MadSci Network: Other Query:

### Re: How do I explain the relationship between 1's and 0's, and 2 to some power?

Date: Wed Sep 27 14:18:42 2000
Posted By: Sarah Tegen, Grad student, Molecular and Cell Biology, UC-Berkeley
Area of science: Other
ID: 967945849.Ot
Message:
```
Hi Robert,
Interesting question.  I asked some of my engineering friends from college to
help with this question.  Here are the answers I got from them:

From Len:
Draw a dot.

Put two lines at about 30 degrees from straight down (two leaves)

Draw dots at the end and draw two more from each dot, and repeat.

At each level, going down, you have 1,2,4,8,16, etc. dots.

This is a binary "tree" and the final row is filled with "leaves"

Explain that each extra digit multiplies the maximum by two, and the final level
may have zero or one or two or 2 to the number of
levels leaves.
--------------------------------
From Victor:

Sure, have the student write out all the possible combinations that a single bit
can yield (2: 0,1), then all the combinations that two bits can yield (4:
00,01,10,11) then all the combinations that three bits can yield
(8: 000, 001, 010, 011, 100, 101, 110, 111) and so on until you're
bored.
It should become pretty obvious that with every bit you add, you get twice
as many total possibilities. Restated, with N bits you get 2^N combinations.

Why does the first bit (on the right) always stand for 1 and then next (to
the left) for 2 and the third for 4 and so on?

For the same reason we write left to right; Its just convention. Our numbers get
bigger to the left, like 7 becomes 27 (and not 72) when you add 20, and binary
works the same way. The smallest "one's place" is on the right and they grow
leftward.

But why can you only count to 2^N-1??

Basically, it's because of 0.

You have 2^N combinations (because each bit you add gives you twice as many
possibilities)  and you can use the right to left system to put the combinations
in a "smallest to largest" order.

BUT, like the fingers on your hands (10 of them) which can count to 10 or 9
depending on whether you make the first finger 1 or 0, binary can go to 2^N (the
number of combinations) if the first combination is 1, or 2^N-1 if you  make the
first combination 0.  The first combination is always all 0 and it would be SO
ugly to call it 1 that we call it 0 and reserve 1 for 00001.

And if THAT doesn't work, make the student keep writing out all the combinations
until it PAINFULLY obvious that there's twice as many with each added bit.
---------------------------
From Yev:
The only additional idea I can come up with is a "tree" because it's pretty easy
to visualize how the number of "possibilities" increases with each level
(although the analogy is better suited to probably and statistics than plain
binary counting).

So you start off with 2^1:

|
/ \
0   1

The two possibilities are 0 and 1 so you have two values that you can represent
with 1 bit.

Now you move to two bits (2^2):
|
/  \
0    1
/ \  / \
0   1 0  1

If you "read" the tree from the root to the leaf you see that you have four
values that can be represented (00 01 10 11)  If you just assign each a decimal
value you can see that you can represent numbers 0-3 this way.

Now as you keep adding more levels to the tree, you will see that you double the
number of leaves each time, and that the number of values you can represent
doubles as well.
|
/      \
0        1
/ \      /  \
0   1    0    1
/\   /\   /\   /\
0  1 0  1 0  1 0  1

(000 001 010 011 100 101 110 111) -> (0 1 2 3 4 5 6 7)

This is really more appropriate to describing outcomes in probability and
statistics classes, but it may help in this application as well.
-------------------
From Arun:

Robert K. may have tried this already, but I have found that clearly explaining
the relationship between digit position and value as powers of 10 in a decimal
digit helps in understanding the same relationship in binary digits.  Since that
last sentence reads pretty incomprehensibly, let me try an example:

10^3  10^2  10^1  10^0

2     3     1     5    = 2 * 10^3 + 3 * 10^2 + 1 * 10^1 + 5 * 10^0 = 2315

In the same way....

2^3  2^2  2^1  2^0

1    1    0    1   = 1 * 2^3 + 1 * 2^2 +  0 * 2^1 + 1 * 2^0  = 1101 binary
= 13 decimal

The second part of this explanation involves explaining why decimal has 10
digits and binary has only two.  This is probably best done by asking and
answering the following questions....

In base 10:

Q: How many 10^0s does it take to have 1 10^1:
A: 10 (counting through these is usually helpful)

Q: How many 10^1s does it take to have 1 10^2:
A: 10 (counting through these is usually helpful)

...

Q: How many 10^2s does it take to have 1 10^3:
A: 10 (counting through these is usually helpful)

In base 2:

Q: How many 2^0s does it take to have 1 2^1:
A: 2 (counting through these is usually helpful)

Q: How many 2^1s does it take to have 1 2^2:
A: 2 (counting through these is usually helpful)

...

Q: How many 2^2s does it take to have 1 2^3:
A: 2 (counting through these is usually helpf

So you need 1 (digit representing 0) plus one less than the number of units to
equal the next place.
---------------
From Jeff:
Binary counter
---------------
From Sarah:
So I'm not an engineer, but the first thing I though of was the tree analogy
that a couple people have suggested.  I also really liked Arun's question
and answer thing.  That seemed to help me a bunch.

Please let me know if you need some other suggestions.  I'm sure I (we)
could come up with others.

Good luck!
-Sarah

```

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