I am trying to solve the following using two's complement in binary form.
7.5 - 6.75
7.5 - 6.75 can be rewritten as (7.5)+(-6.75)
taking binary of decimal numbers
7.5 = 111.1
6.75 = 110.11
-6.75= 001.00
Now where should I add 1, to the left of "."(period) or to the right?
To perform the subtraction of signed numbers (M – N) with 2’s complements proceed as follows:
The number of digits of M and N should be the same.
Obtain the 2’s complement of N (including the sign bit).
Add the 2’s complement of N to M (including the sign bit).
If the summation produces an end carry. Discard the end carry. After the end carry is discarded, the leftmost bit is the sign bit.
If the sign bit is 0, the result is positive,
If the sign bit is 1, the result is negative.
If the summation does not produce an end carry, the leftmost bit is the sign bit.
To Produce 2's compliment:
The 2’s complement of the subtrahend 10001 = 01111
1 1 1 1 carry bits
1 0 0 1 1
+ 0 1 1 1 1
1 0 0 0 1 0
Since the summation produces an end carry the result is positive.
Discard the end carry; the result is a positive number.
Related
I'm doing a question about 2's complement and I was just wandering about the place value of 2's complement. For example, in binary, the place value goes like 1 2 4 8 16 32 and doubles by two. but I'm not sure what the place value is for negative numbers.
The question I've got is asking you that is this number 10110001 was a 2's complement integer what would it be. I under than that you flip the numbers and add one but how can I actullay find the value of this?
In the 2's complement system, the most significant place ("sign bit", though this term causes confusion since it isn't purely a sign) has the weight -2k where k is its index, for example for 8 bits:
weights: -128 +64 +32 +16 +8 +4 +2 +1
example: 1 1 1 1 1 1 0 0 (0xFC) = -128+64+32+16+8+4 = -4
I am trying to convert FFAD (hex) to a decimal value and then do one and two's complement on it. FFAD is represented as a 16 bit integer.
When I convert FFAD to base 2 I get 1111111110101101.
My problem is how I know if it is a negative number or not?
I have the binary, now to do ones complement normally I would change the last bit from a 0 to a 1 and then flip all the bits, but as a 16 bit integer I don't have any more available bits. Since the 16th bit is a 1 does that mean it is a negative number? How would I do the complement of that? I am just all around confused with this problem and any pointers would be greatly appreciated.
Its negative because it is left padded with ones. Left padding with zeros would indicate a positive number. For example, here are the decimal values for three bit numbers:
011 = 3
010 = 2
001 = 1
000 = 0
111 = -1
110 = -2
101 = -3
100 = -4
Notice, numbers that are left padded with ones are negative and numbers that are left padded with zeros are positive.
Standard procedure for 2's compliment is to negate your bits, and then add 1.
1111111110101101 becomes 0000000001010010. Then you add 1 to get 0000000001010011.
Or, what is the range of numbers that can be represented on a 4-bit machine using 2s-complement?
That would be -8 to +7
The range is -8 to 7, or 1000 to 0111. You can see the full range here.
4 bits (using 2's complement) will give you a range from -8 to 7.
This should be straightforward to work out yourself.
Range in twos complement will be:
-1 * 2 ^ (bits - 1)
to
2 ^ (bits - 1) - 1
So for 4 bits:
-1 * 2 ^ (4 - 1) = -1 * 2 ^ 3 = -8
to
2 ^ (4 - 1) - 1 = 2 ^ 3 - 1 = 7
Also, if you are interested and for others maybe browsing this question -
twos complement is used for easy binary arithmetic:
to add - you just add the two numbers without conversion and disregard the overflow:
-6 + 7 = 1
is
1010 = -6
0111 = 7
------
(1)0001 = 1 (ignoring the overflow)
...and more yet - to convert a negative binary number to its opposite positive number:
if the sign bit (highest order bit) is 1, indicating negative... reading from least significant to most significant bit (right to left), preserve every bit up through the first "1", then invert every bit after that.
