5 simple unsolvable equations

According to Ramanujan X=X+1 when x= ♾️
♾️=♾️+1

Bro giving inequalities to confuse us with his Brilliant(!!) Moves, (no factorial).

What? Are you tricking me? I dont know about complex numbers but the 3rd one definitely has 1 as its solution!!!!!!

RootX = -1
X= 1
now if we put the value of x=1 in original that is,
Root of 1= +/- 1
Which satisfy the equation isn't this correct..
Please sir if read this reply wanna clear my doubt

Why doesn't this work???
1/x ≠ 0
2^x ≠ 0
Square root of x ≠ 0

Is it possible just no answer yet

The principal branch of square root can’t do that but regular square root can.

No.3 sqrt(x) = -1

Squaring both sides, we get

X = 1 💯💯

1} x = infinity

The third one should be x²=-1 instead of √x=-1😅
I think you had mistaken it.

no.1answer is x=-1

The squareroot of 1 is 1 or -1 so x is 1

For 3rd why can't it be x=1

Most of them were nonsense but 1&2 can be solved using real analysis

root 1 = 1 or -1

When I studied limits 1/ infinity was 0 why not now

2. What about negative infinity

I guess he forgot to write + 1 on RHS of 5th.

For the third one, x can't equal 1?

No.1 is an invalid question and no situation will ever lead you to this equation. If you try solving the equation, you will end up at 0=1 which is just mathematically impossible

The 1/x = 0 has one solution in the Riemann sphere, it's the north pole (i swear it's not infinity pls)

me for the last one= negative infinity

infinity IS NOT a number, it's a limit

3rd is easy as √i =-1

Bro needs to marry his barber

5 is true in the trivial ring

3 does have a solution it is 1
EZ

İn 3 can you use complex numbers ?

Why o like math? Cuz I don't care about language 😊

Last one is normal for programmers 😂😂

But in limits we can have solutions of some of tyem

wow very useful in real life!! 💩💩

2. X = infinity ♾️

3. X=1

For the third equation - the solution/root of the equation will be 1, i.e. X = 1. This is because each number has two square roots, one positive and another negative. The positive square root is called 'principle square root', and the negative one is generally ignored. Therefore, √1 = (-1), 1.
But there is also 1 thing to notice, '√' (radical) sign always returns/gives or talks about positive square roots, which implies that whenever '√' sign is there, it means positive square root. So, I am not quite sure about my explanation for equation 1.... 🤔🧐

For equation 5 and equation 1, the answer could be infinity ♾️ [X = ♾️], but that answer is not accepted here by him. 😅

Rest of the statements aren't any 'equations' or 'statement of equality' according to me.... 🤔🤓🧐

in x=x+1, i think it's 0/0 and maybe 0/0 is infinity?

x≡x+1(mod2)

Subtract
�
x from both sides:

0
≡
1
(
m
o
d
2
)
0≡1(mod2)

This equation has no solutions in the modulo-2 system because
0
≢
1
(
m
o
d
2
)
0

≡1(mod2).

In traditional arithmetic, there is no solution, but in modular arithmetic, we can see that in modulo 2, there is no integer solution for
�
x that satisfies
�
=
�
+
1
x=x+1.

RIO=Reel/Infinity=0

First Equation :
1/x=0
x^-1=0
x=0^-1
x=1/0
x=undefined
😂😅

His asian beard looks like hair in my White butt crack.

Shave

What about x=i fot the third question?

Question: You say no solutions in the complex plane and real plane. Cant you just say complex plane since all real numbers are complex?

The 1st equation an exemple to why irs impossible imagine its possible then if you multiply both side by x youll get 0=1

Yah they are not solvable because they are wrong💀

2 x 0 = 0

Another one would be 3n+1

"√x=-1" one 😭

I have a doute on 2 we can say x=0
Power can be 0

sqrt(-1) = 1

It is a valid equation. Don't start with me man