# What is Base 2 (binary system)?

• Apr 14th 2008, 06:48 PM
Sothi
What is Base 2 (binary system)?
Hi,

I have a question about how the base 2 binary system works.

In base 2 (binary system), which equation is false?

A) 11+1=101 B)1x10=10 C)11x11=1001 D) 100/1=100 E) 1+1+1=11

If anyone does get the answer, could you explain thoroughly how you got it? Thank you very much!
• Apr 14th 2008, 06:51 PM
Mathstud28
Ok
Quote:

Originally Posted by Sothi
Hi,

I have a question about how the base 2 binary system works.

In base 2 (binary system), which equation is false?

A) 11+1=101 B)1x10=10 C)11x11=1001 D) 100/1=100 E) 1+1+1=11

If anyone does get the answer, could you explain thoroughly how you got it? Thank you very much!

To check I will show you how to convert a number in base $\displaystyle n$ back into base ten....you do this say I have a number ABCDEF$\displaystyle _{n}$ then to convert it back to base ten it would be $\displaystyle F+E\cdot{n}+D\cdot{n^2}+C\cdot{n^3}+B\cdot{n^4}+A\ cdot{n^5}$ make sense?
• Apr 14th 2008, 06:59 PM
TheEmptySet
Quote:

Originally Posted by Sothi
Hi,

I have a question about how the base 2 binary system works.

In base 2 (binary system), which equation is false?

A) 11+1=101 B)1x10=10 C)11x11=1001 D) 100/1=100 E) 1+1+1=11

If anyone does get the answer, could you explain thoroughly how you got it? Thank you very much!

The binary system is base 2 instead of base 10.

When you see a number like

143 we associate 1 as being in the hundreds place, 4 in the tens and 3 in the ones or

$\displaystyle 143=1\cdot 10^2+4 \cdot10^1+3\cdot 10^0$

base two uses two instead of ten.

so lets write 14 as base two

so lets list the powers of 2

$\displaystyle 2^0=1,2^1=2,2^2=4,2^3=8,2^4=16,...$

fourteen could be written as

$\displaystyle 14=1\cdot \underbrace{2^3}_{14-8=6}+1 \cdot \underbrace{2^2}_{6-4=2}+1\cdot \underbrace{2^1}_{2-2=0}+0\cdot 2^{0}$

so 14 base two is 1110

I hope this helps.

Good luck
• Apr 14th 2008, 07:02 PM
xifentoozlerix
Remember that decimal (i.e. base 10) notation breaks down into differing place values...

$\displaystyle 123456789_{10} = 1 \times 10^8 + 2 \times 10^7 + $$\displaystyle 3 \times 10^6 + 4 \times 10^5 + 5 \times 10^4 + 6 \times 10^3 + 7 \times 10^2 + 8 \times 10^1 + 9 \times 10^0 For binary (i.e. base 2), you have a similar breakdown... \displaystyle 100110101_2 = 1 \times 2^8 + 0 \times 2^7 + 0 \times 2^6 + 1 \times 2^5 + 1 \times 2^4 + 0 \times 2^3 + 1 \times 2^2 + 0 \times 2^1$$\displaystyle + 1 \times 2^0 = 2^8 + 2^5 + 2^4 + 2^2 + 2^0 = 256_{10} + 32_{10} + 16_{10} + 4_{10} + 1_{10} = 309_{10}$
• Apr 14th 2008, 07:10 PM
angel.white
Quote:

Originally Posted by Sothi
Hi,

I have a question about how the base 2 binary system works.

In base 2 (binary system), which equation is false?

A) 11+1=101 B)1x10=10 C)11x11=1001 D) 100/1=100 E) 1+1+1=11

If anyone does get the answer, could you explain thoroughly how you got it? Thank you very much!

A is wrong.

First, for you, the calculator that comes with windows can convert between binary and decimal (in scientific mode, which can be enabled under the "view" toolbar) but be careful lest it becomes a crutch.

Now, there are any number of ways to do this, when I'm hazy I just convert to decimal and check.

Converting to decimal this becomes:

A)3+1=5
B)1*2=2
C)3*3=9
D)4/1=4
E)1+1+1=3

Clearly A is the false answer. You can simply get good at converting back and forth (just takes a little practice) then do them all this way.

Otherwise, you can try to do it all in binary like this:

A)11+1 = 1(1+1) = (1+1)0 = 100

B)1*x = x so 1*10=10

C)
$\displaystyle \begin{array}{rrrr} &&1&1\\ &*&1&1\\ \hline&&1&1\\ +&1&1&0\end{array}$

and 11+110
1's digit: 1+0 = 1
2's digit: 1+1 = 10 -> 0 with a 1 carried to the 4's digit
4's digit: 1+0+1 (the last 1 is brought up from the 2's digit) = 10-> 0 with a 1 carried to the 8's digit
8's digit: 0+1 (the 1 is brought up from the 4's digit)