Hello.
I'm Professor Von Schmohawk and welcome to Why U.
We have seen that multiplying any decimal number by ten
shifts each digit in the number one column to the left.
In the last lecture,
we used this trick to convert decimal numbers with a finite number of digits into fractions.
In this lecture, we will see how to convert "repeating" decimal numbers into fractions.
Repeating decimal numbers have one or more digits after the decimal point
that repeat forever.
As an example, let’s convert a decimal number with one infinitely repeating digit
into a fraction.
The first step is to eliminate the infinitely repeating digit.
We do this by first multiplying the number by ten to shift it one column to the left,
and then subtracting the original number.
Since both numbers in this example have an infinite number of repeating fives,
this eliminates all the repeating fives after the decimal point.
We know that our original decimal number is equal to some fraction,
but we don’t yet know what that fraction is.
However, we do know that when it is subtracted from ten times itself, we get five.
Let’s write this statement as an equation.
Ten times the fraction,
minus the fraction,
equals five.
It is possible to write this statement more simply.
Subtracting one thing from ten things leaves nine things.
So instead of writing “ten times the fraction, minus the fraction”
we can simply write “nine times the fraction”.
Since both sides of the equation are equal,
we can divide each side by the same number and they will still be equal.
Let’s divide both sides by nine.
Notice that on the left side of the equation,
the fraction is multiplied AND divided by nine.
This leaves its value unchanged,
so we can cancel out the nines on the left side of the equation.
We are left with an equation which says that the fraction is equal to five-ninths.
We have converted our repeating decimal number to the fraction five-ninths.
Let’s try this again with a repeating sequence of threes.
As we did in the previous example, we eliminate the one repeating digit
by multiplying by ten one time, and then subtracting the original number.
This eliminates all the repeating threes to the right of the decimal point.
So ten times the fraction, minus the fraction, equals three.
We write this as an equation which we can simplify to say:
“Nine times the fraction equals three”.
We then divide each side of the equation by nine
so that the nines on the left side cancel.
This equation tells us that our fraction is equal to three-ninths.
Of course the fraction three-ninths can be reduced to one-third.
So we have converted this repeating decimal number to the fraction one-third.
Now instead of a single repeating digit,
let’s convert a decimal number with a repeating sequence of three digits.
Since there are three repeating digits to eliminate,
we must shift the number three columns to the left, by multiplying it by ten three times.
This, of course, is the same as multiplying by one-thousand.
We then subtract our original number, which eliminates the repeating digits.
Now we write this as an equation:
One-thousand times the fraction,
minus the fraction,
equals one-hundred-twenty-three.
One-thousand times the fraction, minus the fraction
is the same as nine-hundred-ninety-nine times the fraction.
And dividing both sides of the equation by nine-hundred-ninety-nine
cancels out the nine-hundred-ninety-nine on the left.
So our fraction is equal to one-hundred-twenty-three nine-hundred-ninety-ninths.
Let’s see what happens with the repeating decimal number
zero-point-one, followed by an infinite number of sixes.
Since there is only one repeating digit, six,
we shift the number one column to the left by multiplying by ten one time,
and subtract the original number.
This eliminates the repeating sixes,
and we are left with one-point-five.
We then write the equation:
"Ten times the fraction, minus the fraction, equals one-point-five",
which we can simplify to
“Nine times the fraction equals one-point-five”.
Dividing both sides of the equation by nine eliminates the nines on the left,
and we see that our fraction is equal to one-point-five over nine.
But we have a problem.
A fraction’s numerator and denominator must be integers.
We can solve this problem by multiplying the top and bottom of the fraction by ten
to create two integers.
This gives us fifteen-ninetieths,
which can be reduced to one-sixth.
So this repeating decimal number is equal to one-sixth.
As a final example, let’s reduce a number that has the repeating sequence of digits,
one-four-two-eight-five-seven.
Since there are six repeating digits,
we shift this number six columns to the left by multiplying it by ten six times,
which is the same as multiplying by one-million.
We then subtract the original number.
This eliminates all but one copy of the repeating sequence.
We are left with the integer, one-four-two-eight-five-seven.
We then write the equation:
"One million times the fraction, minus the fraction,
equals one-four-two-eight-five-seven."
Using the same logic as before,
the left side of this equation can be simplified to
nine-nine-nine-nine-nine-nine times the fraction.
Dividing both sides of the equation by nine-nine-nine-nine-nine-nine,
leaves us with an equation which says that our number is equal to the fraction
one-four-two-eight-five-seven over nine-nine-nine-nine-nine-nine.
Using the factoring techniques we learned in the lecture on reducing fractions,
this scary-looking fraction can be reduced to one-seventh.
So we have discovered that this repeating decimal number
is equivalent to the fraction one-seventh.
This concludes our lectures on fractions.
So far we have learned about the properties of the arithmetic operations
addition, subtraction, multiplication, and division.
In our next lecture, we will introduce the mathematical operation of "exponentiation".