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/sci/ - Science & Math

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2093065 No.2093065 [Reply] [Original]

Math proof time

do it faggots

>> No.2093074

1/3 = .33333 forever
2/3 = .66666 forever
3/3 = .99999 forever
1 = .999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
9

>> No.2093073

1/3 = .3333
.3333 x 3 = .99999

>> No.2093078

shut up dumbfag

>> No.2093075

You've ensured an even infinite number of nines.

This relation only works for an odd infinite number.

Clever troll, OP.

7/10

>> No.2093081

x = 0.99...
10x = 9.99...
10x - x = 9.99.. - 0.99... = 9
9x = 9
x = 1
aggle baggle

>> No.2093109

0.9999... = 9 * sum(1/10^n) = 9 * 1/(1-1/10) = 9 * 1/9 = 1.

>> No.2093959

>>2093081
>>2093081

this

>> No.2093966

0.99... is defined to be the limit of the sequence
0.9, 0.99, 0.999, 0.9999, ....
Which is indeed 1.

>> No.2093978

Next problem:
Prove that there exists a equation of multiple variables multiplied and added together, but no division (for example ab+c=d^2-abcd)
such that there is a solution in the integers if and only if the variable labeled "a" is a prime number.

>> No.2093981

<div class="math">0.999 \dots= \lim_{n\to \infty} 0.\underbrace{99 \dots 9\,}_{n} = \lim_{n\to \infty} \sum_{k=1}^n \frac{9}{10^k} = \lim_{n\to \infty} (1 - \frac{1}{10^n}) = 1 - \lim_{n\to \infty} \frac{1}{10^n} = 1</div>

>> No.2093982

>>2093075

infinity is both even and odd while being neither... mind=blown

>> No.2094025

>>2093978
I can prove that's impossible. Such an equation will have a finite number of roots. There are an infinite number of prime numbers.

>> No.2094027

my teacher told me that it's not actually equal.
if you have high school education you should've learnt something called arithmetic sequences or something.
like how to calculate
3 + 6 + 12 + 24 + ...... 1536
in that example the "ratio" is 2

when the ratio is between -1 and 1
i.e. -1 < d < 1
you can actually find the limit that it approaches
0.999999.... is the same as

0.9 + 0.09 + 0.009 + 0.0009 + .......

Plug it into the formula a/1-r
where a = first number in the sequence (0.9)
r = ratio (1/10)

= 1
The formula calculates what the pattern approaches not what it equals to.

my teacher also gave me this analogy.
suppose you have a person who can only always walk 1/10 of the distance he walked previously. His first "walk" is 0.9m (thus his second "walk is 0.09m, third = 0.009m etc.)
Will the man ever reach and be able to step inside the frame of a door 1m away?

>> No.2094050

>>2093065
So in theory no integers exist.

>> No.2094051

>>2094027
>The formula calculates what the pattern approaches not what it equals to.

The number 0.99.... is defined to be the limit of the pattern. It doesn't mean anything but the limit of the pattern.

There are some number systems where numbers really can be infinitely close to other numbers, but the reals isn't one of them.

>> No.2094055

>>2094051
>approaches
>limit

approaches means what it goes to but never reaches
limit means what it can never reach

>> No.2094057
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2094057

>>2094055
......the fuck?

>> No.2094059

>>2094055
Yes and no.

limit of 1/x as x approaches 0? Sure.

limit of x as x approaches 0? It reaches it.

>> No.2094060

>>2094055
>approaches means what it goes to but never reaches
>limit means what it can never reach

Um, no.
A limit of a sequence is a number which the terms in the sequence become arbitrarily close to. It's a theorem that in the reals there can only be at most one limit for a given sequence.

A sequence can reach its limit. For example, the sequence 1,1,1,1... has limit 1.

>> No.2094063

>>2094059
>limit of x as x approaches 0? It reaches it.

No it doesn't.

>> No.2094064

>>2094059
limit of x as x approaches 0? It reaches it.
>as x approaches 0
states by itself that x will never equal to 0 (thus why the word approach is used)

the limit as x approaches 0 is 0
x will never equal to 0 and f(x) will never equal to 0 too
(btw pick better numbers its easier to explain)

>> No.2094071

>>2094064
My apologies then.

>> No.2094082

so 0.99___________
does NOT equal 1

>> No.2094083

>>2094082
Um, yeah it does.
It's defined to be the limit of the sequence.
And the limit of the sequence is 1.

>> No.2094087

0.999... = 0.999...999 = 1 - 0.000...001 (an infantesimal amount)

>> No.2094097

>>2094087

cool intuition bro

>> No.2094102

>>2094087
>0.999... = 0.999...999

THIS IS INCORRECT, THE REST OF YOUR PROOF IS INVALIDATED.