So, with 8 bit
10011000 .. becomes
01101000 (* -1) = 104 * -1 = -104
and that is why 10000000 is your lowest negative number (or in X bit 1000.all zeroes..000), it translates to unsigned 10000000 * -1 = -128
Maybe a long winded answer but to those without the 1s and 0s background I figure it is useful
Well let's dissect this question.
Firstly, the question should be framed as - "The least*(because it is negative, so biggest is not the right usage)* possible negative number to be stored in 4-bit would be?"
Numbers are of two types -
Signed (holds both + and - numbers )
Unsigned (holds only + numbers)
We will be using binary representation to understand this.
For Unsigned -
4-bit minimum number = 0000 (0)
4-bit maximum number = 1111 (255)
So range would be Range : 0-15
For Signed 4-bits, first bit represent sign(+/-) and rest 3-bits represent number.
4-bit minimum will be a negative number.
So replace the first bit(MSB) with 1 in 0000(least unsigned number), making it 1000.
Calculate decimal equivalent of 1000 = 1*2^3 = 8
So, number would be -8 (- as we have 1 as the first bit in 1000)
4-bit maximum will be a positive number.
So replace the first bit(MSB) with 0 in 1111(largest unsigned number), making it 0111.
Calculate decimal equivalent of 0111 = 1*2^2 + 1*2^1 + 1*2^0 = 7
So, number would be +7 (+ as we have 0 as the first bit in 0111)
Range would be -8 to +7.
What is the binary form of -10? How it is calculated?
To convert -10 (decimal) to binary:
Repeatedly divide the absolute value (|-10| = 10) of the number by 2 until you get 0 in the quotient:
(10 / 2 = 5 R 0)
(5 / 2 = 2 R 1)
(2 / 2 = 1 R 0)
(1 / 2 = 0 R 1) // zero is the value in the quotient so we stop dividing
Place the remainders in order to obtain the binary equivalent:
1010
For an 8-bit cell the answer is 0000 1010, 16-bit cell 0000 0000 0000 1010, and so on.
Take the one's complement by inverting the bits (we will assume an 8-bit cell holds the final value):
0000 1010
1111 0101 // bits are inverted
Now take the 2's complement by adding 1:
1111 0101
+ 1
----------
1111 0110 // final answer
What happens with a 4-bit cell?
The one's complement would be:
1010
0101 // inverted bits
Taking the 2's complement produces:
0101
+ 1
----
0110 // final answer for a 4-bit cell
Since the number should be negative and the result does not indicate that (the number begins with 0 when it should begin with a 1) an overflow condition would occur.
Take the binary form of 10, invert all the bits, and add one.
10 0000 1010
invert 1111 0101
add 1 1111 0110
Follow this link. You can find the binary form of a negative number say -n by finding out the two's complement of n.
Your question has no "right" answer. First, you have to define the representation you want to use. Do you want to use two's complement, ones' complement, sign-magnitude, or something else? Then you have to define how many bits to use.
Let's say we are working with 5-bits wide representations.
+10: 0 1 0 1 0
Let's look at sign-magnitude system. In this system, the most significant bit (bit 5) is 1 if a number is negative, and 0 otherwise. The rest of the bits represent the magnitude (absolute value) of the number.
So we get:
-10: 1 1 0 1 0 (sign-magnitude, 5 bits)
Let's look at ones' complement now. Here, a negative number is represented by just changing 1s to 0s and vice-versa (hence the name ones' complement—you complement a number with respect to a long sequence of 1s).
So we get:
-10: 1 0 1 0 1 (ones' complement, 5 bits)
Finally, let's look at two's complement system. In this, we take a number in its ones' complement system, and then add 1 to it.
So we get:
-10: 1 0 1 0 1
1
---------
1 0 1 1 0 (two's complement, 5 bits)
---------
Thus, the binary representation of a negative number depends upon the system we use, and the number of bits available to us.
Also, you might have noticed the position of the apostrophe in ones' complement and two's complement. Why is is not one's complement or twos' complement? Then answer, from Knuth:
A two's complement number is complemented with respect to a single power of 2, while a ones' complement number is complemented with respect to a long sequence of 1s. Indeed, there is also a "twos' complement notation," which has radix 3 and complementation with respect to (2...22)3
I have seen these mentioned in the context of C and C++, but what is the difference between signed and unsigned variables?
