I have pi memorised to 362 places. 
Pi is infinite (I’ve read books on it!), it can’t end. You can’t say ‘x’ = pi, because we don’t know what pi fully is. Sure, you can say x = 3.14, but that’s nothing spectacular - it proves nothing. You can’t say that x = pi, because pi never ends. Ever.
Anyone know of the Riemann hypothesis? Prime numbers are more intricate than you may think!
I’ve studied the Riemann hypothesis and the theory of the Riemann zeta function extensively for the last six years (since i was 16!)
I can’t say i have any idea about how to prove it 
but i find it fascinating!
i’m pretty sure the proof will be based on totally new maths, just like the proof the Fermat’s Last Theorem.
Okay. Now that I know calculus:
The assumption here is that the derivative 1 + 1 + . . . + 1 (x 1’s) is in fact 0 + 0 + . . . + 0 (x 0’s), but it isn’t–use the product rule or something, if you want. The fact that there’s an “x” in there means you can’t just differentiate the zeros.
Now, don’t say “well, for some constant c, c = x, so it’s 1 + 1 + . . . + 1 (c 1’s) = x, which you can differentiate to 0 + 0 + . . . + 0 (c 0’s) = 1”. In calculus, you’re just not allowed to set something to a constant and then differentiate over it.
Edit: oh, and:
Yes, you can say x = pi: pi is a real number, just like 3.14. Sure, pi’s decimal expansion never ends, but pi does have definitions which do end. An example: pi is the sum as n goes from 0 to infinity of (4/(1+4n) - 4/(3+4n)).
Oh, given that a day has passed, I think making a new post for this is best.
The theory goes that if something involves accomplishment, you can enjoy it. (Actually, that’s just my own theory, but it seems to work fine.) I’m guessing by your belief that you can make 100 cents equal 1 cent that you believe it’s possible to do anything at all within math, correct?
Things such as the 1 cent = 100 cents thing are essentially mathematical jokes or puzzles: the point of those is to figure out what’s wrong with them. Mathematics at its core essentially consists of “if this is true, then this is true”, so it’s all about questions such as “is this true?”, “when is this true?”, and “what’s wrong with this?”.
Now, there’s more to math than just numbers–in fact, the things dealt with in math are best described as being just “things”. If you want to define, say, a real number, first you come up with some basic things you can do with real numbers: you can add them, you can multiply them, and you can compare them to see which is bigger (or if they’re equal). Then you decide on a list of properties you want the real numbers to have. Given enough rules about how +, * and > work, you end up with a definition of a real number, and from then you can define anything about them. Here’s a definition of a whole number (as distinguished from an integer):
This uses the successor function as the only “primitive”, and defines other things, such as zero, based on this primitive. We can then define anything we want that can be done with whole numbers: addition, multiplication, factorials… we can define subtraction, though sometimes subtraction doesn’t work: -2 is not a whole number.
My point, I guess, is that math is all about rules and proving that they imply things–proving something or otherwise finding something out sounds like an accomplishment to me, especially if you can use it. Occasionally I go to Wikipedia and look for interesting things, often mathematical things. There’s set theory, category theory, real analysis, computer science, topology, cryptology… there are probably whole big fields of mathematics I haven’t even heard of. Lots of stuff to it, and it’s certainly more fun if you understand it.
Square roots as in turning x^2 into x aren’t a function, so you can’t do that to both sides. Consider this false proof, which isolates the problem:
4 = 4
2^2 = (-2)^2
2 = -2
It’s true you know 
what does that big line mean?
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. 
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 
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 
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
(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) 
Its just an excersise i found a few days ago, tried solving it and failed miserably .
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 
. Sorry for the letters but they were the best i could do )
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.
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. 
