- #1

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0.99sq = 0.9801

0.999sq = 0.998001

Etc.

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

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- #1

- 8

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0.99sq = 0.9801

0.999sq = 0.998001

Etc.

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

- #2

DaveC426913

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BTW, it works for numbers greater than 1 too.

- #3

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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 contradictionisthe least sig dig, no other digits will affect the last digit in the answer. so...

BTW, it works for numbers greater than 1 too.

- #4

berkeman

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And Welcome to the PF, BTW.

- #5

DaveC426913

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

One squared doesn't end in zero any more than any

One squared ends in one.

.9

1

or

.9

1

Take your pick. But be consistent.

- #6

fresh_42

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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.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

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

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- #9

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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.

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

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##.

- #10

jbriggs444

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

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.

- #11

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Huge LOL!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.

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.

- #12

jbriggs444

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

Both expressions are equally valid.

- #13

Mark44

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

Both expressions are equally valid.

- #14

DaveC426913

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Last edited:

- #15

Mark44

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Er, this is Physics ForThis is Physics Forms.

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Then how do you multiply infinitiesIn 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##.

- #17

jbriggs444

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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)?Then how do you multiply infinities

- #18

fresh_42

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By ##\lim_{n \to \infty} a_n \cdot \lim_{n \to \infty} b_n = \lim_{n \to \infty} (a_n\cdot b_n)\,.##Then how do you multiply infinities

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

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).

- #20

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That seems to result in the same problem.By ##\lim_{n \to \infty} a_n \cdot \lim_{n \to \infty} b_n = \lim_{n \to \infty} (a_n\cdot b_n)\,.##

- #21

fresh_42

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We should distinguish whether we talk about standards analysis or about hyperreals. But, please, not both at the same time.That seems to result in the same problem.

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They should both work right? Neither should break math because of this trivial arithmetic problem.We should distinguish whether we talk about standards analysis or about hyperreals. But, please, not both at the same time.

- #23

fresh_42

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

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The hyperreals are not the subject for a "B" level thread. They are beyond my knowledge, for one thing.They should both work right? Neither should break math because of this trivial arithmetic problem.

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

- #25

Mark44

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"lowest birthday" ??? What does that mean?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.

I'm not an expert on the hyperreals, but to the best of my knowledge, 0.000... ...001 is

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