Does .9-repeating equal 1?

[Lul Thyme]

Originally posted by KrazyHorse
Here's a clever problem I came across the other day:

I have a punch which is very special. If I center it over a point then it removes all the points in the real plane (RXR) which are an irrational distance away from the original point.

How many punches do I need in order to remove the whole real plane?

EDIT
made a mistake assuming you couldnt center at a number already removed....



didnt actually prove it but its probably n+1 in R^n.

[Addled Platypus]

its like asking what is more

few
several
couple

[Cort Haus]

A couple is two. Then few, then several.

[Dracon II]

Wow... another paradox

[loinburger]

Originally posted by Ben Kenobi
They aren't the same.

n(n)^x where n is 1 is always one.

lim x -> inf will be 1.

What is the value for

n(n)^x where n is 0.9 repeater?

what about the lim x -> inf? that should be 0.

That's rubbish. You're saying "assuming that .9-repeating is less than 1, then .9-repeating^x approaches zero as x approaches infinity, therefore .9-repeating is less than 1."

[Donegeal]

Stupid ASCII code...

1 >= .9-repeat

[Lul Thyme]

Originally posted by chegitz guevara


No, it's from high school. At first glance, it doesn't seem to make sense. People want to inherently believe in an infinitesimal, but 0.0...1 doesn't exist. It can't exist. 0.9... must be 1, because there is nothing you could add to the number to make it equal to 1.

You are right that there is no smallest strictly positive real number.
There are number systems that have been created and used where such infinestimals exist and in those systems in fact 1 and 0.9... would be different.
Basically it comes down to wheter you accept infinestimals in your system.

In real numbers though, 1=0.99...

[Eli]

I find this question to be very disturbing.

As nice as all the proofs on wikipedia are, a number cannot equal another number when they do not have exactly the same digits everywhere.

[Lul Thyme]

Originally posted by Eli
I find this question to be very disturbing.

As nice as all the proofs on wikipedia are, a number cannot equal another number when they do not have exactly the same digits everywhere.

½Students are often "mentally committed to the notion that a number can be represented in one and only one way by a decimal." Seeing two manifestly different decimals representing the same number appears to be a paradox, which is amplified by the appearance of the seemingly well-understood number 1.½

http://en.wikipedia.org/wiki/.999...

[Lul Thyme]

Originally posted by Dracon II
Its an asymptote, no? It comes infinitely close, but never reaches... like Zeno's paradox.

That's actually the main mistake I've seen in layppl talking about this.
They assume we are talking about the sequence
0.9
0.99
0.999
...
which never reaches 1

But we are talking about THE UNIQUE NUMBER
which is the limit of the sequence.
We write it 0.99....
and it is EQUAL to 1.

[civman2000]

Originally posted by Lul Thyme

EDIT
made a mistake assuming you couldnt center at a number already removed....



didnt actually prove it but its probably n+1 in R^n.

Spoiler:

No, the solution with 3 points works for all dimensions. Though obviously only 2 works in 1 dimension.

[Drogue]

Originally posted by Kuciwalker
At the point where you put the little bar over the 9.

Damn you American's and your silly notation. Dots show recurring decimals in the real world.

[Drogue]

Originally posted by Eli
I find this question to be very disturbing.

As nice as all the proofs on wikipedia are, a number cannot equal another number when they do not have exactly the same digits everywhere.

No, that breaks down when the word "infinite" is inserted. 1/3 is exactly equal to 0.(an infinite number of 3s). Similarly, 1 is exactly equal to 0.(an infinite number of 9s). It's not even an assymptote, as an assymptote would be saying that 0.99... get's closer to 1 as you add more 9s - it tends to 1. However at the point where the 9s become infinite, it is 1, it doesn't tend to it.

Another relatively undisputable answer that shows that numbers can be the same even if they have different digits, take the equation:

1+1+1+1+1... to infinity. It equals the same as:
9+9+9+9+9... to infinity. It doesn't matter that the latter is 9 times as big at every point other than infinity, when both are declared to be infinite, they both cease to be different.

[Spec]

No. It depends on the application.

Spec.

[RGBVideo]

x = 0.99... ||×10
10x = 9.9... ||-x
10x-x = 9.9... - 0.99...
9x = 9 ||÷9
x = 1 = 0.99... QED