Help!!

sharpiie01

sharpiie01

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Dec 6, 2012
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okay so this is what the problem says...

Represent 110111101001 (BASE TWO) in BASE EIGHT.


I don't know how to do this. Can anyone help me?
 
sharpiie01 said:
okay so this is what the problem says...

Represent 110111101001 (BASE TWO) in BASE EIGHT.


I don't know how to do this. Can anyone help me?
Click to expand...
There is an easy way and a hard way to do this problem.

Can you translate the number in base 2 above into a number in base 10?

What do you get?
 
JeffM said:
There is an easy way and a hard way to do this problem.

Can you translate the number in base 2 above into a number in base 10?

What do you get?
Click to expand...


The thing is I dont know how to do this "BASE" thing...my teacher tried to explain it to us, but we just dont get it!!:confused:
 
sharpiie01 said:
The thing is I dont know how to do this "BASE" thing...my teacher tried to explain it to us, but we just dont get it!!:confused:
Click to expand...
OK Fair enough.

Base 10 means that numbers are represented using 10 distinct digits, 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.

Base 8 means that numbers are respesented using 8 distinct digits, 0, 1, 2, 3, 4, 5, 6, and 7.

Base 2 means that numbers are represented using 2 distinct digits.

That is not hard, is it? The base tells you how many distinct digits are used.

Now, can you explain place value in numbers to base 10? For example, what does 127 mean?
 
JeffM said:
OK Fair enough.

Base 10 means that numbers are represented using 10 distinct digits, 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.

Base 8 means that numbers are respesented using 8 distinct digits, 0, 1, 2, 3, 4, 5, 6, and 7.

Base 2 means that numbers are represented using 2 distinct digits.

That is not hard, is it? The base tells you how many distinct digits are used.

Now, can you explain place value in numbers to base 10? For example, what does 127 mean?
Click to expand...


After doing the to base 10 I get "3651." What do I do next? Or I guess how do I do the next step?:?
 
sharpiie01 said:
After doing the to base 10 I get "3651." What do I do next? Or I guess how do I do the next step?:?
Click to expand...
Well, you did not answer my question, presumably because it was obvious to you.

\(\displaystyle 127 = (1 * 100) + (2 * 10) + (7 * 1) = (1 * 10^2) + (2 * 10^1) + (7 * 10^0).\)

So not only do numbers in base ten use ten distinct digits, but the number as a whole represents a sum, where each term of the sum is a digit times a power of the base. I think you understand this. Do you?

When we look at a number represented in base two (or binary), we have two digits (0 and 1) and we are working with powers of two.

To keep things simple, I am going to show binary numbers in green and decimal numbers (base 10) in blue

110111101001 =

1 * 2^0 + 0 * 2^1 + 0 * 2^2 + 1 * 2^3 + 0 * 2^4 + 1 * 2^5 + 1 * 2^6 + 1 * 2^ 7 + 1 * 2^8 + 0 * 2^9 + 1 * 2^(10) + 1 * 2^(11)

= 1 + 8 + 32 + 64 + 128 + 256 + 1024 + 2048 = 9 + 96 + 384 + 3072 = 105 + 3456

3561.

So you made a mistake in addition or transcription, but I am pretty sure that you understood the basic process. Is that correct?

OK You now know how to translate into decimal. Let's see how to translate from decimal into a different base. The base we want to use is 8. And we know that the number will represent a sum involving powers of 8.

What is the largest power of 8 that will divide into 3561 and not have a quotient less than 1.

8^4 = 4096 so the highest power we shall need to deal with is 8^3 = 512.

I am going to write numbers in base 8 (or octal numbers) in red. 3561 / 512 = 6.95... so the leftmost digit of our octal number will be 6 = 6.

Now 6 * 512 = 3072 so the remaining octal terms must sum to 3561 - 3072 = 489. How much of that 489 can be represented by 8^2 times some number? If you can figure that out you, you will have the next digit in your octal number. Can you get the whole octal number now?
 
Last edited:
JeffM said:
Well, you did not answer my question, presumably because it was obvious to you.

