# Prove that an improper fraction with a finite binary expansion

• Jamin2112
In summary, the homework statement is to prove that an improper fraction with a finite binary expansion can also be written as a decimal. However, the homework does not state whether or not this is the converse (i.e., that not all finite base 10 expansions have finite base 2 expansions).
Jamin2112

## Homework Statement

I'm supposed to prove that an improper fraction with a finite binary expansion also can be written as a decimal.

## Homework Equations

Obviously my fraction a/b, where a>b, will look like p1/21 + p2/22 + ... + pn/2n

## The Attempt at a Solution

And I have no idea where to go with this.

Can you write $\frac{1}{2^n}$ as a decimal for any n? Maybe try to do it by hand and see if there's a pattern. Then maybe use induction to formally establish that pattern.

hgfalling said:
Can you write $\frac{1}{2^n}$ as a decimal for any n? Maybe try to do it by hand and see if there's a pattern. Then maybe use induction to formally establish that pattern.

1/21 = 1/2 = .5
1/22 = 1/4 = .25
1/23 = 1/8 = .125
1/24 = 1/16 = .0625
1/25 = 1/32 = .03125
1/26 = 1/64 = .015625
.
.
.
Hmmmm ... No pattern detected. Hint?

Are you going to get any non-terminating or repeating decimals out of this? Why or why not?

hgfalling said:
Are you going to get any non-terminating or repeating decimals out of this? Why or why not?

Since the last digit is 5, and 5/2 is a decimal, you will always be able to get a decimal answer.

OK, so now you have a number with a finite binary representation (ie coefficients of negative powers of 2). And you have terminating decimals for each negative power of 2. Prove that you have won. :)

hgfalling said:
OK, so now you have a number with a finite binary representation (ie coefficients of negative powers of 2). And you have terminating decimals for each negative power of 2. Prove that you have won. :)

I'll take a shot at it.

Proof:

An improper fraction can be represented a/b, where a < b, and a, b $$\in$$ $$Z$$. Suppose a/b can be represented as a finite binary expansion. Then a/b = m1/21 + m2/22 + m3/23 + ... + mn/2n.

Where should I go from here? Hence a/b = $$\delta$$1/101 + $$\delta$$2/102 + $$\delta$$3/103 + ... + $$\delta$$k/10k.

OK, so now you need the $\delta_i$ values.

So suppose I give you 1011.101 in binary. Can you convert it to decimal?

I presume you mean can be written as a finite decimal expansion- any number can be "written as a decimal". A number can be written as a finite binary expansion if and only if it is a fraction that, reduced to lowest form, has a power of 2 as denominator. A number can be written as a finite decimal expansion if and only if it is a fraction that, reduce to lowest form, has denominator which is a product of powers of 2 and 5 only.

hgfalling said:
OK, so now you need the $\delta_i$ values.

So suppose I give you 1011.101 in binary. Can you convert it to decimal?

Sure.

(1x23)+(0x22)+(1x21)+(1x20)+(1x2-1)+(0x2-2)+(1x2-3) = 8 + 0 + 2 + 1 + 1/2 + 0 + 1/8 = 11 5/8 = 11.625.

All that matters is how I went about making the (1/2 + 0 + 1/8) into a decimal, i.e., I need to show that 100/2, 100/4, 100/8, 100/16, ... are decimals. I suppose some sort of induction would be necessary. My statement P(n): for all integers n≥1, 100/2n is a rational number. Proof. We use induction on n. If n=1, then 100/2n=100/21, a rational number. Thus the base case holds. Now we wish to show that if 100/2n is a rational number, then 100/2n+1 is a rational number. 100/2n is converted to 100/2n+1 simply by dividing by 2. Because halving a rational yields another rational number, 100/2n is a rational number for all n.

Something like that?

HallsofIvy said:
I presume you mean can be written as a finite decimal expansion- any number can be "written as a decimal". A number can be written as a finite binary expansion if and only if it is a fraction that, reduced to lowest form, has a power of 2 as denominator. A number can be written as a finite decimal expansion if and only if it is a fraction that, reduce to lowest form, has denominator which is a product of powers of 2 and 5 only.

I'm supposed to prove that an improper fraction that can be represented with a finite binary expansion can also be written as a decimal (a base 10 expansion, in other words), but not the converse (show that not all finite base 10 expansions have finite base 2 expansions)

So ...

1/21 = .5 -----
1/22 = .25 -----
1/23 = .125 ----- ends in 125
1/24 = .0625 ----- ends in 625
1/25 = .03125 ----- ends in 125
1/26 = .0015625 ----- ends in 625
1/27 = .00078125 ----- ends in 125
1/28 =.00390625 ----- ends in 625
1/29=.001953125 ----- ends in 125

That's the only patter I can detect: the last 3 digits alternate between 125 and 625.

As I said before, the numbers that can be written as a finite decimal expansion are those rational numbers which, reduced to lowest form, have denominators powers of 2 and 5. A number will have a finite binary expansion if and only if it can be written as a fraction with denominator a power of 2 only.

Obviously if $x= \frac{m}{2^n}$ then it can be written as $x= \frac{m}{2^n5^0}$. 1/5= 0.2 has a finite decimal expansion but its denominator is not a power of 2 and so cannot be written as a finite binary expansion.

## What is an improper fraction with a finite binary expansion?

An improper fraction with a finite binary expansion is a fraction where the numerator is larger than the denominator and can be represented as a terminating binary decimal, meaning it has a finite number of digits after the decimal point.

## How do you prove that a fraction has a finite binary expansion?

To prove that a fraction has a finite binary expansion, you can convert the fraction into a decimal using long division. If the decimal terminates, then it has a finite binary expansion.

## Why is it important to prove that a fraction has a finite binary expansion?

Proving that a fraction has a finite binary expansion is important in computer science and digital electronics, as these fields use binary numbers extensively. It allows for easier and more accurate calculations.

## Can all improper fractions have a finite binary expansion?

No, not all improper fractions can have a finite binary expansion. Only fractions where the denominator is divisible by a power of 2 (such as 2, 4, 8, etc.) can have a finite binary expansion.

## What is the significance of a fraction having a finite binary expansion?

A fraction having a finite binary expansion means that it can be represented accurately in the binary number system. This is important in fields such as computer science and digital electronics, where binary numbers are used for calculations and data storage.

Replies
7
Views
1K
Replies
3
Views
2K
Replies
1
Views
1K
Replies
1
Views
4K
Replies
9
Views
2K
Replies
5
Views
2K
Replies
7
Views
3K
Replies
5
Views
2K
Replies
5
Views
3K
Replies
3
Views
2K