would it not be _ i -i = 0_
because when you add or substract a number from_ i_ it's value still stays the same
but when you substract the same value throughout i
it will become nothing aka 0
what?
what i did is subtract y from both ends of the equation to make y-y=whatever
and when i did that i became infinity minus infinity equals 2 which is wrong which is why this mo fo has no solution
infinity minus infinity is in fact not 0, it can be anything you like, really.
lets say âˆž +2 = âˆž, which is true
> âˆž - âˆž = 2
> 2*(âˆž - âˆž) = 2*(2)
> âˆž - âˆž = 4
therefore we can also say that infinity minus infinity equals infinity. because infinity plus infinity is still infinity. Its an irrational number.
im not serious, i said that because i actually wasn't learn math on the internet, just nonsense math, my use of reverse sarcasm was displaying my false stupidity, hence with the hoped result of creating laughter. Its quite complex humor really, im not surprised you are baffled.
Your lame undertaking to confound me is evidence enough that you are not as of yet fully aware of my country of origin.
I am an Englishman, you see.
While the plashet of melatonin that once was known as my cerebrum has indeed been encumbered with your nebula of deceit, I see now the ironic sarcasm in your statement and stand up fully to accept the extent of my deserved ignominy.
My affluence of intellect has been stripped from me and I am left indigent of cerebral power in your eyes.
I only hope my considerable essay at making my vocabulary as eclectic as your own will make up for my shortcoming.
I extol your innovative use of ironic sarcasm, and I bid you good day.
I too am a brit, and thank you for such a timely response, i admire the flair in your writing.
That second statement was also a joke, due too my humorous and clearly false explanation. I hope you had a jolly good laugh.
I am quite the punster today.
Good day.
LOL nothing.
it has some nice shops. Its a melting pot of humans from all cultures, ( yet a few white people still look at me and immediatly cross the road, then cross again when i turn around to look).
thats about it.
london is so overrated.
I just realised IOW is a quarter of the size of london alone.. thats ******* awesome.
I did have such an inkling, it is good to see!
And I thank you most profoundly for your complements.
Pfffffffffffffffffenough.
Y'right m8?
Nice jokes, and yeh I did catch the second one, but as I said (if you caught it, something to do with melatonin) I'm pretty ******* shattered and not up to much more.
'Ave a good one! ^^
Are you really really sure about that? Would you please provide proof for that statement? You know, even an infinity + an infinity = an infinity. And then there are infinities of different sizes as well. For instance, an infinity made of all rational numbers is smaller than an infinity made of all irrational numbers, but all infinities composed only of rational numbers are equally big. Therefore, an infinity + 2 (rational) = an infinity.
but if we change your equation to the following: "2 = an infinity - an infinity"
so, no matter what these infinitys are they both have to be the same and therefore if you subtract them you'll get 0 and then the equation is "2 = 0" which isn't true so the answer is that there is no solution.
Infinity is the concept that a number is growing larger, so infinity/2 or infinity*2 or infinity+25 or infinity-1351321698654 are still equal to infinity. But nowhere in the original equation do we ever talk about infinity...
I just learned a bit of code as a personal project once, I have no idea what any of the sensical things there mean, let alone the (I presume) nonsensical parts.
Yes. Usually loops require something to be transformed into a boolean value {0,1}. When sober=0, the program may chose to stop looping "beer". Though it doesn't work for all compilers that way. Some end up in an alcoholic coma in hospital.
Except that didn't really solve it, since you didn't even adress the equation itself.
You didn't even try to adress it as an equation. Equations are solved by finding the values of the variable(commonly x, but here it's y) for which both sides are equal.
Functions on the other hand is finding the output for different values of x.
An example of an equation is 2+x=2x with the solution being x=2. Equations can exist for which there is multiple solutions. An example is 1=x^2 with both x=1 and -1 being valid answers
An example of a function is f(x)=3x+7 with there being a different output for each value of x. Functions can exist where certain values of x is invalid. An example is f(x)=1/x with x=0 being invalid.
As a sidenote f(x) means function of x, so for your invalid attempt you should've written f(y).
Also my "attempt" at solving this.
y+2=y
2=y-y
2=0
No valid answer.
I know that equations don't use that kind of notation, but just because the mistake would've been removed if he correctly adressed it as an equation instead of a function, does not mean that the mistake shouldn't be pointed out.
A derivative, put simply, is how fast an equation is changing. The derivative of y=3x is dy/dx=3 because y is changing by 3 every time x changes. The derivative doesn't say anything about the actual solution to the original equation the way you're doing it. You're just proving that the rate at which y is changing is equal to the rate at which y is changing. Plus the way you did it was wrong because you don't know the derivative of y, so what you really would have is y'=y'.
1y+2=1y (the reason I put 1y instead of y is because it is easier to solve) Subtract 1y from both sides
2=0y Divide by 0 on both sides
2/0=y since you can't divide by zero it isn't possible
so y=Undefined
you know in order to properly do this you separate it into left side and right side that way at no point will you have anything equal to something it isn't.
essentially:
LS. | RS.
y+2 | y
therefor LS =/= RS
Q.E.D.
At least in his method he put that y = 2/0 which is the next most correct answer.
so youre telling me y+2 = y, that the right side IS equal to the left? which isn't actually possible except in the imaginary (so in this context its not possible because we're working as elements of the real number system)
forgive me, my brain went to mush, I thought you just meant in a proper equation, the left side does not equal the right side. I didn't know you meant this specifically and forgot that when solving things algebraically you do separate the sides.
Here we go. This goes for any of the commonly known sets of numbers in mathematics, namely the naturals, integers, rationals, reals and complex. For any element in the set (we'll call it x, and in this example we'll just deal with naturals,) there exists a unique solution to the equation x+a=x=a+x, meaning a is the identity element in respect to addition, or is the additive identity. In all of these sets, the additive identity is 0. That means that x+a=x=a+x is only true when a=0. Since this is a case where a=2, it can no longer hold true for any x in any of the above sets.
It's a common fallacy in differential equations that a solution to the derivative implies a solution to the original equations. Such is not the case. Think about f(x)=x^2+3 and g(x)=x^2+5. f `(x) and g `(x) are equal, but f(x) =/= g(x). There are cases in which y=y+2 does yield a solution, but only when 0=2. One example of this is under modulo arithmetic, namely (z mod 2.) In this case, y is congruent to y+2, but as modulo arithmetic is an arithmetic on the integers, you only have integer solutions. Typically when people are discussing problems like these, the assumption is that we're working under elementary arithmetic in the reals.