,1) - \zeta(3,2) - \zeta(2,3) + \zeta(4,1) + \zeta(5) \\
&= \zeta(5) - \zeta(3,2) - \zeta(2,3). \square
\end{align}

For the proof of (3), we need the following additional symmetry result:
\begin{equation}
\zeta_N(a,b,c) + \zeta_N(a,c,b) + \zeta_N(c,a,b) = \zeta_N(c) \zeta_N(a,b) - \zeta_N(a,b+c) - \zeta_N(a+c,b) \tag{4}

\end{equation}

Proof of (4):
\begin{align}
&\zeta_N(a,b,c) + \zeta_N(a,c,b) + \zeta_N(c,a,b) \\
&=\sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1}\frac{1}{x^a y^b z^c} + \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1}\frac{1}{x^a y^c z^b} + \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1}\frac{1}{x^c y^a z^b} \\
&= \sum_{x=1}^N \sum_{z=1}^{x-1} \sum_{y=z+1}^{x-1} \frac{1}{x^a y^b z^c} + \sum_{y=1}^N \sum_{x=y+1}^N \sum_{z=1}^{y-1}\frac{1}{x^a y^c z^b} + \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1}\frac{1}{x^c y^a z^b}, \\
&\:\:\:\:\:\text{ swapping order of summation on the first two sums} \\
&= \sum_{z=1}^N \sum_{x=z+1}^N \sum_{y=z+1}^{x-1} \frac{1}{x^a y^b z^c} + \sum_{y=1}^N \sum_{x=y+1}^N \sum_{z=1}^{y-1}\frac{1}{x^a y^c z^b} + \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1}\frac{1}{x^c y^a z^b}, \\
&\:\:\:\:\:\text{ swapping order of summation on the first sum} \\

&= \sum_{x=1}^N \sum_{y=x+1}^N \sum_{z=x+1}^{y-1} \frac{1}{x^c y^a z^b} + \sum_{x=1}^N \sum_{y=x+1}^N \sum_{z=1}^{x-1}\frac{1}{x^c y^a z^b} + \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1}\frac{1}{x^c y^a z^b}, \\
&\:\:\:\:\: \text{ relabeling variables on the first two sums} \\
&= \sum_{x=1}^N \sum_{y=x+1}^N \sum_{z=1}^{y-1} \frac{1}{x^c y^a z^b} - \sum_{x=1}^N \sum_{y=x+1}^N \frac{1}{x^{b+c} y^a} + \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1}\frac{1}{x^c y^a z^b}, \\
&\:\:\:\:\: \text{ combining the first two sums} \\
&= \sum_{x=1}^N \sum_{y=1}^N \sum_{z=1}^{y-1} \frac{1}{x^c y^a z^b} - \sum_{x=1}^N \sum_{z=1}^{y-1} \frac{1}{x^{a+c} z^b} - \sum_{y=1}^N \sum_{x=1}^{y-1} \frac{1}{x^{b+c} y^a}, \\
&\:\:\:\:\:\text{ combining the first and third sums and swapping the order of summation on the second} \\
&= \zeta_N(c) \zeta_N(a,b) - \zeta_N(a+c,b) - \zeta_N(a,b+c). \square
\end{align}

Proof of (3):

\begin{align}
\zeta_N(2,2,1) &= \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1} \frac{1}{x^2 y^2 z} \\
&= \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1} \frac{1}{x^2 y^2 (y-z)}, \text{ reindexing on the third sum} \\
&= \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1} \left( -\frac{1}{x^2 y z^2} - \frac{1}{x^2 y^2 z} + \frac{1}{x^2(y-z)z^2} \right), \\
&\:\:\:\:\: \text{ by partial fractions decomposition} \\
&= - \zeta_N(2,1,2) - \zeta_N(2,2,1) + \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1} \frac{1}{x^2(y-z)z^2} \tag{5}. \\
\end{align}
Now, let's look at the third expression in (5).
\begin{align}
&\sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1} \frac{1}{x^2(y-z)z^2} \\

