What is a proof simple enough for 6th graders but an undeniable one too, that .999...=1.?

Question

Don't put .111...=1/9, etc. and only answer if you believe me.

Don't put .111...=1/9, etc. and only answer if you believe me.

Edit: Don't use the x=.999..., 10x=9.999..., etc. either

Answer 1

x = 0.9999....
10x = 9.9999....
10x - x = 9.9999...-0.9999... = 9
9x = 9
x = 1

OK, if you're gonna be that way:

If you have any two real numbers a < b, you can always find a third number c with a
So suppose 0.999...< 1. Then there's a number that's greater than each of .9, .99, .999, .9999, ... etc., but less than one. Say it's x = 0.9999....99abc..., where a occurs in the nth place after the decimal point, and a is a digit not equal to 9. Then a<9, and so 0.9999...999, with n "9s" after the decimal point, is clearly greater than x. Contradiction. So 0.999... can't be less than 1, and clearly (I hope) it can't be greater, so it must be equal.

Answer 2

Before you are ready to accept a proof in mathematics, you need to have a clear understanding of what the terms involved mean.

What I want you to focus on is, what *is* .9999...? Or any decimal expansion, like 3.1415... or whatever.

A number doesn't change. It's not 0.9 one moment and then 0.99 the next, as if some outside person is changing it by contemplating it. It's stuck somewhere on the number line, no matter how you describe it.

And if I ask, what is the difference between 1 and 0.9999...., it's going to have to be a fixed number too, not some "infinitely small" thing. There has to be an answer; you can always subtract numbers. And it's easy to see, as many have pointed out, that the difference, if positive, would have to be smaller than any other positive number. If not positive, it must be zero.

Can it be that there is a positive number smaller than any other positive number? No, then it would be smaller than itself (or half itself). Contradiction.

Here's another perspective: do you believe that 1/2 + 1/4 + 1/8 + 1/16 + ...=1? (Classic Achilles & the Hare problem.)
That's the same as saying 0.11111.... in binary is equal to 1. Your problem is the same idea base ten.

Answer 3

You are wrong. Every 6th grader knows that 0.999.. is not the same as 1 and so do you.
It is ALMOST 1. And that is your proof. Is is so almost 1 it is better to write it as 1.

Tear a very small corner of a newspaper or remove the dot of one 'i' from a book. Does it matter to the book it is missing that small piece? No, it is still one book. Even the 'i' is still an 'i'.

This is not so much about math as it is about concept. How small is small? But then again.. In some math it is important to have these very small numbers.

Answer 4

Puppy_Zwolle is suffering from an affliction in mathematics that I refer to as "Zenophobia": the fear of converging sequences.

.999...=(9/10)+(9/100)+...=lim n->oo [SUM i=1 to n: 9/(10^n)]. And this is equal to {lim n->oo [SUM i=0 to n: 9/(10^n)]}-9=1/(1-(9/10))-9=1.

Answer 5

Though the difference is negligible, they are not equal.
.999 can only be rounded to 1.

Answer 6

Don't know much about 6th graders ...

If the difference between two numbers is zero, then the two numbers have the same value. Is that too complex for 6th grade? Then, if they can accept that, follow up with this:

0.0000 ... ... 00001 = 0.0

Answer 7

c = 0.999999....

10c = 9.9999999....

10c - c = 9

9c = 9

c = 1

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