The Most Controversial Equation Ever #maths

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[aazam]

Infinity is not a number
It is just imagination

[CalculusIsFun]

Not as controversial as 1 + 2 + 3 + 4 + 5… = -1/12

[pieceofwaterofficial]

"911, what's your emergency?"
"My math isn't mathing"

[benjamincaneca]

It is not possible since this not a convergent sum, it is divergent, meaning that it does not tend to a defined value, but to infinite, which is undefined. The solution he used is only appliable for convergent sums (0<q<1)

[LwinMinKhit]

The biggest battle between Maths and Logic 💀💀

[Leonhard_Euler]

It's not controversial. It's just incorrect.

[clouds]

"If we rearrange the terms then original sum will change"
Bernhard Riemann
In Reimann paradox

[daonastar]

Bro sum to infinity in gp formula= a/1-r

[ASH.-K]

S - 2S is not equals to -S😂
Infinite cannot be subtract with infinite🗿

[MoeLesteruc]

When you take 2 common, you ignore the fact that there is one term less than the original S. Assume there are n terms
S=1+2(S-n) ==> S=2n-1
Now put n as infinity or anything

[tommasoantonelli]

This equation actually makes some sense!

Not in the way it's presented here, but if you're interested check out the concept of p-adic numbers. In particular in the 2-adic number system this infinite series converges and the convergence value is precisely -1 !!!

[tjfighter]

This question was in beginner exercise of my ALLEN MODULE😂

[ankitjukaria]

What you are doing is wrong this approach can be only used for converging series .

[Stepanchicko]

the distributive law doesn't work on infinite sums because the numbers just don't end and you'll be distributing it forever
the answer is infinity itself

[reader_me]

That's wrong: S=1+2+4+8+..+.2n; S=1+2(1+2+4+8+...+n); here S is not equal to 1+2+4+8+...n.

[lightraventravel]

Convergent Series in the back be like: 😶‍🌫️

[EndlessFantasies]

A few defects here, Infinity can't be subtracted with infinity, and if you change the placement like 1 + 2 to 2 + 1 then this won't work, this is why never mess with infinity, its illogically logical and has many rules to be considered

[laxmiaruna]

TIP:you can use a similar method to solve problems with x on one side and a root on the other side with infinite progression

[ananthgaming]

Infinite GP sum = a/1-r
Where a is initial term and r is common ratio (2nd term divided by first term)
So a=1, r=2
Now a/1-r= 1/1-2
=1/-1
=-1

[Percycorbero]

The moment you assume that infinity is "S" you can prove anything. That is incorrect.