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7.2: Bases

  • Page ID
    96221
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    As I mentioned, a base is simply a number that’s an anchor for our place value system. It represents how many distinct symbols we will use to represent numbers. This implicitly sets the value of the largest quantity we can hold in one digit, before we’d need to “roll over" to two digits.

    In base 10 (decimal), we use ten symbols: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Consequently, the number nine is the highest value we can hold in a single digit. Once we add another element to a set of nine, we have no choice but to add another digit to express it. This makes a “ten’s place" because it will represent the number of sets-of-10 (which we couldn’t hold in the 1’s place) that the value contains.

    Now why is the next place over called the “hundred’s place" instead of, say, the “twenty’s place"? Simply because twenty — as well as every other number less than a hundred — comfortably fits in two digits. We can have up to 9 in the one’s place, and also up to 9 in the ten’s place, giving us a total of ninety-nine before we ever have to cave in to using three digits. The number one hundred is exactly the point at which we must roll over to three digits; therefore, the sequence of digits 1-0-0 represents one hundred.

    If the chosen base isn’t obvious from context (as it often won’t be in this chapter) then when we write out a sequence of digits we’ll append the base as a subscript to the end of the number. So the number “four hundred and thirty-seven" will be written as \(437_{10}\).

    The way we interpret a decimal number, then, is by counting the right-most digits as a number of individuals, the digit to its left as the number of groups of ten individuals, the digit to its left as the number of groups of hundred individuals, and so on. \(5472_{10}\) is just a way of writing \(5 \times 1000 + 4 \times 100 + 7 \times 10 + 2 \times 1\).

    If we use exponential notation (remember that anything to the \(0^{\text{th}}\) power is 1), this is equivalent to: \[5472_{10} = 5 \times 10^3 + 4 \times 10^2 + 7 \times 10^1 + 2 \times 10^0.\]

    By the way, we will often use the term least significant digit to refer to the right-most digit (2, in the above example), and most significant digit to refer to the left-most (5). “Significant" simply refers to how much that digit is “worth" in the overall magnitude of the number. Obviously 239 is less than 932, so we say that the hundreds place is more significant than the other digits.

    All of this probably seems pretty obvious to you. All right then. Let’s use a base other than ten and see how you do. Let’s write out a number in base 7. We have seven symbols at our disposal: 0, 1, 2, 3, 4, 5, and 6. Wait, you ask — why not 7? Because there is no digit for seven in a base 7 system, just like there is no digit for ten in a base 10 system. Ten is the point where we need two digits in a decimal system, and analogously, seven is the point where we’ll need two digits in our base 7 system. How will we write the value seven? Just like this: 10. Now stare at those two digits and practice saying “seven" as you look at them. All your life you’ve been trained to say the number “ten" when you see the digits 1 and 0 printed like that. But those two digits only represent the number ten if you’re using a base 10 system. If you’re using a base 34 system, “10" is how you write “thirty-four."

    Very well, we have our seven symbols. Now how do we interpret a number like \(6153_7\)? It’s this: \[6153_{7} = 6 \times 7^3 + 1 \times 7^2 + 5 \times 7^1 + 3 \times 7^0.\] That doesn’t look so strange: it’s very parallel to the decimal string we expanded, above. It looks weirder when we actually multiply out the place values: \[6153_{7} = 6 \times 343 + 1 \times 49 + 5 \times 7 + 3 \times 1.\] So in base 7, we have a “one’s place," a “seven’s place," a “forty-nine’s place," and a “three hundred forty-three’s place." This seems unbelievably bizarre — how could a number system possibly hold together with such place values? — but I’ll bet it wouldn’t look funny at all if we had been born with 7 fingers. Keep in mind that in the equation above, we wrote out the place values as decimal numbers! Had we written them as base-7 numbers (as we certainly would have if base 7 was our natural numbering system), we would have written: \[6153_{7} = 6 \times 1000_7 + 1 \times 100_7 + 5 \times 10_7 + 3 \times 1_7.\] This is exactly equivalent numerically. Because after all, \(1000_7\) is \(343_{10}\). A quantity that looks like an oddball in one base system looks like the roundest possible number in another.


    This page titled 7.2: Bases is shared under a not declared license and was authored, remixed, and/or curated by Stephen Davies (allthemath.org) via source content that was edited to the style and standards of the LibreTexts platform.

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