&= \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{x-y-1} \frac{1}{x^2(x-y-z)z^2}, \text{ reindexing the second sum} \\
&= \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=y+1}^{x-1} \frac{1}{x^2(x-z)(z-y)^2}, \text{ reindexing the third sum} \\
&= \sum_{x=1}^N \sum_{z=1}^{x-1} \sum_{y=1}^{z-1} \frac{1}{x^2(x-z)(z-y)^2}, \text{ swapping the order of summation} \\
&= \sum_{x=1}^N \sum_{z=1}^{x-1} \sum_{y=1}^{z-1} \frac{1}{x^2(x-z)y^2}, \text{ reindexing the third sum} \\
&= \sum_{x=1}^N \sum_{z=1}^{x-1} \sum_{y=1}^{z-1} \left(-\frac{1}{x y^2 z^2} - \frac{1}{x^2 y^2 z} + \frac{1}{(x-z)y^2 z^2} \right), \text{ by partial fractions decomposition} \\
&= - \zeta_N(1,2,2) - \zeta_N(2,1,2) + \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1} \frac{1}{(x-y)y^2 z^2} \tag{6}, \text{ relabeling variables}.
\end{align}
Let's look at the third expression in (6).
\begin{align}
&\sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1} \frac{1}{(x-y)y^2 z^2} \\

&= \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1} \frac{1}{(x-y)y^2 z^2} + \sum_{x=1}^N \sum_{y=N+1-x}^N \sum_{z=1}^{y-1} \frac{1}{x y^2 z^2} - \sum_{x=1}^N \sum_{y=N+1-x}^N \sum_{z=1}^{y-1} \frac{1}{x y^2 z^2} \\
&= \left(\sum_{x=1}^N \frac{1}{x}\right) \left(\sum_{y=1}^N \sum_{z=1}^{y-1} \frac{1}{y^2 z^2} \right) - \sum_{x=1}^N \sum_{y=N+1-x}^N \sum_{z=1}^{y-1} \frac{1}{x y^2 z^2}, \\
&\:\:\:\:\: \text{ via the finite sum version of the Cauchy product} \\
&= \zeta_N(1) \zeta_N(2,2) - e_N(1,2,2), \tag{7} \\
\end{align}
where
$$e_N(1,2,2) = \sum_{x=1}^N \sum_{y=N+1-x}^N \sum_{z=1}^{y-1} \frac{1}{x y^2 z^2}.$$
Putting (5), (6), and (7) together, we have
\begin{align}
\zeta_N(2,2,1) =& - \zeta_N(2,1,2) - \zeta_N(2,2,1) - \zeta_N(1,2,2) - \zeta_N(2,1,2) + \zeta_N(1) \zeta_N(2,2) \\

&- e_N(1,2,2), \\
\zeta_N(2,2,1) + \zeta_N(2,1,2) &= - \zeta_N(1) \zeta_N(2,2) + \zeta_N(2,3) + \zeta_N(3,2) + \zeta_N(1) \zeta_N(2,2) \\
&- e_N(1,2,2), \text{ by (4)} \\
=& \zeta_N(2,3) + \zeta_N(3,2) - e_N(1,2,2). \\
\end{align}
All that remains to complete the proof of (3) is to show that $e_N(1,2,2) \to 0$ as $N \to \infty$. We have
\begin{align}
e_N(1,2,2) &= \sum_{x=1}^N \sum_{y=N+1-x}^N \sum_{z=1}^{y-1} \frac{1}{x y^2 z^2} \\
&\leq \sum_{x=1}^N \sum_{y=N+1-x}^N \sum_{z=1}^N \frac{1}{x y^2 z^2} \\
&= \zeta_N(2) \sum_{x=1}^N \sum_{y=N+1-x}^N \frac{1}{x y^2} \\

&= \zeta_N(2) \sum_{y=1}^N \sum_{x=N+1-y}^N \frac{1}{x y^2}, \text{ swapping the order of summation} \\
&\leq \zeta_N(2) \sum_{y=1}^N \frac{1}{y^2} \sum_{x=N+1-y}^N \frac{1}{N+1-y} \\
&= \zeta_N(2) \sum_{y=1}^N \frac{1}{y^2} \frac{y}{N+1-y} \\
&= \zeta_N(2) \sum_{y=1}^N \frac{1}{y (N+1-y)}\\
&= \zeta_N(2) \frac{1}{N+1}\sum_{y=1}^N \left(\frac{1}{y} + \frac{1}{N+1-y} \right), \text{ by partial fractions decomposition} \\
&= \zeta_N(2) \frac{2}{N+1} \zeta_N(1),
\end{align}
which goes to $0$ as $N \to \infty$, since $\zeta_N(1) = O(\log N)$ and $\zeta_N(2) = O(1)$. $\square$

combinatorics - Notation for writing multinomial coefficient as sum of smaller multinomial coefficients

