0.999 … = 1

OldTrout, Going Postal

It really does.

Many people have a problem accepting this equality.

The ellipsis denotes that the nines continue forever; there is no last nine.

Jen the Blue brought this up one night recently.  I didn’t see it at the time and was asked by Upset if it was true.

 

We can go through some arguments.  There are many.

 

\(1)\) An Arithmetical Argument.

Do you accept that \(1/3 = 0.333 …\)?

Do you accept that \(1/9 = 0.111 …\)?

Multiply (an arithmetic operation) the first by \(3\) :

Thus, \(1 = 0.999 …\).

Multiply the second by \(9\) :

Thus, \(1 = 0.999 …\).

Still not convinced?

 

\(2)\) An Algebraic Argument.

Define :

Let \(x = 0.999 …\)

Multiply both side by \(10\) :

\(10x = 9.999 …\)

Separate the RHS :

\(10x = 9 + x\)

Subtract \(x\) from both sides :

\(9x = 9\)

Divide both sides by \(9\) :

\(x = 1\)

 

\(3)\) A Pre-Calculus Argument

A series is the sum of the terms of an infinite sequence of numbers.

A series is convergent if the sequence of its partial sums tends to a limit.

We can express \(0.999 …\) as

\(0.999 … = \frac{9}{10} + \frac {9}{100} + \frac{9}{1000} + \frac{9}{10000} +…\)

A geometric series is of the form

\(a + ar + ar^2 + ar^3 + …\)

where \(a\) is the first term and \(r\) is the common ratio.  When \(-1 < r < 1\) the series converges to a limit given by the formula  \(\displaystyle \frac{a}{1-r}\).

Our number \(0.999 …\) is a geometric series with first term \(\frac{9}{10}\) and common ratio \(\frac{1}{10}\).

Substituting these into the formula we have

\(\displaystyle \frac{9/10}{1-1/10} = \frac{9/10}{9/10} = 1\)

 

\(4)\) A Wolfram Alpha Argument.

Type \(0.999 …\) into Wolfram Alpha and it returns a value of \(1\).

OldTrout, Going Postal
 

© OldTrout \(2018\)
 

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