Multiplication by Power of 10 by Moving Decimal Point

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Theorem

Let $n \in \R$ be a real number.


Let $n$ be expressed in decimal notation.

Let $10^d$ denote a power of $10$ for some integer $d$


Then $n \times 10^d$ can be expressed in decimal notation by shifting the decimal point $d$ places to the right.

Thus, if $d$ is negative, and so $10^d = 10^{-e}$ for some $e \in \Z_{>0}$, $n \times 10^d$ can be expressed in decimal notation by shifting the decimal point $e$ places to the left.


Proof

Let $n$ be expressed in decimal notation as:

$n = \sqbrk {a_r a_{r - 1} \dotso a_1 a_0 \cdotp a_{-1} a_{-2} \dotso a_{-s} a_{-s - 1} \dotso}$

That is:

$n = \ds \sum_{k \mathop \in \Z} a_k 10^k$


Then:

\(\ds n \times 10^d\) \(=\) \(\ds 10^d \times \sum_{k \mathop \in \Z} a_k 10^k\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop \in \Z} 10^d \times a_k 10^k\) Distributive Property
\(\ds \) \(=\) \(\ds \sum_{k \mathop \in \Z} a_k 10^{k + d}\) Product of Powers
\(\ds \) \(=\) \(\ds \sum_{k \mathop \in \Z} a_{k - d} 10^k\) Translation of Index Variable of Summation

The effect of presenting digit $a_{k - d}$ in position $k$ of $n$ is the same as what you get having moved the decimal point from between $a_0$ and $a_{-1}$ to between $a_{-d}$ and $a_{-d - 1}$.

Thus:

if $d$ is positive, that is equivalent to moving the decimal point $d$ places to the right

and:

if $d$ is negative, that is equivalent to moving the decimal point $d$ places to the left.

$\blacksquare$


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