\(\displaystyle 127 = (1 * 100) + (2 * 10) + (7 * 1) = (1 * 10^2) + (2 * 10^1) + (7 * 10^0).\)

So not only do numbers in base ten use ten distinct digits, but the number as a whole represents a sum, where each term of the sum is a digit times a power of the base. I think you understand this. Do you?

When we look at a number represented in base two (or binary), we have two digits (0 and 1) and we are working with powers of two.

To keep things simple, I am going to show binary numbers in green and decimal numbers (base 10) in blue

110111101001 =

1 * 2^0 + 0 * 2^1 + 0 * 2^2 + 1 * 2^3 + 0 * 2^4 + 1 * 2^5 + 1 * 2^6 + 1 * 2^ 7 + 1 * 2^8 + 0 * 2^9 + 1 * 2^(10) + 1 * 2^(11)

= 1 + 8 + 32 + 64 + 128 + 256 + 1024 + 2048 = 9 + 96 + 384 + 3072 = 105 + 3456

3561.

So you made a mistake in addition or transcription, but I am pretty sure that you understood the basic process. Is that correct?

OK You now know how to translate into decimal. Let's see how to translate from decimal into a different base. The base we want to use is 8. And we know that the number will represent a sum involving powers of 8.

What is the largest power of 8 that will divide into 3561 and not have a quotient less than 1.

8^4 = 4096 so the highest power we shall need to deal with is 8^3 = 512.

I am going to write numbers in base 8 (or octal numbers) in red. 3561 / 512 = 6.95... so the leftmost digit of our octal number will be 6 = 6.

Now 6 * 512 = 3072 so the remaining octal terms must sum to 3561 - 3072 = 489. How much of that 489 can be represented by 8^2 times some number? If you can figure that out you, you will have the next digit in your octal number. Can you get the whole octal number now?
Click to expand...

Oh, that's awesome and makes more sense... :-D
 
JeffM said, initially, "There is an easy way and a hard way to do this problem". Changing from base 2 to base 10 and then changing that to base 8 is the "hard way". And, there is an easier way to go from base 10 to base 8 than directly using the definition. For example, if I divide, say, 733 by 8 I find that it divides 91 times with remainder 5: 733= 8(91)+ 5. Now divide 91 by 8. The quotient is 11 with remainder 3. 91= 8(11)+ 3 so \(\displaystyle 733= 8(8(11)+ 3)+ 5= 8^2(11)+ 8(3)+ 5\). Now divide 11 by 8. The quotient is 1 with remainder 3. 11= 8(1)+ 3 so \(\displaystyle 733= 8^2(8(1)+ 3)+ 8(3)+ 5= 8^3(1)+ 8^2(3)+ 8(3)+ 5\). That is, since the "digits" in base 8 are the coefficients of the powers of 8, 733 in base 8 is \(\displaystyle 1335_8\). That is, the digits are just the remainders in those divisions.

We can do the same thing with base 2: 2 divides into 733 366 times with remainder 1, 2 divides into 366 183 times with remainder 0, 2 divides into 183 91 times with remainder 1, 1 divides into 91 45 times with remainder 1, 2 divides into 45 22 times with remainder 1, 2 divides into 22 11 times with remainder 0, 2 divides into 11 5 times with remainder 1, 2 divides into 5 2 times with remainder 1, 2 divides into 2 1 time with remainder 0, and, finally, 2 divides into 1 0 times with remainder 1. Reversing the order of those, 733 in base 2, 1011011101.


But even simpler is the fact that 8 is a power of 2: \(\displaystyle 8= 2^3\). 1, 10, 11, and 100, in base 2, are, of course, 1, 2, 3, and 4 in base 10. Continuing 101, 110, 111, and 1000 are 5, 6, 7, and 8. Notice that when we get to powers of two we have a new digit just as happens, in base 10, when we have powers of 10: \(\displaystyle 10= 10^1\), \(\displaystyle 100= 10^2\), \(\displaystyle 1000= 10^3\), etc. So, separating those binary digits ("bits") into groups of 3, we have 001 011 011 101. Separately, those would be 1 3 3 5. But since they are all less than 8, they are also digits in base 8. And, using the fact that \(\displaystyle 8= 2^3\), we can include the powers of 8 as \(\displaystyle (1\times 8^3)+ (3\times 8^2)+ (3\times 8)+ 5\). That is, 1011011101, in base 2, is the same as 1335 in base 8.
 
