Converting from base b to base 10
The next natural question is: how do we convert a number from another base into base 10? For example, what does 42015 mean? Just like base 10, the first digit to the left of the decimal place tells us how many 50’s we have, the second tells us how many 51’s we have, and so forth. Therefore:
42015 = (4.53 + 2.52 + 0.51 + 1.50)10
= 4.125 + 2.25 + 1
=55110
From here, we can generalize. Let x =(anan-1 … a1a0)b be an n + 1 -digit number in base b. In our example (274610) a3 = 2, a2 = 7, a1 = 4 and a0 = 6. We convert this to base 10 as follows:
x = (anan-1 … a1a0)b
= (bn.an + bn-1 . an-1 + … + b.a1 + a0)10
Converting from Base 10 to Base b
It turns out that converting from base 10 to other bases is far harder for us than converting from other bases to base 10. This shouldn’t be a suprise, though. We work in base 10 all the time so we are naturally less comfortable with other bases. Nonetheless, it is important to understand how to convert from base 10 into other bases.
We’ll look at two methods for converting from base 10 to other bases.
Method 1
Let’s try converting 1000 base 10 into base 7. Basically, we are trying to find the solution to the equation 1000 = a0 + 7a1 + 49a2 + 343a3 + 2401a4 + …
where all the are digits from 0 to 6. Obviously, all the ai from a4 and up are 0 since otherwise they will add in a number greater than 1000, and all the terms in the sum are nonnegative. Then, we wish to find the largest a3 such that 343a3 does not exceed 1000. Thus, a3 = 2 since 2a3 = 686 and 3a3 = 1029.This leaves us with 1000 = a0 + 7a1 + 7a1 + 49a2 + 343(2) ↔ 314 = a0 +7a1 + 49a2.
Using similar reasoning, we find that a2 = 6, leaving us with 20 = a0 + 7a1.
We use the same procedure twice more to get that a1 = 2 and a0 = 6.
Finally, we have that 100010 = 26267.
An alternative version of method 1 is to find the “digits” a0, a1,… starting with a0. Note that is just the remainder of division of 1000 by 7. So, to find it, all we need to do is to carry out one division with remainder. We have 1000:7 = 142(R6). How do we find a1, now? It turns out that all we need to do is to find the remainder of the division of the quotient 142 by 7: 142:7 = 20(R2), so a1 = 2. Now, 20:7 = 2(R6), so a3 = 6. Finally, 2:7 = 0(R2), so . We may continue to divide beyond this point, of course, but it is clear that we will just get 0:7 = 0(R0) during each step.
Note that both versions of this method use computations in base 10.
It’s often a good idea to double check by converting your answer back into base 10, since this conversion is easier to do. We know that 26267 = 343.2 + 6.49 + 2.7 = 1000, so we can rest assured we got the right answer.
Method 2
We’ll exhibit the second method with the same problem used to exhibit the first method.
The second method is just like how we converted from other bases into base 10. To do this, we pretend that our standard number system is base 7. In base 7, however, there is no digit 7. So 7 is actually represented as “10.” Also, the multiplication rules we know do not hold. For example, 3.3 ≠ 9 (in base 7). For one, there is no 9 in base 7. Second, we need to go back to the definition of multiplication to fully understand what’s happening. Multiplication is a shorthand for repeated addition. So, 3.3 = 3 + 3 + 3 = 127.
In base 7, we have that 10 (the decimal number 10) is 13. Thus, if we view everything from base 7, we are actually converting 100013 to base 10. So, this is just 133. Remember that we aren’t doing this in our regular decimal system, so 133 ≠ 2197. Instead, we have to compute 13 x 13 x 13 as (13 x 13) x 13 = 202 x 13 = 2626.
This method can be very confusing unless you have a very firm grasp on the notion of number systems.
Read more below-
SS 2 Mathematics Third Term: Problem Solving on Number Bases Expansion, Conversion and Relationship
