say that o.999999 repeating equals x

10x = 9.9999999999 repeating

subtract 1x from both sides and you get

9x = 9.000000000

divide both sides by 9 and you get

x = 1

weird huh?

]]>say that o.999999 repeating equals x

10x = 9.9999999999 repeating

subtract 1x from both sides and you get

9x = 9.000000000

divide both sides by 9 and you get

x = 1

weird huh?

]]>are you aware that 0.99999999 repeating is equal to 1 ? watch this

say that o.999999 repeating equals x

10x = 9.9999999999 repeating

subtract 1x from both sides and you get

9x = 9.000000000divide both sides by 9 and you get

x = 1

weird huh?

no,

if you subtract 1 from 9.9 repeating, you get 8.9 repeating.

are you aware that 0.99999999 repeating is equal to 1 ?

Yes, they are equal. It’s like saying the limit of 1-x as x approaches 0 equals 1.

]]>It’s kind of like the thing where you can get 1=2.

Let x=y, then

xy=x^2

xy-y^2=x^2-y^2

y(x-y)=(x+y)(x-y)

y=x+y

y=y+y

y=2y

1=2

:lol:

1/9 in decimal is 0.1111…

Thus, 0.999… is 9*1/9 = 1

not too suprising

@Dirt:

Let x=y, then

…

y(x-y)=(x+y)(x-y)

y=x+y

…

The good old “divide by zero” trick

nice is alos when you use only the positive solution of a square root and forget about the negative one.

]]>nice is alos when you use only the positive solution of a square root and forget about the negative one.

I didn’t forget, I just…decided not to include it. Yeah. :roll:

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