HallsofIvy said:
JeffM said, initially, "There is an easy way and a hard way to do this problem". Changing from base 2 to base 10 and then changing that to base 8 is the "hard way". And, there is an easier way to go from base 10 to base 8 than directly using the definition. For example, if I divide, say, 733 by 8 I find that it divides 91 times with remainder 5: 733= 8(91)+ 5. Now divide 91 by 8. The quotient is 11 with remainder 3. 91= 8(11)+ 3 so \(\displaystyle 733= 8(8(11)+ 3)+ 5= 8^2(11)+ 8(3)+ 5\). Now divide 11 by 8. The quotient is 1 with remainder 3. 11= 8(1)+ 3 so \(\displaystyle 733= 8^2(8(1)+ 3)+ 8(3)+ 5= 8^3(1)+ 8^2(3)+ 8(3)+ 5\). That is, since the "digits" in base 8 are the coefficients of the powers of 8, 733 in base 8 is \(\displaystyle 1335_8\). That is, the digits are just the remainders in those divisions.

We can do the same thing with base 2: 2 divides into 733 366 times with remainder 1, 2 divides into 366 183 times with remainder 0, 2 divides into 183 91 times with remainder 1, 1 divides into 91 45 times with remainder 1, 2 divides into 45 22 times with remainder 1, 2 divides into 22 11 times with remainder 0, 2 divides into 11 5 times with remainder 1, 2 divides into 5 2 times with remainder 1, 2 divides into 2 1 time with remainder 0, and, finally, 2 divides into 1 0 times with remainder 1. Reversing the order of those, 733 in base 2, 1011011101.


But even simpler is the fact that 8 is a power of 2: \(\displaystyle 8= 2^3\). 1, 10, 11, and 100, in base 2, are, of course, 1, 2, 3, and 4 in base 10. Continuing 101, 110, 111, and 1000 are 5, 6, 7, and 8. Notice that when we get to powers of two we have a new digit just as happens, in base 10, when we have powers of 10: \(\displaystyle 10= 10^1\), \(\displaystyle 100= 10^2\), \(\displaystyle 1000= 10^3\), etc. So, separating those binary digits ("bits") into groups of 3, we have 001 011 011 101. Separately, those would be 1 3 3 5. But since they are all less than 8, they are also digits in base 8. And, using the fact that \(\displaystyle 8= 2^3\), we can include the powers of 8 as \(\displaystyle (1\times 8^3)+ (3\times 8^2)+ (3\times 8)+ 5\). That is, 1011011101, in base 2, is the same as 1335 in base 8.
Click to expand...
Halls is quite right about easier ways to do things. Conversion between octal and binary (and conversion between hexadecimal and binary) is super easy. I was going to explain that shortcut once you had the concept because the shortcut will not work if you are asked to convert base 5 to base 7 or base 6 to base 11.

To go from binary to octal, see whether the number of digits in the binary representation is evenly divisible by three. If it is not, add leading zeroes to make it evenly divisible. Now divide the string of binary digits into groups of three contiguous digits. Now convert each sub-string of three binary digits into one octal digit according to this table

000 to 0
001 to 1
010 to 2
011 to 3
100 to 4
101 to 5
110 to 6
111 to 7.

You can convert from octal to binary even more simply by translating the other way round, with each octal number being translated into the corresponding three bit string of binary digits.

316 in octal is 011001110 in binary.

A similar process using strings of 4 binary digits makes conversion between hexadecimal and binary almost as easy.

As I said, however, this shortcut does not work generally for conversion among bases.
 
This is why computer work typically involves hexadecimal (and some times octal) number- they are easy to convert to and from binary, which is the computer's "natural" number system.
 
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