In foundations class

>this isn't even a university I bet

This is a university...

Where else would they be teaching foundations?

1 = 0.999... only if you don't know how to divide.

I don't get this proof. So you let x = 0.44... in one and x = 0.4999.. in the other and then x = 4/9 but x also equals .5?

I think the lower rightmost column stands alone and is read top-down.

this.

>And how are you going to proof this.
By contradiction. Assume 0.999.. < z+0.5 < 1 Consider the first decimal were z+0.5 differs from 0.9999.. that decimal must be different from 9 and therefore lower than 9 and it follows that z+0.5 is smaller than 0.9999..

From my experience the best teachers write on boards instead of just going through slides

On chalkboards, not whiteboards

>On chalkboards, not whiteboards
>Chalkboards
>16th century technology
>Better than Whiteboards invented in the 1950s

Hipster go on /killyourself/

But 0.4999... Is not the same as 0.5 and same applies for 1 and 0.999...

Have you even looked at the proofs? Theyre nonsense

yeah lets have this fucking thread again for the 0.999... millionth time

Nice proof m8 I r8 8/8.

> it's come up with a thousand first way to prove 0.(9) = 1

HOW MANY TIMES FAGGOTS: .(9) = 1 BY DEFINITION OF RATIONALS YOU FUCKING RETARDS

The fuck? The way you construct the reals is literally by identifying sequences like 0.4999... and 0.5. Your teacher should be fired.

This is obvious you buffoon.
Suppose there was number between 0.9... and 1. Call it x, so 0.9... < x < 1.
Clearly, x = 0.x1x2x3... where xi is the number in the ith position following the decimal in the decinal representation of x. If x != 0.9... then there exists an n such that xn != 9. But then x < 0.9, a contradiction. Hence 0.9... = 1.

How can there be number between 0.9.. and 1?

>.(9) = 1 BY DEFINITION OF RATIONALS

By what definition of rationals? Rational numbers are equivalence classes of integers, in the way we defined them.

How does that immediately imply that 0.99... = 1?

In what prehistoric intuitionistic way are you defining rationals?

You obtained a contradiction by assuming that x_n != 0.999, not x_n != 1.

0.999... is equal to 1 in the sense that it would be if we could naturally obtain the infinite sequence of 9s. Consider a string treatment of 0.999...: 0.999... is the infinite-th element in the length-ordered set {0.9, 0.99, 0.999, ...}; we see that, as the index in the order increases to infinity, the value at that index converges to 1. More accurately, rather than saying 0.999...=1, it's more accurate to say that the limit of the sequence s_n= s_{n-1} + 9 * 10^(-n) (s_0 = 0) is 1 as n tends to infinity. From this point of abstraction, you can form your own convenient subjective opinion based on how you feel about infinity.

What the hell is the difference, except that whiteboards are easier to clean?

I thought that rationals were equivalence classes on Z^2, not Z.

Equivalence classes on pair of integers...

>is the infinite-th element in the length-ordered set {0.9, 0.99, 0.999, ...}
please stop talking

Did everyone in Veeky Forums forgot how to math? In math being close to something does not means equals just like limits don't as well.

>4/9 = 0.49...
Maybe you should learn basic arithmetic before trying to do big boy math.

well, they are also cleaner in general, allow better and more understandable drawings, the chalk does not makes your hands dirty, you can rub your eyes after writing without infecting your eyes for two weeks and you can make better multicolor drawings because everyone knows every not-white chalk on black or green boards looks like shit.

Nah, you definitely are fucked if your teacher isnt readily aware of how to convert repeating decimals to fractions. Good luck learning.

Coloured chalk on blackboards is fine. The problem is that the majority of blackboards are actually green and coloured chalk on greenboards is ass cancer

>Suppose... a contradiction

reread my post moron, you completely misunderstood it

>chalkboard are more advanced than whiteboards
Dare I say more?

You must go the absolute shittest of shit-tier universities.

Fess up, OP. What university is this so I can avoid it like the plague?

when we run through cantor's diagonalization argument and produce a new decimal expansion that differs from every element in a supposed list of all real numbers, how do we know that it's genuinely not on the list, i.e. not just an alternative representation of a number already on the list like in this 0.9999 and 1 case?

this is something that has to be explicitly dealt with and it usually isn't.

one thing you can do is just say "if the digit is 1-7, then increase by 1, else make it 2," or similar. Then there are no 9s in the new number.

Different user here that's barely following.
Won't the problem remain with any repeating numbers? Even if they're not 9.

