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[Felissan]

Alternate proof. The definition of "..." isn't what you used to think it is. QED

I've always thought 0.999... = 1 was a lame result to present as a curiosity because of that, it exclusively relies on a notation being unintuitive for most people.

[ChairmanPoo]

But that's just it, it becomes infinitely small, not "no difference at all". I can easily accept that it's "close enough" or "effectively the same", but that does not mean it is exactly equal. This is mathematics, not engineering.

An infinitely large fraction is as much the whole number as an infinite number of zeros can equal anything other than zero. It doesn't matter that it's infinite, the fact that it can continue shortening the gap between 0.999... and 1 (I.e., being infinite) proves that it isn't exactly 1. If it were exactly 1, it wouldn't be able to be any more 1 than it already was.

So, 0.999... = 1? Sure. 0.999... == 1? No.

https://en.wikipedia.org/wiki/0.999...#Skepticism_in_education

Students 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.[35]
Some students interpret "0.999..." (or similar notation) as a large but finite string of 9s, possibly with a variable, unspecified length. If they accept an infinite string of nines, they may still expect a last 9 "at infinity".

I think this is you

[bloop_bleep]

As ChairmanPoo says, it's important to realize that the decimal representation of a number is not actually complete and unambiguous, but is instead a convenient visual aid to help us understand the number itself.

Therefore, it's important to rely on sound mathematical relations rather than decimal manipulations.

I'd like to present my proof of 0.99999... = 1. First thing we have to do is define what 0.999999.... is. We will define it as the sum

Code: [Select]

S = 9 x 10^-1 + 9 x 10^-2 + 9 x 10^-3 + ...

This is nice because it uses solely integers instead of fractional decimals, so there's no ambiguity about the value of any number in the expression. Now, we find an expression for the value of S/10.

Code: [Select]

S/10 = (9 x 10^-1 + 9 x 10^-2 + 9 x 10^-3 + ...)/10 = 9 x 10^-2 + 9 x 10^-3 + 9 x 10^-4 + ...

We then subtract the second equation from the first, in this way:

Code: [Select]

     S = 9 x 10^-1 + 9 x 10^-2 + 9 x 10^-3 + 9 x 10^-4 + ...
- S/10 =             9 x 10^-2 + 9 x 10^-3 + 9 x 10^-4 + ...
------------------------------------------------------------
 9S/10 = 9 x 10^-1

(Note: this technique is often called "shifty algebra", since we're shifting one of the equations in order to align with the other.)

Or if you want to be more rigorous, you could do it using normal notation just as easily:

Code: [Select]

S - S/10 = (9 x 10^-1 + 9 x 10^-2 + 9 x 10^-3 + 9 x 10^-4 + ...) - (9 x 10^-2 + 9 x 10^-3 + 9 x 10^-4 + ...)
         = 9 x 10^-1 + 9 x 10^-2 + 9 x 10^-3 + 9 x 10^-4 + ... - 9 x 10^-2 - 9 x 10^-3 - 9 x 10^-4 - ...
         = 9 x 10^-1 + (9 x 10^-2 - 9 x 10^-2) + (9 x 10^-3 - 9 x 10^-3) + (9 x 10^-4 - 9 x 10^-4) + ...
         = 9 x 10^-1

We have 9S/10 = 9/10. This is a simple linear equation, to which the only solution is 1. QED.

Note how we didn't use any intuition or any other questionable things (even the shifty algebra could be quite easily explained using normal notation.) This uses only basic algebraic properties, like the distributive and associative laws. In fact, this proof is a special case of how the formula for the sum of an infinite geometric series is deduced; we could've applied that formula to this problem and have proven it immediately, but I wanted to write out the whole proof in order to show you exactly what I'm talking about.

[Magistrum]

Do it like a physicist and consider both values of the same magnitude and thus can be considered equal.

[bloop_bleep]

Do it like a physicist and consider both values of the same magnitude and thus can be considered equal.

