Does Squaring 0.999... Always Result in 1?

[Fala483]

I am trying to prove that all numbers of the form 0.999... Squared end in a decimal value of 1. For example

0.99sq = 0.9801
0.999sq = 0.998001

Etc.

Is it possible to prove for all 0.999... ?

 

[DaveC426913]

Well, in every case, the least sig dig is always 9, which, when squared is 81. And since it is the least sig dig, no other digits will affect the last digit in the answer. so...

BTW, it works for numbers greater than 1 too.

 

[Fala483]

Well, in every case, the least sig dig is always 9, which, when squared is 81. And since it is the least sig dig, no other digits will affect the last digit in the answer. so...

BTW, it works for numbers greater than 1 too.

So since 0.999... Squared ends in a decimal value of 1 and 1 squared ends in a decimal value of 0, 0.999... = 1 is a contradiction

 

[berkeman]

Have you done a forum search about this? I'm pretty sure it's been addressed here many times...

And Welcome to the PF, BTW.

 

[DaveC426913]

So since 0.999... Squared ends in a decimal value of 1 and 1 squared ends in a decimal value of 0...

Wait. What?

One squared doesn't end in zero any more than any other number squared ends in zero - including .999.
One squared ends in one.

.9² = .81
1² = 1

or

.9² = .810
1² = 1.0

Take your pick. But be consistent.

 

[fresh_42]

So since 0.999... Squared ends in a decimal value of 1 and 1 squared ends in a decimal value of 0, 0.999... = 1 is a contradiction

No, because you only squared numbers with finite digits: $0.\underbrace{9...9}_{n \text{ times }}$ which is not the same as if it doesn't end. If it doesn't end, the $1$ at the end will never be reached.

 

[Rada Demorn]

The problem with this thread and others discussing the same proof ground on a common misunderstanding:

Are the expressions: 0.999999... and 0.99999...9 to be considered equal ? Here "..." represent an infinite string of 9's.

 

[hilbert2]

The "0.9999..." would in principle be as correct a decimal expansion of number 1 as "1.0000...", but by convention we generate the decimal expansions of positive numbers by approaching them from the right instead of left side.

 

[PeroK]

The problem with this thread and others discussing the same proof ground on a common misunderstanding:

Are the expressions: 0.999999... and 0.99999...9 to be considered equal ? Here "..." represent an infinite string of 9's.

In your second expression, the last 9 has no meaning. In general, the nth digit is that digit times $10^{-n}$. And, the expression $0.999 \dots$ represents a number that has a $9$ in every place.

Formally, this is $\Sigma_{n = 1}^{\infty} 9 \times 10^{-n}$.

This can be evaluated by anyone who has learned about infinite geometric series.

Your last 9 does not represent $9 \times 10^{-n}$ for any number $n$.

 

[jbriggs444]

In your second expression, the last 9 has no meaning.

To amplify on this: One can attempt to create a system of arithmetic that works on unconventional strings with infinite sequences of digits "in the middle". But it is harder than it looks. The conventional scheme which only permits infinite sequences "at the end" is used for good reason.

One way of understanding the standard system is that decimal strings are a convenient notation to express exemplars of equivalence classes of Cauchy sequences of rationals. A course in real analysis would show that equivalence classes of Cauchy sequences of rationals are one of the standard ways to construct the Reals.

 

[Rada Demorn]

In your second expression, the last 9 has no meaning. In general, the nth digit is that digit times 10−n10−n10^{-n}. And, the expression 0.999…0.999…0.999 \dots represents a number that has a 999 in every place.

Formally, this is Σ∞n=19×10−nΣn=1∞9×10−n\Sigma_{n = 1}^{\infty} 9 \times 10^{-n}.

This can be evaluated by anyone who has learned about infinite geometric series.

Your last 9 does not represent 9×10−n9×10−n9 \times 10^{-n} for any number nnn.

Huge LOL!

What about 0.9999...99 ? What about 0.9999...999 etc?

Both expressions are equally valid. The only difference is that you construct 0.9999...9 without the addition: ie. instead of 9/10+9/100+9/1000... you just compute 9/10, 99/100, 999/1000 etc. 0.9999...9 is just a backward construction.

 

[jbriggs444]

What about 0.9999...99 ? What about 0.9999...999 etc?

Both expressions are equally valid.

Indeed they are equally invalid. No such sequences have any valid interpretation as real numbers that gives meaning to the finite trailing sequence of digits.

 

[Mark44]

What about 0.9999...99 ? What about 0.9999...999 etc?

Both expressions are equally valid.

