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Express the fractions , for several small values of , in base 6. Determine which rational numbers have terminating expressions in base 6.

I'm not really sure where to begin.

allie
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2 Answers2

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In number system $6$ we have $6$ digits: $0,1,2,3,4,5$. The sign $10$ now refers to six, and so on.

What are the fractions $0.1,\ 0.2,\ 0.3,\ 0.4,\ 0.5$?

Now $5$ plays the role of 'last digit'. Can you guess what will be $0.1111\dots$?

What is the 'half digit' ($1/2$), so, what will be $1/4$?

Berci
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  • So 0.1=1/6, 0.2=2/6, 0.3=3/6, and so on. Right? and the half digit is 0.3, 1/4=0.13. I don't see the connection between the 'last digit' and the repeating decimal 0.1111... and also how do you reduce fractions in base 6? e.g. 1555/4435? – allie Nov 22 '13 at 06:10
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What fractions have terminating expansions in base $10$? if we have a number: $a$ which has a decimal representation which ends after $k$ digits then it means $10^k*a$ is an integer right? in other words when they are represented in fraction form the denominator is a product of only $2$'s and $5$'s.

Similarly for base $64$ only fractions which when reduced have denominators which are only divisible by $1,2$ and $3$ have terminating expansions: can you prove it?

Suppose we have a number b which has k digits after the decimal pont. Then $b*10^k$ is an integer. since b is terminating we knoe it is rational and can be expressed as a simplified fraction $\frac{p}{q}$. Therefore $ b*10^k=\frac{(p*10^k)}{q}$ is an integer so $q$ divides $p*10^k$, but p and q are relatively prime because the fraction has been simplified, that means q divides $10^k$.

Can you use this argument for base 6?

Asinomás
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  • I'm still confused. How do you know the fractions in base 10 terminate if the denominator is a product of 2's and 5's? I'm not sure I understand your explanation – allie Nov 22 '13 at 05:59
  • since $10^ka$ is an integer then if $a$=\frac{p}{q} where the fraction has been simplified that means $q=2^n5^m$ for non-negative n and m. – Asinomás Nov 22 '13 at 15:21