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\title{\bf Announcement 247: The gradient of y-axis is zero and $\tan (\pi/2) =0$ by the division by zero $1/0=0$}
\author{{\it Institute of Reproducing Kernels}\\
\date{September 22, 2015}
\maketitle
In Announcement 246, we stated:
\medskip
Consider the lines $y = ax$ with gradients $a$ through the origin $ 0$. Consider the two limits that $a \quad (>0)$ tends to $ + \infty$ and $a \quad (<0)$ tends to $- \infty$, respectively. As their limits, we see that the limiting lines are $y$ — axis. Note that the gradient of the $y$ axis is zero, not infinity.
This example shows as in the graph of the function $y = f(x) = 1/x$ at $x = 0$ as $f(0) =0$, that was introduced by the division by zero $1/0=0$ mathematically (\cite{s,kmsy,ttk,ann}).
\medskip
For this announcement, Professor H. Begehr kindly referred to the gradient of the $y$ axis in the above: If the gradient of the imaginary axis is $0$ this would mean $\tan (\pi/2)=0$,
right? Of course this would be a consequence of $1/0=0$!
\medskip
We had sent the e-mail, soon as follows:
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For the gradient of $y$ axis, we can define it as zero, very naturally and in the intuitive sense; of course, we can give its definition precisely.
However, as you stated, we can derive it formally by the division by zero $1/0=0$; this deduction will be very interested in itself, because, the formal result $1/0=0$ is coincident with the natural sense.
\medskip
The gradients of y axis and x axis are both zero.
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Surprisingly enough, this would mean $\tan (\pi/2)=0$,
right?
THIS IS RIGHT for our sense; we gave the definition of the values for analytic functions at an isolated singular point:
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{\bf Theorem :} {\it Any analytic function takes a definite value at an isolated singular point }{\bf with a natural meaning.} The definite value is given by the first coefficient of the regular part in the Laurent expansion around the isolated singular point (\cite{ann}).
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As the fundamental results, we would like to state that
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{\huge \bf I) The gradient of the y axis is zero,}
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and
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{\huge \bf II) $\tan \frac{\pi}{2} = 0,$}
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in the sense of the division by zero in our sense.
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Note that the function $y = \tan x$ is similar with the function $y = 1/x$ around $x = \frac{\pi}{2}
$ and $ x = 0$, respectively.
\footnotesize
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\begin{thebibliography}{10}
\bibitem{s}
S. Saitoh, Generalized inversions of Hadamard and tensor products for matrices, Advances in Linear Algebra \& Matrix Theory. Vol.4 No.2 (2014), 87-95. http://www.scirp.org/journal/ALAMT/
\bibitem{kmsy}
M. Kuroda, H. Michiwaki, S. Saitoh, and M. Yamane,
New meanings of the division by zero and interpretations on $100/0=0$ and on $0/0=0$,
Int. J. Appl. Math. Vol. 27, No 2 (2014), pp. 191-198, DOI: 10.12732/ijam.v27i2.9.
\bibitem{ttk}
S.-E. Takahasi, M. Tsukada and Y. Kobayashi, Classification of continuous fractional binary operators on the real and complex fields, Tokyo Journal of Mathematics (in press).
\bibitem{ann}
Announcement 185: Division by zero is clear as z/0=0 and it is fundamental in mathematics,
Institute of Reproducing Kernels, 2014.10.22.
\end{thebibliography}
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