# Thread: A more concise and intuitive demonstration that 0.9999...=1

1. ## A more concise and intuitive demonstration that 0.9999...=1

Most of the explanations I've seen of why 0.999...repeating equals 1.0 have seemed a little esoteric, but I just thought of a really simple demonstration showing that they're equal

Subtract 1 (leaving you with 0.1111...repeating)

and you see immediately that it gives you back the original number  Reply With Quote

2. ## Re: A more concise and intuitive demonstration that 0.9999...=1

Esoteric?

0.333.... = 1/3.

3 x 1/3 = 1, so 3 x 0.333.... = 1, so 0.999.... = 1.  Reply With Quote

3. ## Re: A more concise and intuitive demonstration that 0.9999...=1 Originally Posted by Bohandas Most of the explanations I've seen of why 0.999...repeating equals 1.0 have seemed a little esoteric, but I just thought of a really simple demonstration showing that they're equal

Subtract 1 (leaving you with 0.1111...repeating)

and you see immediately that it gives you back the original number Originally Posted by Peelee Esoteric?

0.333.... = 1/3.

3 x 1/3 = 1, so 3 x 0.333.... = 1, so 0.999.... = 1.

It's worse than that, it's wrong Jim.

0.1111 + 0.9999 = 1.1111. Which may be what was meant by "the original number", but it wasn't immediately clear to me that that was what was meant.  Reply With Quote

4. ## Re: A more concise and intuitive demonstration that 0.9999...=1 Originally Posted by halfeye 0.1111 + 0.9999 = 1.1111. Which may be what was meant by "the original number", but it wasn't immediately clear to me that that was what was meant.
Yes. It starts with 1.1111.... and ends with 1.1111....  Reply With Quote

5. ## Re: A more concise and intuitive demonstration that 0.9999...=1 Originally Posted by Bohandas Subtract 1 (leaving you with 0.1111...repeating)

and you see immediately that it gives you back the original number
I don't think this is a good proof. It looks to me that you're assuming what you're trying to prove on line 3. If .9 repeating doesn't equal 1, then adding it wouldn't result in the original number.  Reply With Quote

6. ## Re: A more concise and intuitive demonstration that 0.9999...=1 Originally Posted by Bohandas Most of the explanations I've seen of why 0.999...repeating equals 1.0 have seemed a little esoteric, but I just thought of a really simple demonstration showing that they're equal

Subtract 1 (leaving you with 0.1111...repeating)

and you see immediately that it gives you back the original number
From what I can tell, proving that 0.9999... + 0.1111.... = 1.1111.... requires*** some way of working with sums of infinite geometric series, and if you can do that it's more straight forward to just directly prove that 0.9999... = 1.

***Maybe it doesn't but I suspect other methods would be even more esoteric  Reply With Quote

7. ## Re: A more concise and intuitive demonstration that 0.9999...=1

(X × 10 - X)/9 = X
Correct?

Now use 0.9999... for X

0.9999... × 10 = 9.9999...
9.9999... - 0.9999 = 9
9 / 9 = 1

Simple, right?  Reply With Quote

8. ## Re: A more concise and intuitive demonstration that 0.9999...=1 Originally Posted by Rogan (X × 10 - X)/9 = X
Correct?
Uncertain. Need proof.   Reply With Quote

9. ## Re: A more concise and intuitive demonstration that 0.9999...=1 Originally Posted by Rogan (X × 10 - X)/9 = X
Correct?

Now use 0.9999... for X

0.9999... × 10 = 9.9999...
9.9999... - 0.9999 = 9
9 / 9 = 1

Simple, right?
It may be because it's been more than a decade since I've done algebra, but I'm getting x=x  Reply With Quote

10. ## Re: A more concise and intuitive demonstration that 0.9999...=1

Between any two distinct numbers, there are infinite other numbers. (e.g. between 1.3 and 1.4 is 1.31, 1.32, 1.34593875834754985398475, etc)

If .999... does not equal 1, you should be able to name at lest one of the numbers in between.

You can't, therefore they must be the same number, written in two different ways.

Nothing esoteric about that, I feel.

Grey Wolf  Reply With Quote

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