>> No.2094105

>>2093065
for all practical purposes 0,999....... = 1

fuck your theoritical shit

>> No.2094109

>>2094105
Um the theoretical shit says that too.
Everything says that.
It's just a little counterintuitive at first.

>> No.2094113

>>2094102
the final ...999 is at the infinite decimal place. Maybe you misunderstood that part.

>> No.2094116

>>2094113
Infinity is not a number.

>> No.2094117

>>2094116
If infinity is not a number, then you can't have an infinite number of decimal places, then.

>> No.2094118

>>2094113

NO, I UNDERSTAND WHAT YOU MEANT, THE FACT IS THOSE TWO NUMBERS AREN'T EQUAL, BECAUSE ONE ENDS (ARBITRARILY LONG) AND ONE DOESN'T (INFINITELY LONG).

.999[...]999 IS ARBITRARILY LONG, BUT DOES END, AS SHOWN BY THE LAST 9. .999[...] IS INFINITELY LONG, HAVING NO END.

.999[...] - .999[...]999 == .000[...]000999999

THOSE 0S STOP WHERE THE ARBITRARILY LONG NUMBER ENDED.

SEE HOW THEY CAN'T BE EQUAL?

>> No.2094121

>>2094118
You are putting zeros after the infinite decimal place. That is an error because infinity+1=infinity.

>> No.2094124

>>2094117
You can only ever have one decimal place.

>> No.2094125

>>2094121

I CAN PUT ZEROES AFTER IT NO MATTER HOW LARGE EACH [...] IS BECAUSE THERE IS A LAST 9 IN .999[...]999, AS ILLUSTRATED BY IT STOPPING.

[...] COULD BE OF ANY LENGTH (ARBITRARILY LARGE), BUT IT CANNOT BE INFINITELY LONG, BECAUSE IT ENDS.

>> No.2094129

>>2094125
It ends at infinity. That's the definition of infinitely long. I'm sorry you're having such a hard time understanding.

>> No.2094130

>>2094125
Oh I get it now. Thanks for the input.

>> No.2094131

>>2094129
Infinity doesn't end.

>> No.2094132

>>2094129

NO, THAT ISN'T THE DEFINITION OF INFINITELY LONG. OKAY, I'M GOING TO TRY A METAPHOR.

I HAVE TWO MOUNTAINS. ONE IS ARBITRARILY TALL, ONE IS INFINITELY TALL.

THE ARBITRARILY TALL MOUNTAIN HAS A PEAK, IT'S SIMPLY WHATEVER NUMBER I CHOOSE, BUT IT DOES INDEED HAVE A PEAK, AND THIS PEAK IS VISIBLE BECAUSE IT EXISTS.

THE INFINITELY TALL MOUNTAIN HAS NO PEAK, BECAUSE NO MATTER WHAT POINT I DECLARE TO BE THE PEAK, THERE IS A POINT HIGHER THAN THAT, AND A POINT HIGHER THAN THAT ONE, AND SO ON. THERE IS NO PEAK, BECAUSE THERE WILL ALWAYS BE A POINT HIGHER THAN ANY OTHER POINT. SINCE THERE IS NO PEAK, IT'S NOT VISIBLE.

>>2094130

I'M GLAD AT LEAST SOMEONE DOES.

>> No.2094135

>>2093081

This is a very common method of proving that 0.999... = 1.0 but it seems to me that one of the lines in this proof contains an error based on a misconception. I can point it out if anyone's interested.

>> No.2094138

>>2094135
SHOOT, BABY!

>> No.2094142

>>2094132
The infinitely tall mountain would have infinity as its height. Therefore its peak is at h=infinity.

>> No.2094147

>>2094142

NO, YOU IMBECILE, INFINITY IS NOT A NUMBER. IT IS AN ERROR.

IF YOU ARE USING INFINITY, THEN WHAT YOU ARE DOING IS NOT VALID MATHEMATICS.

>> No.2094148

>>2094142
you can't have an end to something if it's infinite

>> No.2094152
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2094152

I love you MR. RAGE will you be my waifu

>> No.2094153

>>2094148
No, but you can have a limit, and that limit is at infinity.

>> No.2094154

>>2094142

YOU'RE GETTING CLOSE, BUT IT'S MEANINGLESS TO STATE THAT A VARIABLE IS EQUAL TO INFINITY, SUCH AS H=INFINITY, BECAUSE ARITHMETIC MANIPULATIONS HAVE NO EFFECT AT ALL ON INFINITY AND CANNOT BE PERFORMED ON IT.

NOW, IT IS MEANINGFUL TO SAY THAT THE HEIGHT OF THE SECOND MOUNTAIN GOES TO INFINITY, FOR IT IS EVER-INCREASING AND HAS NO ARBITRARY MAXIMUM, BUT IT IS NOT EQUAL TO IT.