As @jbriggs444 already said, both expressions are invalid. If you write 0.9999..., that means that an infinite number of 9 digits follow, but in the meaningless expression 0.9999...9, at which position is that final 9?

 

[DaveC426913]

Huge LOL!

This is Physics Forums.
We don't LOL here.

 

[Mark44]

This is Physics Forms.

Er, this is Physics Forums...

 

[Fala483]

In your second expression, the last 9 has no meaning. In general, the nth digit is that digit times $10^{-n}$. And, the expression $0.999 \dots$ represents a number that has a $9$ in every place.

Formally, this is $\Sigma_{n = 1}^{\infty} 9 \times 10^{-n}$.

This can be evaluated by anyone who has learned about infinite geometric series.

Your last 9 does not represent $9 \times 10^{-n}$ for any number $n$.

Then how do you multiply infinities

 

[jbriggs444]

Then how do you multiply infinities

Are you asking for how to a prescription for multiplication of normal one-way infinite digit strings? (The sort that do not end with ...9)?

 

[fresh_42]

Then how do you multiply infinities

By $\lim_{n \to \infty} a_n \cdot \lim_{n \to \infty} b_n = \lim_{n \to \infty} (a_n\cdot b_n)\,.$

 

[Fala483]

As @jbriggs444 already said, both expressions are invalid. If you write 0.9999..., that means that an infinite number of 9 digits follow, but in the meaningless expression 0.9999...9, at which position is that final 9?

The number won't be a real number. It might still be a more general kind of number.

Now there is the corresponding problem in surreal numbers. If we allow also transfinite integers as the integers in rational numbers of the form a/b does some "middle infinite" decimal expansion correspond to such a transfinite rational. In fact the transfinite integer with the lowest birthday ω={1,2,3,4...|} can give raise to 1/ω = ε which might be argued to have the decimal expansion of 0.000... ...001. However with the relevant definition of rationals this is a rational number. It might be a hint that any infinite end would only be a finite real multiple of ε. But then there is the trouble that 1/ε is also a valid surreal. Thus for those kind of expanded decimal notions you migth need a notion of size of the infinities to make things work out (that is adding countably many infinite digits might be a different thing than adding uncountably many).

 

[Fala483]

By $\lim_{n \to \infty} a_n \cdot \lim_{n \to \infty} b_n = \lim_{n \to \infty} (a_n\cdot b_n)\,.$

That seems to result in the same problem.

 

[fresh_42]

That seems to result in the same problem.

We should distinguish whether we talk about standards analysis or about hyperreals. But, please, not both at the same time.

 

[Fala483]

We should distinguish whether we talk about standards analysis or about hyperreals. But, please, not both at the same time.

They should both work right? Neither should break math because of this trivial arithmetic problem.

 

[fresh_42]

They should both work right? Neither should break math because of this trivial arithmetic problem.

It is not about right and wrong, it is about a common basis for debate. Otherwise this is a cheap trick to reason and change cloths whenever suited.

 

[PeroK]

They should both work right? Neither should break math because of this trivial arithmetic problem.

The hyperreals are not the subject for a "B" level thread. They are beyond my knowledge, for one thing.

And, they are not an easy way out of the hard realities of standard analysis.

 

[Mark44]

In fact the transfinite integer with the lowest birthday ω={1,2,3,4...|} can give raise to 1/ω = ε which might be argued to have the decimal expansion of 0.000... ...001.

"lowest birthday" ? What does that mean?

I'm not an expert on the hyperreals, but to the best of my knowledge, 0.000... ...001 is meaningless, whatever the number system happens to be. The most significant problem with it is that you can't say exactly where (what index) the 1 digit is. You can't simply say that it is an infinite number of places out in the decimal representation.

 

[Mark44]

This was post #1 in the thread, marked as "B" (basic, or high school level)

I am trying to prove that all numbers of the form 0.999... Squared end in a decimal value of 1. For example

0.99sq = 0.9801
0.999sq = 0.998001

Etc.

Is it possible to prove for all 0.999... ?

Yes, using limits on finite sequences of decimal fractions.

From post #19:

The number won't be a real number. It might still be a more general kind of number.

From post #1, the context was that you were dealing with real numbers, and there was no mention that you weren't talking about any other kinds of numbers.

Based on your other thread, https://www.physicsforums.com/threads/solve-this-math.932820/, which is closed, it appears that you are trying to prove that $(0.999...9)^2$ is something other from 1, with the aim at showing that $0.999 \dots \ne 1$. If so, you're barking up the wrong tree.

Thread closed.