The Wikipedia says:
"To convert from a base-10 integer numeral to its base-2 (binary) equivalent, the number is divided by two, and the remainder is the least-significant bit. The (integer) result is again divided by two, its remainder is the next least significant bit. This process repeats until the quotient becomes zero."
http://en.wikipedia.org/wiki/Binary_number#Conversion_to_and_from_other_numeral_systems
But my students ask, "Why? What do we actually do when repeatedly dividing by two? Why is the first bit LSB?"
I have an explanation but it is interesting for me to see your opinion.
To reveal the algorithm idea, I use first a little strange example - convert a decimal number (e.g., 345) to... a decimal number (345) again:)
You have suggested a clever trick but my purpose is to show what the idea of the classic division-by-2 method is.
By dividing a number by 2, you are forcing the least bit (in binary format) to the remainder. By doing this again, you are forcing the next least significant bit to the remainder. You repeat this until you have forced all bits out to the remainder, one bit by one bit. By aligning the last remainder as the most significant bit, and aligning the remaining bits accordingly, you get the binary, bit by bit.
It is sort of difficult to express this in text only, but I think you get the idea.
Similar expression also applies to multiply-by-2 to get the fractional binaries. Only this time, you force the most significant bit into the whole part, bit by bit.
Hope this helps.
Hello Cyril. I like your questions because I always have hard time like you trying to explain to myself in a way as to approach the students. On this issue, I tell the students that, in analogy with decimal numbers with 1st digit on the right is for ones, 2nd for 10's, 3rd for 100's,...etc, the binary 1st on the right is ones, 2nd should be 2's, 3rd for 4's, 8's, 16's,..etc. So in converting decimal to binary, we are decomposing, same as we decompose a decimal number by dividing by the base (10) first to get how many 10s with the remainder naturally the ones, and we divide again by a 10 (which is as if we divided by 100) to get the remaining 10s (after the hundreds), again divide by 10 (as we already divided by 1000) to get the remaining 100's digit,...etc.
Similarly, we do a repeated division by 2 (the base). First division by 2 to get how many 2s are in the number and the remaining ones (which can be only a 1 or 0), Second division by 2 as if we already divided by 4 to get the remaining 2s in the 2's digit, and third division by 2 (or 8 total) to get the remaining 4s in the 4's (third digit), ...and so on. This way, I can proceed to converting from decimal to for example base 8 (ones, 8's, 64's, ...etc) or Hex (ones, 16's, 256's,...etc) or whatever smoothly with no objection. Interesting. Thanks. @AlDmour.
Thank you, Ismat! I think and present the procedure exactly in the same manner. I only extend the explanation with additional thoughts as follows.
In the example above, I suggest to my students to imagine that we have 345 objects (e.g., a sack of 345 nuts-:) what is an example of a number written in a non-positional numerical system, and they have to convert (by grouping) them into a number of a positional decimal system. And we have notated this amount of objects with a number (345) of a decimal system only because we have somehow to notate it (thus the paradox of this decimal-to-decimal conversion).
For this purpose, as you do, we have first to find the number of the whole 10's - the remaining is an incomplete 10, i.e., a few ones. Then, by dividing the whole 10's by 10, we find the number of the whole 100's - the remaining is an incomplete 100, i.e., a few 10's... Presented in this manner, the procedure resembles an analog-to-digital procedure... I always make a connection between these abstract methods and the material analog-to-digital and digital-to-analog conversions...
By the way, I have asked another, even more interesting question - "What is the idea of the multiplication-by-2 method for a fraction decimal-to-binary conversion?, but it has not appeared and I abandoned it then. It is interesting that in this case, the first digit is the most significant digit (MSB in the binary representation)...
Regards, Cyril.
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