This question is an attempt to extend the Pascal triangle's hockey stick identity to multinomial coefficients as asked in question Hockey-Stick Theorem for Multinomial Coefficients.

Consider the following recursive relation:

$$\binom{n_1+n_2+\cdots+n_t}{n_1,n_2,\cdots,n_t}=\sum_{\text{For all nonzero $x_j$ except last}}\binom{n_1+n_2+\cdots+n_t-1}{n_1,\cdots,n_j-1,\cdots,n_t}+
\binom{n_1+n_2+\cdots+n_t-1}{n_1,n_2,\cdots,n_t-1}_{\text{$n_t$ being last non-zero $x_j$}}$$

where
$$\binom{n_1+n_2+\cdots+n_t}{n_1,n_2,\cdots,n_t}=\frac{(n_1+n_2+\cdots+n_t)!}{n_1! n_2! \cdots n_t!} $$
Example:

\begin{eqnarray}
\binom{6}{3,1,2}&=&\binom{5}{2,1,2}+\binom{5}{3,0,2}+\binom{5}{3,1,1}\\
&=&\binom{5}{2,1,2}+\binom{5}{3,0,2}+\left\{ \binom{4}{2,1,1}+\binom{4}{3,0,1}+\binom{4}{3,1,0} \right\}\\
&=&\binom{5}{2,1,2}+\binom{5}{3,0,2}+\binom{4}{2,1,1}+\binom{4}{3,0,1}+
\left\{\binom{3}{2,1,0}+\binom{3}{3,0,0} \right\}\\
&=&\binom{5}{2,1,2}+\binom{5}{3,0,2}+\binom{4}{2,1,1}+\binom{4}{3,0,1}+
\binom{3}{2,1,0}+\left\{\binom{2}{2,0,0} \right\}\\
\end{eqnarray}
How may I write the following line in compact sigma notation?

$$\binom{6}{3,1,2}=\binom{5}{2,1,2}+\binom{5}{3,0,2}+\binom{4}{2,1,1}+\binom{4}{3,0,1}+
\binom{3}{2,1,0}+\binom{2}{2,0,0}$$

How to write it for general form?

Answer

$$\binom{n_1+n_2+\cdots+n_t}{n_1,n_2,\cdots,n_t}=1+\sum_{i=2}^t \sum_{j=1}^{i-1} \sum_{k=1}^{n_i} \binom{ n_1+n_2+\cdots+n_{i-1}+k }{n_1,n_2,\cdots,n_j-1,\cdots,n_{i-1},k }$$

functions - Is 4.99999......... exactly equal to 5?

I'm a student of 10th std. Recently our teacher asked a Question that

"Is 4.999...equal to 5 or not?"
Everyone said that is isn't equal or it is approximately equal. Teacher too agreed to that. But I did't agree. I opposed the teacher as I think it is precisely equal to 5. I also prove but then too she isn't satisfied. Can you please explain what is right and what is wrong in this case?Please Justify.

   Proof that I showed : let x=4.999...  (a)
                           therefore, 10x = 49.999...
                                      10x -x = 49.999... -4.999...

                                      9x     =  45
                                      x = 45%9
                                      X= 5     (b)
                         hence (a) =(b).
                                    And therefore, ***4.999... = 5***

soft question - Best Fake Proofs? (A M.SE April Fools Day collection)

In honor of April Fools Day $2013$, I'd like this question to collect the best, most convincing fake proofs of impossibilities you have seen.

I've posted one as an answer below. I'm also thinking of a geometric one where the "trick" is that it's very easy to draw the diagram wrong and have two lines intersect in the wrong place (or intersect when they shouldn't). If someone could find and link this, I would appreciate it very much.