This is a very good question and one not adequately addressed by most presentations of Cantor's argument, imo.

Probably the cleanest way to do is to note that Cantor's proof implies without any problem that the set T of all binary sequences (never mind real numbers for now) is uncountable. Let [math] f: T\to [0,1] [/math] be the function that takes a binary sequence to the corresponding real number (more precisely, f(a1, a2, ...)=0.a1a2...).

This function is at most two-to-one (because a real number can have at most two different decimal expansions).

Now we can show that [0,1] is not countable. Suppose for the sake of contradiction we could enumerate the elements x1, x2, ... We can write [math] T=f^{-1}(x_1)\cup f^{-1}(x_2)\cup\dots [/math]. But each [math] f^{-1}(x_i)[/math] has at most two values, and in particular is finite. Thus we have expressed T as a countable union of finite sets, which implies T is countable, a contradiction.

Diagonal arguments are unfortunately quite subtle.

Try this paper user:
maa.org/sites/default/files/images/upload_library/22/Ford/Gray819-832.pdf

use binary. it can be shown that any number which differs from another by exactly one digit is not that other number. since this is the exact number constructed in the diagonal argument, we are done.

This does not address that user's question at all.

¿A qué universidad vas user?

>maa.org/sites/default/files/images/upload_library/22/Ford/Gray819-832.pdf

why can't we do the same thing with natural numbers and use the diagonal method to construct a transcendental?

Consider that our usual rules are
1) you have a finite number of symbols, read left to right
2) you have a radix point
3) you have a possibly-infinite number of symbols, read left to right

This is just a convention. You could have
1) a finite number of symbols, read right to left
2) a radix point
3) a possibly-infinite number of symbols, read right to left

This would be a p-adic type construction. Then indeed the diagonal argument works on p-adic integers.

It abuses much; but, if you carry out such a construction on the naturals in base 10 you can construct the "number" [math]\overline{9}.0[/math]. If you consider 10-adics (please don't), then this would be -1. -1 is not a natural number, so of course it doesn't appear on your list.

What thr fuck is your math background to think thst is a meme?

>4/9=0.499..
wow

Also since you guys wanted to know

I go to UC Irvine

topkek that is an american university and you can clearly see spanish written in the board, fag.

You are not OP. I know because I am OP and I am NOT fucking telling because I am not retarded. You think I want to turn my university into a meme? Fuck that. I work hard for my shit, y'all boys are just fucking whack

I think he meant by definition of decimal system.

Nice try, I've taken classes there

Actually it works fine in any base OTHER than binary, lol.

Agreed, he should be precise: 0.(9) is the [math]\omega[/math]th element in the ordered set {0.9, 0.99, 0.999, ...}.

>mexican intellectuals

Now prove that the only ambiguous decimal representations are those involving infinite strings of 9s.

>list is 0.1000..., 0.1010..., 0.1100..., 0.0110..., ...
>"new" element generated is 0.0111...
Binary isn't immune to this either, user.

MIT

>He doesn't know in 2016 MIT stands for MEMES AND IT

retard

You're korrect.

>Some proofs that 0.999… = 1 rely on the Archimedean property of the real numbers: that there are no nonzero infinitesimals. Specifically, the difference 1 − 0.999… must be smaller than any positive rational number, so it must be an infinitesimal; but since the reals do not contain nonzero infinitesimals, the difference is therefore zero, and therefore the two values are the same.
>However, there are mathematically coherent ordered algebraic structures, including various alternatives to the real numbers, which are non-Archimedean.
You're literally like ancient greeks realizing there is no natural number that corresponds to the square root of 2 and then insisting that this means the square root of 2 doesn't exist.

Did you even read the post lol

Why do autists get so excited when they correct the professor?

>that is an american university and you can clearly see spanish written in the board, fag.
Sounds about right.

t. Donald 'Builddewall' Trump

While this seems to be generally true, I've always thought that presentation slides have so much more potential for laying out examples and graphs in a clearer or more detailed fashion, and allow more material to be covered in a shorter time because you won't be spending as much time waiting for the guy to finish writing things down or spend half a minute wiping the board.

OP started it

Nice, thank you.

Is this some epic meme that people pretend to be stupid by refusing to believe that 0,999... = 1? I don't get it. Like, haha, you made me believe that you're an idiot. Joke's on me I guess.

This is correct, however, in terms of notation
[eqn]0.999... := \lim_{n\to\infty}(1-\frac{1}{10^n}):=1[/eqn]