Show that decimal number expansion is finite

[AllRelative]

Homework Statement

Taken from a Ring theory class:

Let $A$ be the ring of decimal number:$$ A= \{ n10^k|n \in Z, k \in Z \}$$

Show that $x \in A$ if and only if it's decimal expansion is finite. Here we assume that $x \in R$.

Homework Equations

I have no relevant equations really except all the basic definitions of rings which I don't think is useful here.

The Attempt at a Solution

I see it intuitively why but I have a hard time proving it. My attempt is something like:

$\Rightarrow$ :
Since $x \in A$, we can write $x = n * 10^k$ for $k,n \in Z$.Therefore the decimal expansion will have at most a term of $k$th power. Therefore, the decimal will have at most $k + 1$ terms which means it is finite.

$\Leftarrow$ :
Since the decimal expansion of x has a finite number of terms, we can multiply $x$ by a power of 10, $(10^h)$, high enough that $x 10^h \in N$.

Is that a sufficient enough proof? If feel like the right implication is close but not there for some reason.

Thanks

 

[fresh_42]

What is $R$?
What is $A$?

How do you add and multiply in $A$?

 

[AllRelative]

What is $R$?
What is $A$?

$A$ is definitely no ring, because for $a=10^k\; , \;b=10^l\; , \;l\neq k$ we don't have $a+b \in A$.

$N$, $R$ and $Z$ are respectively Natural numbers, Real numbers and Integers.

It is given from the assignment that $A$ is the ring of decimal numbers with the definition I stated before. Wikipedia says It's a unitary ring. It is stable under addition, multiplication and 0 and 1 belongs to it.

I found some proofs but they are in french (language of my course) and the only difference is that they define $A = \{ \frac{n}{10^p}, n \in Z$ and $p \in N \}$.

 

[fresh_42]

Again, write down how the ring operations are defined: $a=10^k\; , \;b=10^m$ and now what is $a-b$ and $a\cdot b\,$?

$A$ is defined as a set, which is normally not a ring. I would have expected that $A$ generates the ring. So by saying "given the ring $A$" there is already a strong statement implicitly made, namely, that all ring elements look a certain way. This can be used in the proof. One way to show the sum is finite, and the other way that an infinite representation for $x \in \mathbb{R}$ cannot be in $A$.

 

[Mark44]

Let $A$ be the ring of decimal number:$A= \{ n10^k|n \in Z, k \in Z \}$

Show that $x \in A$ if and only if it's decimal expansion is finite. Here we assume that $x \in R$.

$\Rightarrow$ :
Since $x \in A$, we can write $x = n * 10^k$ for $k,n \in Z$.Therefore the decimal expansion will have at most a term of $k$th power. Therefore, the decimal will have at most $k + 1$ terms which means it is finite.

Based on the definition of A above, it looks to me like there will be only one term. Did you mean "digit" instead of "term"? Terms are subexpressions that are added in some larger expression.
Some elements of set A are $2\cdot 10^2 = 200$ and $17\cdot 10^{-3} = .017$ and so on. By $\mathbb Z$, I'm assuming the usual definition, which includes both positive and negative integers.

 

[AllRelative]

Based on the definition of A above, it looks to me like there will be only one term. Did you mean "digit" instead of "term"? Terms are subexpressions that are added in some larger expression.
Some elements of set A are $2\cdot 10^2 = 200$ and $17\cdot 10^{-3} = .017$ and so on. By $\mathbb Z$, I'm assuming the usual definition, which includes both positive and negative integers.

Sorry if I used ambiguous terms. What I mean is $A$ is the set of all decimal numbers. They are the numbers written in the form $n 10^k$ where $n$ and $k$ are integers. This would give us this relation $\mathbb{Z} \subset A \subset \mathbb{Q}$. ($\mathbb{Z}$ are rationals and $\mathbb{Z}$ are integers)

For example, let the element $x \in A$ and $x = 3.251$.

We can write $x$ as $3251 * 10^{-3}$

It's decimal expansion is: $$3 * 10^0 + 2*10^{-1} + 5*10^{-2} + 1 * 10^{-3}$$

Which would make this decimal expansion have 4 terms. I need to prove that every element of A have a finite decimal expansion. So it can be represented as a finite sum.

 

[Ray Vickson]

Homework Statement

Taken from a Ring theory class:

Let $A$ be the ring of decimal number:$$ A= \{ n10^k|n \in Z, k \in Z \}$$

Show that $x \in A$ if and only if it's decimal expansion is finite. Here we assume that $x \in R$.

Homework Equations

I have no relevant equations really except all the basic definitions of rings which I don't think is useful here.

The Attempt at a Solution

I see it intuitively why but I have a hard time proving it. My attempt is something like:

$\Rightarrow$ :
Since $x \in A$, we can write $x = n * 10^k$ for $k,n \in Z$.Therefore the decimal expansion will have at most a term of $k$th power. Therefore, the decimal will have at most $k + 1$ terms which means it is finite.

$\Leftarrow$ :
Since the decimal expansion of x has a finite number of terms, we can multiply $x$ by a power of 10, $(10^h)$, high enough that $x 10^h \in N$.

Is that a sufficient enough proof? If feel like the right implication is close but not there for some reason.

Thanks

Are you sure you do not mean to say that
$$A= \{ n/10^k | n, k \in Z \} \; ? $$
For $A$ the way you wrote it, all entries are integers, so of course have finite decimal expansions; in fact, they do not really have decimal expansions at all, at least not in the usually-understood meaning of that term.

 

[WWGD]

Assuming it is $a/10^k$ as Ray suggested ( and only way I see it making sense), consider a string:

a.a_0a_1...a_k = a.a_0a_1...a_k000000...

Can you see how to do the converse?