Is this mathematical equation correct, if so how is this possible?

Question

(Would prefer a Math professors opinion)

1.Lets say variable x = 0.9999......[meaning its a repeating number]

2.Now lets times both sides by ten.
10x = 9.9999.......

3.Now lets subtract x from both sides.
(10x) - x = (9.9999.......) - x

4. Since x = 0.9999......[repeating]
(10x) - x = (9.9999......) - (0.9999.....)

5.Substract 9.9999..... from 0.9999......
(10x) - x = 9

6. Now substract x from 10x
9x = 9

7. Divide both sides by 9
x = 1

Does this mean 0.9999..... equals 1
I would think that 0.9999..... is a number that is infinitely smaller than 1 by 0.1111.....[repeating] or is there some sort of mathematical phenomenon that I'm unaware of.

CORRECTION: to my last sentence, I now realize thats a worng statement. It would require a infinite amount of 0's after the decimal and before a number "1" inorder to make that statement true.

My thought process from stpe 4 to 5 is this:
9.9999.... = 9 + 0.9999......

Therefore:
(9 + 0.9999.....) - 0.9999..... = 9

Im starting to realize now that there other concepts related to this equation. Don't just mention them, elaborate please.

Answer 1

Yes, 0.9999...(recurring FOREVER) is mathematically IDENTICALLY equal to 1.

Here's a shorter way to show it, using just the properties of decimal numbers:

1/9 = 0.11111111... recurring; therefore 9/9 = 0.99999999... recurring. But 9/9 = 1. Therefore 0.99999999... recurring = 1. QED.

(I think you'll agree that this is a much shorter way of showing the same thing?)

I could stop here, but yours is in fact rather an interesting, ingenious and amusing, alternative way of showing that it must be so. (However, there's a written error near the end; instead of "...smaller than 1 by 0.1111.....[repeating]" you mean "... smaller than 1 by 0.0000.....[repeating]" [implicitly with a "1" somewhere infinitely down the line!])

Although I'm content with my own proof from considering 9/9 above, more solid mathematical reasoning related to an important method of arguing limits and convergence etc. is this:

If two numbers are NOT equal, you can write down their FINITE difference, agreed? O.K., so consider several successive subtractions of the following decimal numbers from 1.000... :

0.90000... gives 0.10000... ,
0.99000... gives 0.01000... ,
0.99900... gives 0.00100... ,
0.99990... gives 0.00010... ,
*
*
0.99999999999999999990... gives 0.00000000000000000010... , etc.

But this table extends INFINITELY FAR to both the bottom and the right, with NO LAST ENTRY. If you nevertheless try to imagine a last entry on the right-hand side (the DIFFERENCE between 1 and the 0.9999... etc.), it must start with an INFINITE NUMBER OF 0's after the decimal point.

Therefore it's NOT a NEGATIVE number, yet it's SMALLER than ANY POSITIVE NUMBER YOU CAN NAME BEFOREHAND. You name any positive number as small as you please, and I can tell you how many 9's to write down so that the difference from 1 will be SMALLER than that. YOU set the target small difference, but I can beat it --- EVERY TIME. Then it must be zero --- there's no where else it can hide!

There is in fact an interesting and important branch of mathematics, called ANALYSIS, in which showing that something or some difference can be made LESS THAN ANY SMALL, PREVIOUSLY CHOSEN TARGET NUMBER plays the major role in determining things like convergence, continuity and differentiability, etc. The idea I've expressed above uses essentially that same argument.

Nevertheless, your slightly flawed way was very good and stimulated me to think about something I first learned about 50 years ago!

Live long and prosper.

P.S. By the way, numbers like 0.9999... [recurring] have lots of fascinating connections. They go back to topics like "The Hare and the Tortoise," or Xeno's Paradox, i.e. the idea (very strange to most Ancient Greeks) that the sum of an infinite geometric series could nevertheless converge to a finite sum. (Think about, and write down, the formula for summing 0.9 + 0.09 + 0.009 + ... . Clearly this sum is also 0.9999...[recurring]. However, you'll find once again that the standard method for summing that series also gives you --- just exactly : 1).

Other connections are to such diverse questions as to when the hands of an analogue clock overlap, the relationships between Synodic and Sidereal Periods of the Moon and the Planets, and ultimately the great flowering of modern science in the 17th century and the rise of our present day technological society. Truly quite a vista of apparently disjoint topics, all connected together by a great historical web of related reasoning!

Answer 2

That's correct. That's one of two ways to show it. The misconception comes from your statement
'infinitely smaller than 1'. The decimal 0.9999.... just looks like it's a little smaller than one. Consider what it is within the context of our place value number system. It's an infinte sum of

9(1/10 + 1/100 +
1/1000 + 10^(-4) + ...+ 10^(-k) + .... on to k-> inf )

This is a geometric series which can be summed. This results in the second way to prove it.

This question gets asked about once every 10 days or so on this site.

Answer 3

0.9 repeating is equal to 1.

I warn you that your above work it not proof of this. In going from step 4 to 5 you actually make some assumptions that are not obviously true.

The real proof that 0.9 repeating = 1 deals with real analysis and the concept of a limit.

Answer 4

you made your mistake in step 5
it should be 9x = 9

remember, an equation is just a statement.
10 x 0.999999999 - 0.99999999 = 9.9999999 - 0.99999999

since 9.9999999 = 10 x 0.99999999
they are equal

by the way, you can always find a number between 0.999999999 and 1.000000000 there is actually an infinite amount of number... such as 0.999999999999999999999999923232432

Answer 5

0.99999... is only .00000000000000000....infinite zeros...1
less than 1.
And, I think that you cannot correctly subtract a reapeating number from another.

Answer 6

you will never be able to find a number between 0.99999.... and 1, therefore we put 0.99999.... similar to 1