- #1

pseudogenius

- 7

- 0

The problem is: Write 5 in base 3/2

5*(2/3)= 3 remainder 1

6*(2/3)= 2 remainder 0

2*(2/3)= 1 remainder 1

1*(2/3)= 0 remainder 2

Take 2101 and divide each digit by 2.

5 in base 3/2 = 2/2 1/2 0/2 1/2

Check:

(2/2)*(3/2)^3+(1/2)*(3/2)^2+(0/2)*(3/2)^1+(1/2)*(3/2)^0=

27/8+9/8+0+1/2= 5

It worked.

Is it okay to extend digits to rational digits?

Is this a valid way to put integers in rational bases?

I have worked out a way to put rational numbers into rational bases but the process is complicated.