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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?

sidoshi
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    Because you wrote it in base 10. – Parcly Taxel Oct 07 '18 at 11:53
  • I don't understand, 3 could be in many other base? I am having a hard time getting my head around this – sidoshi Oct 07 '18 at 11:55
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    In which base are expressed the $2$s on the LHS? Because your conversion is expressed with $3$ in any base greater than $2$ – N74 Oct 07 '18 at 11:57
  • Yes, in base 3, "3" is just 10 which we would interpret as $1(3)+ 0(1)= 3$. The point is that we are doing that arithmetic [b]in[/b] base 10 because that is our usual base for arithmetic. – user247327 Oct 07 '18 at 11:58
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    The use of base $10$ "by default" dates back when ten-fingered people started counting. This has not always been the case. For instance, the early Babylonians seem to have had one six-fingered hand and one ten-fingered one. –  Oct 07 '18 at 12:31

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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.

David K
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  • I notice you used the subscript "two" to indicate base 2. An example of our inherent base 10 bias is that it is common to indicate the base using a subscript in base 10. A neutral but unhelpful convention would be to indicate the base in its own base but then it would always look like $_{10}$. – badjohn Oct 07 '18 at 12:31
  • Thanks. I understand it clearly now. So the value 3 received in the question is assumed to be base 10 automatically. But it could have been any base > 3 as explained by @Yves. What was made clear from your answer is that when I am calculating (2^x), I am writing the value in base10 and hence I am getting the final answer in base 10. Hope I am not misunderstanding here but your answer helped clear a lot of things up. Thanks again. – sidoshi Oct 07 '18 at 12:42
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$$11_2=3_b$$ 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),

$$11_2=3.$$

Note that a single digit is never ambiguous.