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It has to do with our choice of system. But what is it exactly? I don't seem to grasp it or isn't there anything to grasp?

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- Thread starter Lorentz
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It has to do with our choice of system. But what is it exactly? I don't seem to grasp it or isn't there anything to grasp?

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matt grime

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Thanks for clearing that up... though I still think it's weird.

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There isn't anything to grasp, depending on the system you use, you will not be able to accurately display certain fractions, like for example 1/3 in both binary and decimal system you would need an infinite amount of numbers. That is an entire issue when doing math in the computer and one that has already frequently caused me problems. It is not weird, just plain annoying.Lorentz said:It has to do with our choice of system. But what is it exactly? I don't seem to grasp it or isn't there anything to grasp?

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[tex]\frac{1}{3} = 0.\bar{3}[/tex]

Aren't both ways 100% accurate?

- #6

Integral

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Or you could write it

[tex] .1_3 [/tex]

That's base 3. The existence of the real number line is independent of the any representation of points on the line. In a Real Analysis course, where the properties of the Real Line are explored in depth, such representations are not used, thus the results apply to all representations.

[tex] .1_3 [/tex]

That's base 3. The existence of the real number line is independent of the any representation of points on the line. In a Real Analysis course, where the properties of the Real Line are explored in depth, such representations are not used, thus the results apply to all representations.

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TALewis said:

[tex]\frac{1}{3} = 0.\bar{3}[/tex]

Aren't both ways 100% accurate?

Actually, I don't think that's true. If you multiply your equation there through by 3, you'll see why.

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matt grime

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[tex]1=0.\bar{9}[/tex]

I think that's an often asked question. But it's true. Or that's what they tell me.

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Zurtex

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Before you go there, it is true and there are many many proofs that as stated above [itex]1=0.\bar{9}[/itex]. If you are interested please try and learn them and see why they are right then just trying to argue them wrong illogically.Stevo said:Actually, I don't think that's true. If you multiply your equation there through by 3, you'll see why.

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Ok, I apologise for making that comment. I found a proof, I'm convinced!

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TALewis said:[tex]1=0.\bar{9}[/tex]

Does this notation mean: 0.999999... (going on infinately) ?

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Triangle with two 1cm sided sides and a hypotonuse the Hypotonuse will be exactly = to Sqrt of 2

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Lorentz said:Does this notation mean: 0.999999... (going on infinately) ?

Yes, that's the so-called "bar" notation. The line indicates a repeating sequence of numbers. Here is another:

[tex]\frac{1}{11} = 0.\overline{09}[/tex]

Here two digits 09 repeat continuously (0.09090909...).

- #15

Integral

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[tex].1_{10} = .000 \overline {1100}_2 [/tex]

The implication of this is that any time .1 is used by a computer in a calculation it MUST be rounded off.

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TALewis said:

[tex]\frac{1}{3} = 0.\bar{3}[/tex]

Aren't both ways 100% accurate?

Of course, you could write 1/3 as .3 with a bar over the three to represent an unending decimal, but this is again just a mater of "what looks pretty," and has nothing to do with the calculations in a computer. You see that 0, and 5 are divisable in base 10, and thus, 1/2 = .5. Now if you want to do math to the base 3, then we have 1/3=.1 (base 3) same as we have in base 10, 1/10 = .1 (base 10). By changing bases we solve such problems!?

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jcsd

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[tex]0.\bar{3} = 0.\dot{3}[/tex]

[tex]0.\overline{30} = 0.\dot{3}\dot{0}[/tex]

[tex]0.\overline{307} = 0.\dot{3}0\dot{7}[/tex]

I think it's preferable to use dots when writi git out by hand as it makes mistakes less likely.

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