>> No.2094156

>x = 0.99...
>10x = 9.99...
>10x - x = 9.99.. - 0.99... = 9
>9x = 9
>x = 1

Specifically,
>10x = 9.99...

This step seems to be based on the assumption that you can multiply by ten just by "moving the decimal point" to the right. But that's just a convenient trick. It's not how multiplication actually works.

For instance, anytime you actually multiply a number by ten, the result will end in zero, even decimals. And as MR. RAGE points out, 9.999... has no end. It's not a multiple of ten.

>> No.2094158

>>2094153
You're mixing unrelated concepts.

>> No.2094159

>>2094158
>he doesn't know what a limit is

>> No.2094161

>>2094147
This thread is retarded, but mathematicians use infinite sets all the time. That is not an error.

>> No.2094162

>>2094159
let's say that the function of a mountain's height is x

the mountain's height is equal to x

as x approaches infinity, the limit of the function is infinity

it gets close to infinity, it does not equal infinity

the limit of a function at x is not the same as the function's value at x

>> No.2094163

>>2094159
The precise definition of a limit is that the argument approaches the point. It can never be equal to the point.

>> No.2094164

>>2094156
0.333... x 10 = 1/3 x 10
= 10/3
=3 + 1/3
=3.333...
WHICH DOES NOT END IN A ZERO!!!

>> No.2094165

>>2094159

LIMITS AND EQUALITY ARE DISTINCT, DIFFERENT THINGS.

THE LIMIT OF THE SECOND MOUNTAIN IS INDEED INFINITY, BUT ITS HEIGHT DOESN'T EQUAL ITS LIMIT.

IT *IS* POSSIBLE TO HAVE A SEQUENCE/FUNCTION WHOSE LIMIT IT CAN EQUAL.

FOR EXAMPLE THE SIMPLE SEQUENCE {1, 1, 1, 1, 1, [...], 1} HAS A LIMIT OF 1 AND, IN EVERY STEP, EQUALS 1.

BUT HAVING A LIMIT OF 1 AND EQUALING 1 ARE DISTINCT CONCEPTS, AS ARE EQUALING INFINITY (IMPOSSIBLE) AND HAVING AN INFINITE LIMIT (POSSIBLE).

>> No.2094169

>>2094163

YOU, TOO, ARE CONFUSING LIMITS WITH ASYMPTOTES.

AN ASYMPTOTE IS A SPECIAL TYPE OF LIMIT, THAT A SEQUENCE/FUNCTION APPROACHES BUT NEVER EQUALS.

>> No.2094173

Series of partial sums of the form .9 + .09 + .009 + .... + 9 x 10^(-n)

Difference between each partial sum and 1 goes to zero and n goes to infinity.

>> No.2094174

>>2094147
What the hell is this nonsense? Infinity is not a number, so it doesn't belong in math at all? That's absurd. Mathematics deals with all sorts of things that aren't numbers.

>> No.2094175

>>2094173

as* n goes to infinity.

>> No.2094178

>>2094174

YOU ARE CORRECT. ONE OF THE /sci/ENTISTS CRACKED MY TRIP A LONG WHILE AGO, I GUESS I COAXED HIM OUT.

>> No.2094179

>>2094173
But it GOES TO zero. It never EQUALS zero. It only has zero as its LIMIT. We should write

lim(0.99...) = 1

>> No.2094180

>>2094174
he said that treating infinity as a number is not valid mathematics

>> No.2094183

>>2094179

That's what .99..... means in the first place. The limit of the series of partial sums I described. It's a shorthand for a fairly complex and delicate concept.

>> No.2094186

>>2094183
But you can't write this shorthand because then your notation means two different things

.99... = a number with infinite digits
.99... = the limit of a number as digits approach infinity

>> No.2094190

>>2094180
He said infinity was an error, which is just nonsense. Although I suppose treating infinity as a number is an error. Maybe that's what he meant.

>> No.2094193

>>2094186

In pure mathematics, .99... means the limit of particular a sequence of a terminating decimals, that is .9, .99, .999, .9999 etc. It has no other meaning.

There is no formally defined concept of a infinite decimal. The decimal expansion of pi for example, is a series of approximations, each of finite length.

>> No.2094213

>>2094164

Okay, but if we think about what multiplication really is, then we're talking about adding a number to itself several times. So:

(times four, for instance)
With two digits, .33 + .33 + .33 + .33 = 1.32
With five digits, .33333 + .33333 + .33333 + .33333 = 1.33332
With infinite digits, .333... + .333... + .333... + .333... =
1.333...

Where's the two? When multiplied by four all the answers end in two. With infinite digits, what happens? It doesn't fit the pattern. Likewise, when multiplied by ten, all the answers would end in zero.

So .333... times ten wouldn't be 3.333...