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Reduce the base $11$ fraction $\dfrac{587}{749}$ to its lowest terms.

$(\dfrac{587}{749})_{11}=\dfrac{5\times 11^2 + 8\times 11 + 7}{7\times 11^2 + 4\times 11 + 9}$

But $\dfrac{...+7}{...+9}$ can't be simplified any further, so I'm not sure how else to approach this problem.

ronald christenkkson
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2 Answers2

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Reducing a fraction to lowest terms means dividing the numerator and denominator by their greatest common factor. There are a couple ways to do this. First, you could use the Euclidean algorithm, which could get an answer without being able to factor either number. The alternative is, of course, to factor numerator and denominator. We're actually pretty lucky in this case that it's very easy to check if a number in base 11 is divisible by 10. For example, in the numerator, we have

$$5(11^2)+8(11)+7\equiv5(1^2)+8(1)+7\equiv 20\equiv 0\pmod{10}$$

Similar to the divisibility test for 9 in base 10 numbers, all you need to do is sum the digits and see if the sum is divisible by 10 (or 5 or 2). In this case, both numerator and denominator sum to 20 and are therefore divisible by 10. Dividing by ten, we get $\left(\frac{587}{749}\right) _{11}=\left(\frac{64}{82}\right) _{11}$. Again, both numerator and denominator sum to a multiple of ten, so we can divide both again to get $\left(\frac{64}{82}\right) _{11}=\left(\frac79\right) _{11}$

Mike
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If you want to simplify directly, then factorise in base 11 as $587 = 299 \times 2 = 14(ten) \times 2 \times 2 = 32 \times 5 \times 2 \times 2 = 7 \times 5 \times 5 \times 2 \times 2,$ and then cancel the common factors. However, this is very tedious to do manually, so change to base 10 first : $587_{11}$ is 700 and $748_{11}$ is 900 in base 10. Simplify, and go back to base 11.

Dhawal Patil
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  • I'm interested in doing it directly like you said. But I don't understand your calculations. For example how can $587=299 \times 2$? Thank you for your help. – ronald christenkkson Feb 13 '24 at 05:25
  • @ronaldchristenkkson: you have to work in base $11$ with carrying there so $9 \times 2=18_{10}=17_{11}$ so you write down $7$ and carry $1$. Then $2 \times 9 +1=18_{11}$ so write $8$ and carry $1$. Finally $2 \times 2+1=5$. Just like in base $10$ a number may have factors that are not apparent in the base expansion. $57_{10}=5 \times 10+7$ has a factor $3$ that is not apparent in the base expansion. – Ross Millikan Feb 13 '24 at 05:29
  • I understand $299 \times 2 = 587$ in base $11$. But how did you know, doing the reverse operation, just from looking at $587_{11}$ alone that it is $299\times 2$? I don't understand this part. – ronald christenkkson Feb 13 '24 at 06:03