How does 0=infinity?

A)First, let's start with $\frac{0}{0}$ =x

A1. multiply by 0
0=0*x

B)Now we do $\frac{0}{∞}$

B1. Let's say $\frac{0}{∞}$ =y multiply by ∞ 0=y*∞

B2. divide by 0 multiply by ∞ [ 0= $\frac{y*∞}{0}$ ]

C) Now $\frac{∞}{0}$ =z

C1. multiply by 0 0*z=∞

C2. divide by z 0= $\frac{∞}{z}$

Take all the bolded equations that have 0

x* 0=y*∞= $\frac{∞}{z}$ divide by 0 x= $\frac{y*∞}{0}$ = $\frac{∞}{0}$

Take the [equation with square brakets] and the above that have $\frac{y * ∞}{0}$ x= $\frac{∞}{0}$ = $\frac{0}{0}$ multiply by 0 x*0= $\boxed{∞=0}$

#Algebra

Easy Math Editor

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Comments

lol, u violated a lot of math rules.
what is your intention? to create a new set of rules?
nice idea, but still u violated your own new rules.

num IC - 8 months, 1 week ago

I am ASSUMING that dividing by 0 and infinity is possible.

JW C - 8 months, 1 week ago

ok. but to violate math rules dont prove this assumption

num IC - 8 months, 1 week ago

if u multiply an equation with 0, u will always get 0 = 0. thats why multiplication with 0 is not used to solve equalitys.
else u could prove 3 = 4 by multiply with 0: 3 * 0 = 4 * 0 -> 0=0

num IC - 8 months, 1 week ago

i can add more steps where maths violations happened, if it is helpful for u.

num IC - 8 months, 1 week ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$