I admit that I haven't learned much in physics, past the basics, but I'm not entirely sure what you're getting at here. Are you talking about the magnitudes of vectors?

[Egan_BW]

Three doors. One of them has the prize. You pick a door, doesn't matter which because you don't know which one has the prize.
At this point, there's a 33% chance you've chosen the prize and a 66% chance you haven't. Because 66% is bigger than 33%, it's the better bet that you haven't picked the prize, so let's assume that.
Now, having assumed that the prize isn't behind you door you chose, it must be behind one of the others. The host knows which one has the prize and shows you by picking the other one. Because there's only two doors and we're assuming that the prize is behind one of them, when the host shows us which one it isn't behind, we know it's behind the other.

The host just gave you useful information in 66% of cases, where you picked a goat first, and didn't give you any useful information in 33% of cases, where you picked a car first. Your first choice is pretty much saying "okay, I bet the car is behind one of the two doors that I'm not picking". If that guess is correct, the host will show you which door the car is behind. In the less likely case that the car is behind neither of those doors, the host just picks a door at random.

[pisskop]

If you are wondering: yes, it actually plays out that way.  If you do a test where you either stay with your first picked door or swap over, you'll win more often by swapping.  Its testable.

[TD1]

Spent all day on essay.

375 words added.

...
..
.

[hector13]

Better than 375 words lost brah.

[TD1]

To be fair, I did write a lot more. It's just I mostly discarded it/rewrote it. And I did also come up with some ideas for future paragraphs which may be useful.

Still doesn't seem like much.

[Tack]

"The Goddamn Airplane on the Goddamn Treadmill"

Don't planes require airflow?
I dun get it.

[IcyTea31]

Still doesn't seem like much.

But is it good text?

In an essay, focus on what you're writing, not how much you're writing. You're not a politician (at least to my knowledge), so you shouldn't go on a filibuster. If it's hard to come up with meaningful text, instead of writing meaningless text you should make sure you've fully understood the subject matter and see if there's more you can learn about it and write about.

Don't planes require airflow?
I dun get it.

"Airflow" is the wrong word here, but in essence yes. A wing won't generate lift unless it's moving relative to the surrounding air. A fixed-wing plane won't work on a speed-matching treadmill (but a helicopter might).

The problem is also that the question is vague about what "on a treadmill" means, as explained behind that link.

[Tack]

Well I mean it's the same as if a fixed wing aircraft wasn't moving, but happened to be in a wind tunnel. It'd still take off. The wheels are unimportant to the equation, unless you bring normal physics into it.

It's the issue I have between those and the monty hall problem.
With the blue-eyes brown-eyes puzzle, once it's worked out and explained I can see all of the pieces and understand how it works. When it comes to doors... I'm kind of somehow still operating on faith. I believe that it's true (barring savvy and belligerent gameshow hosts), but it hasn't really 'clicked'.

[Greiger]

Didn't mythbusters test that and find the plane still took off?  SOmething about how the propeller was moving the plane so the plane still moved forward and took off and the wheels were just spinning faster than usual?

[wierd]

But the turbines in the engines will force airflow under the wing, even if the plane is otherwise stationary.  Granted, this is NOT the intended kind or pattern of airflow the wing's designers had in mind, so this is not very likely to produce useful lift.  What it will do, however, is produce thrust.  The conserved momentum generated by the turbine+fuselage is exerted against the air itself, not the runway. Even a small lift force will eventually lift the wing up, and lift the plane off the runway, and then the effective drag of the inverse directional movement of the runway stops being effective at resisting the vehicle's forward velocity increase. The plane *WILL* lift off, just not efficiently.

That's just a very verbose way of saying that the pressure differential caused by the engines being on will cause a prevailing draft over the wing, which will cause at least some lift, even if the plane is not moving forward, due to the conveyor.

Compare:  Tied down craft with engines off lifts off runway due to headwind until the tether goes taught and holds it down. It does this even though it is not moving forward.