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How to draw a square with 0.999... area?

gvitalie shared this question 3 years ago
Needs Answer

Good morning, Dear GeoGebra and GeoGebrians.

How could I draw a square with area 0.999... (zero whole and nine periodic digits) or a segment with length equal 0.999... (zero whole and nine periodic digits).

I couldn't do that because GeoGebra is rounding numbers.


Thank You.

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23d000f8873a93c78edd8b79a95b0a29

2ced9781d6e0c561389cd437b2e4626d

5e6bae7a4d7c76ac985926b35f861900

Comments (4)

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2

quote: How to draw a square with 0.999... area?

.

try with 1 - 10⁻¹⁵

see attachment

Note: in the classic algebra is (a/b) b = a

see also here.

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1

Good morning, Dear @rami.

I did it graphically, but numerically not rounding correctly I think, if I'm not wrong.

According to Pythagoras Theorem, we should sum areas of two squares something like that.

It should be 0.999..., if 2/9 + 7/9 = 0.222... + 0.777... = 0.999...


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Good morning, Dear @rami.

I've understood what's the explanation: it's difference of numerical representation with finite precision and infinite precision.


Example:

- with FINITE precision we'll always have: 1/9 + 8/9 = 0.999... until maximum precision resolution

and when we do rounding, we compensate that finite precision resolution.

- with INFINITE precision we'll always have 1/9 + 8/9 = 1

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2

Don't you know, that 0.99999....(periodic) =1 ?

Otherwise: Tell me a number between 0.99999.... and 1.


Btw: Last digit of 6/9:

"0,66666666667 is wrong

... must be 0.66666666666"

Both results are wrong, because 0.6666666666 < 6/9 < 0.6666666667.

But 0.6666666667 is a better appoximation than 0.6666666666.

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