Where do 1.0000001 and 0.9999999 come from?

Plug a large number and you will have an idea (a pattern) what is going on in \(\displaystyle \infty\). Say \(\displaystyle 9999999\).

\(\displaystyle \frac{\ln(9999999 + 1)}{\ln(9999999)} \approx 1.000000006 \approx 1.00000001\)

\(\displaystyle \frac{(9999999 + 6)}{(9999999 + 7)} \approx 0.9999999\)

I should have written \(\displaystyle 1.00000001\) (7 zeros) instead of \(\displaystyle 1.0000001\) (6 zeros). But it does not matter much as we are here to compare:

\(\displaystyle 0.99999990\)

\(\displaystyle 1.00000001\) (We care only for the last digit here \(\displaystyle \geq\) 1)

\(\displaystyle 1.00000001 \times 0.99999990 < 1\)

If the numbers were, for example,

\(\displaystyle 0.99999999\)

\(\displaystyle 1.00000001\)

As long as the last digit is \(\displaystyle 1\) or \(\displaystyle 0\),

\(\displaystyle 1.00000001 \times 0.99999999 < 1\)

If the last digit is not \(\displaystyle 1\), we look at the digit above it. If it is \(\displaystyle 9\), then we have an expression \(\displaystyle > 1\). If the last digit is not \(\displaystyle 1\) and the digit above it is \(\displaystyle 0\), then we have an expression \(\displaystyle < 1\).

These are the most frequent cases that you will encounter:

a. (\(\displaystyle < 1\))

0.99999999

1.00000001

b. (\(\displaystyle > 1\))

0.99999999

1.00000002 (or any number above 1)

c. (\(\displaystyle < 1\))

0.99999990

1.00000001 (or any number above 1)

The position of the digit \(\displaystyle 1\) is not the last. To understand this, I will give an example below.

d. (\(\displaystyle > 1\))

0.99999999

1.00010000 (or any number above 1)

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Let us flip the series upside down and see what happens.

\(\displaystyle \frac{\ln(9999999)}{\ln(9999999+1)} \approx 0.99999999\)

\(\displaystyle \frac{(9999999 + 7)}{(9999999 + 6)} \approx 1.0000001\)

We are comparing:

0.99999999

1.00000010 (the position of the digit \(\displaystyle 1\) is not the last)

We have an expression \(\displaystyle > 1\).