Signed variables, such as signed integers will allow you to represent numbers both in the positive and negative ranges.
Unsigned variables, such as unsigned integers, will only allow you to represent numbers in the positive and zero.
Unsigned and signed variables of the same type (such as int and byte) both have the same range (range of 65,536 and 256 numbers, respectively), but unsigned can represent a larger magnitude number than the corresponding signed variable.
For example, an unsigned byte can represent values from 0 to 255, while signed byte can represent -128 to 127.
Wikipedia page on Signed number representations explains the difference in the representation at the bit level, and the Integer (computer science) page provides a table of ranges for each signed/unsigned integer type.
While commonly referred to as a 'sign bit', the binary values we usually use do not have a true sign bit.
Most computers use two's-complement arithmetic. Negative numbers are created by taking the one's-complement (flip all the bits) and adding one:
5 (decimal) -> 00000101 (binary)
1's complement: 11111010
add 1: 11111011 which is 'FB' in hex
This is why a signed byte holds values from -128 to +127 instead of -127 to +127:
1 0 0 0 0 0 0 0 = -128
1 0 0 0 0 0 0 1 = -127
- - -
1 1 1 1 1 1 1 0 = -2
1 1 1 1 1 1 1 1 = -1
0 0 0 0 0 0 0 0 = 0
0 0 0 0 0 0 0 1 = 1
0 0 0 0 0 0 1 0 = 2
- - -
0 1 1 1 1 1 1 0 = 126
0 1 1 1 1 1 1 1 = 127
(add 1 to 127 gives:)
1 0 0 0 0 0 0 0 which we see at the top of this chart is -128.
If we had a proper sign bit, the value range would be the same (e.g., -127 to +127) because one bit is reserved for the sign. If the most-significant-bit is the sign bit, we'd have:
5 (decimal) -> 00000101 (binary)
-5 (decimal) -> 10000101 (binary)
The interesting thing in this case is we have both a zero and a negative zero:
0 (decimal) -> 00000000 (binary)
-0 (decimal) -> 10000000 (binary)
We don't have -0 with two's-complement; what would be -0 is -128 (or to be more general, one more than the largest positive value). We do with one's complement though; all 1 bits is negative 0.
Mathematically, -0 equals 0. I vaguely remember a computer where -0 < 0, but I can't find any reference to it now.
Signed variables use one bit to flag whether they are positive or negative. Unsigned variables don't have this bit, so they can store larger numbers in the same space, but only nonnegative numbers, e.g. 0 and higher.
For more: Unsigned and Signed Integers
Unsigned variables can only be positive numbers, because they lack the ability to indicate that they are negative.
This ability is called the 'sign' or 'signing bit'.
A side effect is that without a signing bit, they have one more bit that can be used to represent the number, doubling the maximum number it can represent.
Signed variables can be 0, positive or negative.
Unsigned variables can be 0 or positive.
Unsigned variables are used sometimes because more bits can be used to represent the actual value. Giving you a larger range. Also you can ensure that a negative value won't be passed to your function for example.
unsigned is used when ur value must be positive, no negative value here,
if signed for int range -32768 to +32767
if unsigned for int range 0 to 65535
Unsigned variables are variables which are internally represented without a mathematical sign (plus or minus) can store 'zero' or positive values only. Let us say the unsigned variable is n bits in size, then it can represent 2^n (2 power n) values - 0 through (2^n -1). A signed variable on the other hand, 'loses' one bit for representing the sign, so it can store values from -(2^(n-1) -1) through (2^(n-1)) including zero. Thus, a signed variable can store positive values, negative values and zero.
P.S.:
Internally, the mathematical sign may be represented in one's complement form, two's complement form or with a sign bit (eg: 0 -> +, 1-> -)
All these methods effectively divide the range of representable values in n bits (2^n) into three parts, positive, negative and zero.
This is just my two cents worth.
I hope this helps.