0.89 recurring as a fraction

Been playing about and tried to get 0.899999999999...... as a fraction

so x = 0.8999................
10x = 8.999...................
100x=89.999.................

90x = 81

x=9/10=0.9

is this because 10x should actually equal 9 as 0.999.......... is equal to 1
and 100x = 90 for the same reason

by this logic 0.0999999999999............ is equal to 0.1 and my original x value is 0.9 (0.8+0.1)

Edit. So any decimal number ending with an infinite repeating 9, although on first glance does not seem to have an exact value, actually does have an exact value.
Please show me a glaring error in my working out.

(edited 12 years ago)

Reply 1

ccfcadam36

1

8/9

EDIT: this is wrong

(edited 12 years ago)

Reply 2

Joinedup

20

89/90 you made a typo I think.

Oops misread the question - ignore me.

(edited 12 years ago)

Reply 3

PennineAcute

OP

3

cannot see as 10x-100x=-90x and 8.9.... - 89.9.. = -81 so x = 81/90

Reply 4

L. Serafy

8

x = 0.09999
100x = 9.99999
10x = 0.99999

x = 9/90 = 0.1

In response to your question, 0.0999999 = 0.1

Reply 5

SsEe

13

Original post by PennineAcute

Been playing about and tried to get 0.899999999999...... as a fraction

so x = 0.8999................
10x = 8.999...................
100x=89.999.................

90x = 81

x=9/10=0.9

is this because 10x should actually equal 9 as 0.999.......... is equal to 1
and 100x = 90 for the same reason

by this logic 0.0999999999999............ is equal to 0.1 and my original x value is 0.9 (0.8+0.1)

Edit. So any decimal number ending with an infinite repeating 9, although on first glance does not seem to have an exact value, actually does have an exact value.
Please show me a glaring error in my working out.

There's no error. 0.899999... = 0.9 and 0.9999... = 1.

Also, any decimal whether it's 0.9999.... or 0.3 or 3.1415926535... has an "exact value" in the sense that it corresponds to some real number (though given a random decimal expansion, it almost certainly won't have a neat representation). However, some numbers have more than one decimal expansion and this is the reason why. 0.999... and 1 both represent the same number.

Reply 6

Bobifier

13

This all stems from the common misconception regarding the decimal 0.9999999... It is in fact equal to 1. Once you accept that this is true, you can divide by ten appropriately to get whaveter other numbers you are concerned with.

Reply 7

lamprey5

1

Original post by PennineAcute

Been playing about and tried to get 0.899999999999...... as a fraction

so x = 0.8999................
10x = 8.999...................
100x=89.999.................

90x = 81

x=9/10=0.9

is this because 10x should actually equal 9 as 0.999.......... is equal to 1
and 100x = 90 for the same reason

by this logic 0.0999999999999............ is equal to 0.1 and my original x value is 0.9 (0.8+0.1)

Edit. So any decimal number ending with an infinite repeating 9, although on first glance does not seem to have an exact value, actually does have an exact value.
Please show me a glaring error in my working out.

Consider the sum of a series

0.8 + 0.09 +0.009 + 0.0009 +0.00009 +....
Which from 0.09 onward is geometric with common ratio r=1/10, which is less than one, therefore it is convergent. The sum s is
s=a/(1-r) where a is the first term (=0.09).

The total sum is

8/10 + 0.09/(1-1/10) = 9/10

Reply 8

raheem94

13

Original post by PennineAcute

Been playing about and tried to get 0.899999999999...... as a fraction

so x = 0.8999................
10x = 8.999...................
100x=89.999.................

90x = 81

x=9/10=0.9

is this because 10x should actually equal 9 as 0.999.......... is equal to 1
and 100x = 90 for the same reason

by this logic 0.0999999999999............ is equal to 0.1 and my original x value is 0.9 (0.8+0.1)

Edit. So any decimal number ending with an infinite repeating 9, although on first glance does not seem to have an exact value, actually does have an exact value.
Please show me a glaring error in my working out.

A number with an infinite repeating number doesn't has an exact value but you get an exact value in calculator because calculator can only display a limited numbers.
For example write 0.7777777777777777777777 in calculator and press = sign. Calculator will display 0.777.........8. The last number is rounded off.

In the case when the repeating number is 9 than rounding it off means that an exact value is obtained.

