The Maths Topic

Eh, no. Leave it to the math freak. I have not found the value of x, I have “found” the value of pi. Essentially I haven’t, because the first equation is completely correct; I simply made incorrect transformations in the process to “prove” that pi was 3. Basically, I have incorrectly elmininated our unknown variable x. You see, I can make whatever statement mathematically to assign (or in this case, correlate) the variable x. pi in this case is some constant.

The incorrect part lies when I take the square root out of both sides of the equation; the value within the paranthesis may as well be it’s negative correspondance; and when you actually add that, you don’t get pi = 3 anymore. It may as well be x-pi = ±(x-3).

Alright, if you still want to argue about it, try replacing pi with “4” everywhere in the “proof”. The first equation will be correct, yet the end result (that 4=3) will trivially be incorrect.

Where does he ever say that π=x?

Hi, I’m studying Statistics 1 and I’m stuck on a question on Discrete Random Variables. I have the question and the answer from the mark scheme but I don’t understand how they answered it. If anyone can please explain what I have to do, to answer it would be greatly appreciated.

799×510

The first table shows the probablity that x is smaller than or equal to a certain value.
The second table will show the probablity that x is precicely equal to a certain value.

Starting with the first entry, from the first table we see that the probablity that x is smaller than or equal to 1 is 0.1. The question then is, what is the probablity that x is precicely 1? Since the probablity that x is smaller than 1 is 0 (1 is the lowest value it can take) the probability that x is precicely 1 must also equal 0.1.

For the second entry, we see from table 1 that the probability that x is smaller than or equal to 2 is 0.2. This means that we have
(probability that x is smaller than 2) + (probability that x is precicely 2) = 0.2
The unknown we are looking for here is the probability that x is precicely 2. The probability that x is smaller than 2 must be equal to the probability that x is precicely 1, since the only smaller value than 2 x can take is 1. Thus we have
0.1 + (probability that x is precicely 2) = 0.2
and we see that the probability that x is precicely 2, and thus the second entry in the solution table, is 0.1.

For the third entry, we carry on the same way. We have
(probability that x is smaller than 3) + (probability that x is precicely 3) = 0.25
The probability that x is smaller than 3 must be the probability that x is precicely 1 + the probability that x is precicely 2, since 1 and 2 are the only lower values x can take. Thus we have
(probability that x is smaller than 3) = 0.1 + 0.1 = 0.2 and
0.2 + (probability that x is precicely 3) = 0.25
and thus (probability that x is precicely 3) = 0.05 - this is the third entry in the solution table.

For the rest, the same general pattern (which I’ve hopefully made clear by now) is followed. Hope this helps.

Proof that .999999… equals 1:

.(9)/3=.(3)

.(3) = 1/3
1/3 · 3 = 3/3
.(3) · 3 = .(9)
So…
.(9) = 3/3 = 1

well, the error lies in that .(3)=1/3 that’s not true

1/3=.333333333333333…

so therefore it’s not a proof of .9999999… equals 1

Mag, .(3) = .33333333…, it’s just a different notation.

No, that just isn’t necessarily the case, and here is why. Lets just use x^2=9.

sqrt(x^2)=sqrt(9)
x=3
so, thats all fine and dandy…but, you’re missing an answer

x=-3 also works as a solution.

Maths topic? Hmm… Well I just learned how to card count a Level 3 advanced strategy for the wonderful game of Blackjack. I’m up $203 this month.

I have o friend who is able to prove that 1=0,999…, but I forgot how did he do this. Neverthless, I am a maths genius in my parents’ eyes, but reality shows different facts. I don’t think I’m even good.
On the other side… the humanities are also too complicated. They wouldn’t be, if I hadn’t got to read antique books written with totally out-of-date language which barely anyone can nowadays understand…
And one-page long text from these books to memorize, of course

Hmm…
Let x=0.999…
10x=9.999…
Subtract x from both sides, and
9x=9
x=1
0.999…=1

Something seems funny here:

Seems like your x value has changed

Wouldn’t you need to get a 1=1 type of thing to prove?

Transcendental property of equality: if a=b and b=c, then a=c. That .999…=x was given. From that I proved that x=1. Thus, since .999…=x=1, .999…=1.

Can anyone here explain how the delta-epsilon definition of a limit applies when the limit is infinity? I tried doing it, but everything went kablooie.

Okay, I get that…I wasn’t thinking. I am, however, still confused on this:

I only see x on the left side?

Oh, and delta-epsilon definition - is that:

lim f(x) = L
x > c

= goes to

alright, lambda calculus and combinatory logic. anyone else kinda drools every time they realise once again how exactly the Y combinator works?

Right side is 9.999…
X=0.999…
9.999…-X=9.999…-0.999…=9.

I admit that it’s not at all obvious that that’s allowed.

And for the limit, thanks, but that’s not exactly what I was looking for.

I know. I was trying to confirm that what I posted is the delta-epsilon “version” of the limit.

The delta epsilon definition says that “the limit as x approaches c of f(x)=L” means that for any epsilon (e) > 0, there is a delta (d) > 0 such that if 0<|x-c|<d, then |f(x)-L| < e.

I need help!

Probability

1. You have a deck of cards (52 cards), and pick three.
   What’s the probability to pick 3 different colors?

2. A family has 4 kids.
   What’s the probability that there are 2 boys?
   (By the way: P(boy) = 0.516), and P(girl) = 0.486)

I know the answeres, but not how to calculate it