Difference between revisions of "Logarithm of 1"

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where $\log$ denotes the [[logarithm]].
 
where $\log$ denotes the [[logarithm]].
 
==Proof==
 
==Proof==
 +
By the definition,
 +
$$\log(z) = \displaystyle\int_1^z \dfrac{1}{\tau} \mathrm{d}\tau.$$
 +
Plugging in $z=1$ and using the [[integral from a to a]],
 +
$$\begin{array}{ll}
 +
\log(1) &= \displaystyle\int_1^1 \dfrac{1}{\tau} \mathrm{d}\tau \\
 +
&= 0,
 +
\end{array}$$
 +
as was to be shown.
  
 
==References==
 
==References==
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[[Category:Theorem]]
 
[[Category:Theorem]]
[[Category:Unproven]]
+
[[Category:Proven]]

Latest revision as of 12:24, 17 September 2016

Theorem

The following formula holds: $$\log(1)=0,$$ where $\log$ denotes the logarithm.

Proof

By the definition, $$\log(z) = \displaystyle\int_1^z \dfrac{1}{\tau} \mathrm{d}\tau.$$ Plugging in $z=1$ and using the integral from a to a, $$\begin{array}{ll} \log(1) &= \displaystyle\int_1^1 \dfrac{1}{\tau} \mathrm{d}\tau \\ &= 0, \end{array}$$ as was to be shown.

References