You can make fraction for numbers with less repeating numbers such as 0.89999999. The fraction for this number will be

\frac{0.8(10^8) + 10^7 - 1 }{10^8}[br]

I hope you will know how did i made the fraction if not then you can ask me.

Reply 9

RichE

15

Original post by raheem94

A number with an infinite repeating number doesn't has an exact value ...

That's just wrong - for example, 0.333333.... (recurring for ever) equals 1/3.

The decimal expansion for 1/3 can't just depend on what your calculator says the answer is (not least because other people will have different calculators and still more so because the decimal expansion of a 1/3 was known well before the advent of electronic calculators).

Reply 10

raheem94

13

Original post by RichE

That's just wrong - for example, 0.333333.... (recurring for ever) equals 1/3.

The decimal expansion for 1/3 can't just depend on what your calculator says the answer is (not least because other people will have different calculators and still more so because the decimal expansion of a 1/3 was known well before the advent of electronic calculators).

1/3 has a decimal expansion of 0.333333333333333333333...........
Calculators usually round off the last displayed digit so in the expansion of 1/3, 3 will be the last diplayed digit so the value will remain as it is.

Reply 11

TenOfThem

18

Original post by raheem94

A number with an infinite repeating number doesn't has an exact value but you get an exact value in calculator because calculator can only display a limited numbers.

This is incorrect

All recurring decimals are rational ... the method used by the OP is the standard method for writing recurring decimals as fractions

Your example of

0.7^.

is in fact

\frac{7}{9}

(edited 12 years ago)

Reply 12

raheem94

13

Original post by TenOfThem

This is incorrect

All recurring decimals are rational ... the method used by the OP is the standard method for writing recurring decimals as fractions

Your example of

0.7^.

is in fact

\frac{7}{9}

Your example of

0.7^.

is in fact

\frac{7}{9}

I am not understanding your above statement.

Reply 13

FredrickTrott

6

Original post by PennineAcute

Been playing about and tried to get 0.899999999999...... as a fraction

so x = 0.8999................
10x = 8.999...................
100x=89.999.................

90x = 81

x=9/10=0.9

is this because 10x should actually equal 9 as 0.999.......... is equal to 1
and 100x = 90 for the same reason

by this logic 0.0999999999999............ is equal to 0.1 and my original x value is 0.9 (0.8+0.1)

Edit. So any decimal number ending with an infinite repeating 9, although on first glance does not seem to have an exact value, actually does have an exact value.
Please show me a glaring error in my working out.

You could express it as a continued fraction

\dfrac{1}{1 + \dfrac{1}{8 + \dfrac{1}{1 + \dfrac{1}{8 + \dfrac{1}{1 + ...}}}}}

Reply 14

TenOfThem

18

Original post by raheem94

Your example of

0.7^.

is in fact

\frac{7}{9}

I am not understanding your above statement.

Not sure what you do not understand

You suggested that 0.7recurring did not have an exact value

I am pointing out that 0.7recurring =

\frac{7}{9}

All recurring decimals can be written as an exact fraction

Reply 15

RichE

15

I'm not sure how you can consistently believe

Original post by raheem94

A number with an infinite repeating number doesn't has an exact value ...

and

Original post by raheem94

1/3 has a decimal expansion of 0.333333333333333333333...........

Reply 16

ghostwalker

17

On a side note, the standard method of writing a repeating decimal, in my day, was:

If it's the single trailing digit repeating, then put a dot over it, and if there is a repeating sequence, put a dot over the first and last digits that repeat, viz.

\frac{1}{3}=0.\dot{3}

and

\frac{1}{7}=0.\dot{1}4285\dot{7}

indicating that this sequence goes on indefinitely, without terminating.

Me too Ghost :smile:

Isn't that how everyone writes it? :/

Reply 19

EEngWillow

11

Original post by ghostwalker

On a side note, the standard method of writing a repeating decimal, in my day, was:

If it's the single trailing digit repeating, then put a dot over it, and if there is a repeating sequence, put a dot over the first and last digits that repeat, viz.

\frac{1}{3}=0.\dot{3}

and

\frac{1}{7}=0.\dot{1}4285\dot{7}

indicating that this sequence goes on indefinitely, without terminating.

Original post by Dj.Clay

Isn't that how everyone writes it? :/

I would imagine that Ghostwalker is commenting on how people tend to express their recurring decimals are "0.9999999" or "0.999......" here, and is casually pointing out how it should actually be. (And conveniently providing the necessary LaTeX code, too!)

0.89 recurring as a fraction

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