To prove that, we will find a number between 0.9999... and 1.
0.9999... = \sum{infinity|n=0}\frac{9|10\power{\power{}n}}
0.9999... = \frac{10\power{\power{}n}-1|10\power{\power{}n}}
Let 0.TTTT... be the number that satisfies the following:
(In case you do not get it, we have changed radix from 10 to 20.
Which means that T=19, TT = 19 x 20 + 19.)
0.TTTT... = \sum{infinity|n=0}\frac{19|20\power{\power{}n}}
0.TTTT... = \frac{20\power{\power{}n}-1|20\power{\power{}n}}
Now let us take the difference between 0.TTTT... and 0.9999...
0.TTTT... - 0.9999... =
\sum{infinity|n=0}\frac{20\power{\power{}n}-1|20\power{\power{}n}} - \frac{10\power{\power{}n}-1|10\power{\power{}n}} =
\sum{infinity|n=0}\frac{20\power{\power{}n}-1|20\power{\power{}n}} - \frac{2\power{\power{}n}(10\power{\power{}n}-1)|2\power{\power{}n}10\power{\power{}n}} =
\sum{infinity|n=0}\frac{20\power{\power{}n}-1|20\power{\power{}n}} - \frac{20\power{\power{}n}- 2\power{\power{}n}|20\power{\power{}n}} =
\sum{infinity|n=0}\frac{2\power{\power{}n} - 1|20\power{\power{}n}} > 0
So 1 > 0.TTTT... > 0.9999... So 0.9999... < 1 and all infinite numbers that are pure decimals with no unit will never be equal to one, as you can always find another number closer to one.
0.9999... = \sum{infinity|n=0}\frac{9|10\power{\power{}n}}
0.9999... = \frac{10\power{\power{}n}-1|10\power{\power{}n}}
Let 0.TTTT... be the number that satisfies the following:
(In case you do not get it, we have changed radix from 10 to 20.
Which means that T=19, TT = 19 x 20 + 19.)
0.TTTT... = \sum{infinity|n=0}\frac{19|20\power{\power{}n}}
0.TTTT... = \frac{20\power{\power{}n}-1|20\power{\power{}n}}
Now let us take the difference between 0.TTTT... and 0.9999...
0.TTTT... - 0.9999... =
\sum{infinity|n=0}\frac{20\power{\power{}n}-1|20\power{\power{}n}} - \frac{10\power{\power{}n}-1|10\power{\power{}n}} =
\sum{infinity|n=0}\frac{20\power{\power{}n}-1|20\power{\power{}n}} - \frac{2\power{\power{}n}(10\power{\power{}n}-1)|2\power{\power{}n}10\power{\power{}n}} =
\sum{infinity|n=0}\frac{20\power{\power{}n}-1|20\power{\power{}n}} - \frac{20\power{\power{}n}- 2\power{\power{}n}|20\power{\power{}n}} =
\sum{infinity|n=0}\frac{2\power{\power{}n} - 1|20\power{\power{}n}} > 0
So 1 > 0.TTTT... > 0.9999... So 0.9999... < 1 and all infinite numbers that are pure decimals with no unit will never be equal to one, as you can always find another number closer to one.
