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
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.
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.
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
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.
.
. Sorry for the letters but they were the best i could do 
I was trying to confirm that what I posted is the delta-epsilon “version” of the limit.