Can u prove 2=3 ?

If you take this seriously… well you’re wrong

I like the KISS method using basic math fundamentals and a little fools’ logic. Suppose we have positive values x and y where y =
so xy=1
and x =
and x != y
and y>x,
so <
and 0 < x < 1 < y

we would all agree on those fundamentals, right? If x were and y were , then xy = 1
simple, right?
So if xy = 1
and 2 + 1 = 3,
then 2 + xy = 3

Let’s shift gears for a bit.
1/10 = .1
1/100 = .01
1/1000 = .001
So for j>=1 where we have j-1 zeroes at the end of decimal point and before the final 1
= .0001
And so on.
So if we had k = ,
1-k = .9 to j places
and .9 + k = 1
So = .1, so .9 with one place,
= .01 so .99 with two places and so on…

So for j = 1 to who knows what, we have
1-() = .9 to j places


So if we drive for infinite 9’s we have .

So where k = x = (1 -0.) and since 2 + xy = 3
On principal that 3 + 3 + 3 = 9,

. + . + . = .

but since = .
and + + = 1,

1 = .

and since x = 1 - .,
or x = 1 -1 = 0

x=0
and therefore xy = 0
and 2 + xy = 3
so 2 + 0 = 3
and 2 = 3


…of course I left out the part where now
y is probably somewhere close to ∞ (1 divided by an astronomically small number)
if it was, 0 * ∞ = 1 and/or 0
so and/or = ∞
and to reiterate 0=1 (and to think I wanted to add the astronomically small number to 2, y individual times to get 3)

And my favorite
∞ = 1 (I will now call Guinness and try to break the record for the most memorized digits of ∞)

Laugh all you want – this is my math and you reserve the right to follow my lead or be wrong. ;)