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A typo in the article #3539
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A typo in the article #3539
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not helpful. perhaps:
let mcs = 1e-6; // five zeroes to the left of 1 (to the right of the decimal point)
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There is no such "up to the decimal point" condition here. Besides, literally after this sentence comes the following:
If we count the zeroes in
0.000001, there are 6 of them. So naturally it's1e-6.
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Besides, literally after this sentence
then fix that too:
If we count the zeroes in
0.000001, there are 5 of them after the decimal point. So naturally it's1e-6.
(also note i edited first suggestion to specify to the right of decimal point, i figure you saw original through email or smt.)
shallow-beach
commented
Aug 27, 2024
same confusion as #3541
joaquinelio
commented
Sep 8, 2024
I never liked the "zeroes count" explanation.
It doesn't help to build intuition.
We have decimal notation. It comes with a key property:
Multiplying by 10, 100, 1000, moves the decimal point 1, 2 , 3 places to the right.
Dividing, to the left.
1.23e-9 moves the decimal point 9 places to the left. "Naturally" for real, no need for extra steps
I couldn't figure out how to change the text without ruining it.
I never liked the "zeroes count" explanation. It doesn't help to build intuition.
We have decimal notation. It comes with a key property: Multiplying by 10, 100, 1000, moves the decimal point 1, 2 , 3 places to the right. Dividing, to the left.
1.23e-9 moves the decimal point 9 places to the left. "Naturally" for real, no need for extra steps
I couldn't figure out how to change the text without ruining it.
zeros count makes sense in that X is zero count of multiplier (either of numerator's zeros (5555e-4, that results in 0.5555, which does not have 'usefully' countable zeros.
joaquinelio
commented
Sep 8, 2024
I never liked the "zeroes count" explanation. It doesn't help to build intuition.
We have decimal notation. It comes with a key property: Multiplying by 10, 100, 1000, moves the decimal point 1, 2 , 3 places to the right. Dividing, to the left.
1.23e-9 moves the decimal point 9 places to the left. "Naturally" for real, no need for extra steps
I couldn't figure out how to change the text without ruining it.zeros count makes sense in that e X ≡ 10 X (
Xis zero count of multiplier (either of numerator's zeros ( X ≥ 0 , with denominator=1) or of denominator's zeros ( X < 0 , with numerator=1)), which coincides with count of decimal point moves). trying to apply it directly to a result is confusing - e.g, if we have5555e-4, that results in 5555 ∗ 10 − 4 =0.5555, which does not have 'usefully' countable zeros.
It's easier to think that way, don't you think?applying directly
e-4,
5555. --> 0.5555
4 spaces to the left
Where each space is /power of 10
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Numbers
There are six zeroes to the left from$\large1$ in $\large0.000001$ , not five.