Comments on A mathematical amusement: bidecimal numbers

RRt (2014-11-17T13:30:09Z)

@Hélène :
Un calcul donne par exemple l'approximation
6.55957332…
= 5^677287 / 2^1572609
= 5^2249896 / 10^1572609
Ça fait donc quand même 1.572.609 chiffres après la virgule…

Hélène (2014-11-14T21:19:37Z)

>> Actually, a bidecimal number between 6.559565 and 6.559575 needs quite an impressive number of digits to be written exactly (or its inverse); but it can be done.

Et quel est l'ordre de grandeur du nombre de décimales nécessaires après 6,5595 ?

Karl (2004-01-19T13:34:44Z)

Now here I list all the seven digit bidecimal numbers. Note how unevenly distributed they are.

1 000 000
1 024 000
1 048 576
1 250 000
1 280 000
1 310 720
1 562 500
1 600 000
1 638 400
1 953 125
2 000 000
2 048 000
2 097 152
2 500 000
2 560 000
2 621 440
3 125 000
3 200 000
3 276 800
3 906 250
4 000 000
4 096 000
4 194 304
5 000 000
5 120 000
5 242 880
6 250 000
6 400 000
6 553 600
7 812 500
8 000 000
8 192 000
8 388 608
9 765 625

Karl (2004-01-14T16:59:59Z)

For a number to be Bidecimal its digits ignoring the decimal point (or virgule) must form either a power of two or a power of five.

Examples upto to 4 digits are:
2 4 8 16 32 64 128 256 512 1024 2048 4096 8192
5 25 125 625 3125

The denseness works only because 10 is a composite number. For binary or any other prime base only a power of that base is bi-that-basal.


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