Why does conversion from base happen to base 10 naturally for any base?

Question

Let's take number 11 in base(2) which is 3 in base(10).

If we try to convert this to base 10 using the power and positions,

$$ (1 * 2^1) + (1 * 2^0) = 3 $$ Why is it automatically represented to base 10?

Answer 1

The only thing that's "automatic" about the use of base-ten representation is that the use of base ten is so overwhelmingly prevalent compared to the use of any other base, you can almost always get away with writing something in base ten and having it correctly understood without saying "base ten," whereas if you write something in any other base you often must say explicitly what base you are using to prevent misunderstanding.

The use of base ten as the "default" choice of base is so prevalent, even in places where other bases are in use, that some people are misled into thinking that the base-ten representation of a number is the number.

Our familiarity with base ten is such that if you had a more substantial conversion to do between two other bases, for example converting the number written $1110010$ in base two into its representation in base five, you might find it most convenient to convert the base-two representation to base ten and then convert that to base five.

Think about what it would take to work that example without using base ten. We could write \begin{align} 1110010_\text{two} &= 1\cdot 2^{11} + 1\cdot 2^{10} + 1\cdot 2^{4} + 0\cdot 2^{3} + 0\cdot 2^{2} + 1\cdot 2^1 + 0\cdot 2^0 \\ &= 1\cdot 224 + 1\cdot 112 + 1\cdot 31 + 0\cdot 13 + 0\cdot 4 + 1\cdot 2 + 0\cdot 1\\ &= 224 + 112 + 31 + 2 \\ &= 341 + 33 \\ &= 424, \end{align} in which every number except the base-two number $1110010_\text{two}$ is written in base five. Did you follow all the arithmetic operations on the right-hand side easily, with no temptation to do them incorrectly? If so, this method might work for you. I would not recommend it to most people.

Answer 2

$$11_2=3_b$$ is true in all bases $b>3,$ whereas

$$11_2=10_3.$$

In particular, when the base is implicitly $10_d$ (or $a$ if you prefer),

$$11_2=3.$$

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Note that a single digit is never ambiguous.