The Maths Topic

The one on the far left? That’s an integral sign. You’ll learn about it if you take a math class on calculus. It can mean either finding the antiderivative of a function, or, as in this case, finding the area under the graph of a function. For more info, and probably a better explanation on what an integral is, see Wikipidia’s entry on integrals. :happy:

For anyone who is interested, this integral is so important that it has it’s own name, the Gaussian Integral, and a proof of it can also be found on Wikipedia.

Eh, you’re wrong. You can give any variable any value, even irrational ones. You may protest I give x = pi, but do you protest against me solving an equation and finding x = sqrt(2), which is also an irrational number? No, I don’t think so :content:

This is the major flaw in your proposal.

Essentially, you are letting x = π.

That means that the equation x = (π + 3) / 2 is incomplete. If π=x, then you should not have both x and π in your equation. You need to finish your substitution of x for π.

If you do this, your equation will then look like:
x = (x + 3) / 2
2x = x + 3
x = 3

Now it would seem that you have proved π=3. But no. The equation you began with was completely made up. What you have found is that x=3 in your equation. You need to prove that x=π. This is impossible. An attempt looks like this:

x = π
3 = π
3 = 3.14…

So really, I think that cookie belongs to me :smile:

Can anyone solve this?

a,b > 0 and a+b=z
prove that:
(4/3z)<(1/a+z) + (1/b+z)<3/2z

Your notation isn’t helping much :tongue:

(4/3z) is supposed to mean 4/(3z) or (4/3)z ?

(1/a+z) is (1/a)+z or 1/(a+z)?

Is this some kind of an exersize for school??
Let’s see!

a,b>0 and a*b=z => z>0
so we can multiply z!

4/3< [(b+z)z+(a+z)z]/(a+z)(b+z)< 3/2
4/3 < z[(b+z)+(a+z)]/(ab+az+zb+z^2)/3/2
8/6<[z(a+b+2z)]/(z+az+zb+z^2)<9/6
8/6<az+zb+2z^2/(z+az+zb+z^2)<9/6
we can add z and -z cause z-z=0
8/6<[(az+zb+z^2+z)-z+z^2]/(z+az+bz+z^2)<9/6
8/6<1+ (-z+z^2)/(z+az+bz+z^2)<9/6

1=6/6 and z^2>-z for every z in R

so 8/6<1+(something >0)<9/6

2/6<(z^2-z)/(z+az+bz+z^2)<3/6

I think it goes further but i’m a bit bored right now:P:P
so this is for you:P

Ok i’ll take it a bit further!
2/6<[(ab)^2-ab]/(ab+a^2b+b^2a+(ab)^2)<3/6
2/6< [ab(ab-1)]/ab(1+b+a+ab)<3/6
2/6<(ab-1)/(1+b+a+ab)<3/6

(thank god a,b>0 cause if any one of them was 0 then,this one could not be solved)
If ab=1 then 2/6<0<3/6 which is wrong!
So i might have done a mistake somewhere!If someone else doesn’t solve this,i’ll look at it later:P

a second edit! ab=a*b and so on…

oh heh it supposed to mean 4/(3z) and 1/(a+z) :shy:
Its just an excersise i found a few days ago, tried solving it and failed miserably :tongue: .

EDIT: hold on… i solved the first part → 4/(3z)<1/(a+z) + 1/(b+z)
(i wrote the answer on paint cuz it would take me too long to type it and id probably get things wrong again - not used to typing maths :razz: . Sorry for the letters but they were the best i could do :tongue: )


Eh, no. Leave it to the math freak. :tongue: 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.


It’s supposed to be an easy question but I don’t get it. =/

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. :content:

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 :wink:

1/3=.333333333333333…

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

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

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. :smile:

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 :razz:

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.