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Number Systems - De Montfort Universityjordan/teaching/elec1099/NumberSystems.pdf · Number Systems...
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Number Systems J. Dimitrov [email protected] Software Technology Research Laboratory (STRL) De Montfort University Leicester, UK. J.Dimitorv, STRL, DMU, [email protected] – p. 1
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Transcript of Number Systems - De Montfort Universityjordan/teaching/elec1099/NumberSystems.pdf · Number Systems...
Number SystemsJ. Dimitrov
Software Technology Research Laboratory (STRL)
De Montfort University
Leicester, UK.
J.Dimitorv, STRL, DMU, [email protected] – p. 1
Overview
Octal and Hex
Conversion from base 10
Binary, Octal and Hexadecimal arithmetic
Negative numbers
2’s complement
Real numbers
J.Dimitorv, STRL, DMU, [email protected] – p. 2
Binary Numbers
Let us recall from last week!
Computers use Binary (base 2).
Base b = 2.
Digits dj = 0, 1.
N = dn−1×bn−1+dn−2×bn−2+...+d1×b1+d0×b0.
Example:N = 10011101(2) = 1706051413120110 = 1×27+0×
26+0×25+1×24+1×23+1×22+0×21+1×20 = 157
However, these are the positive numbers only!
J.Dimitorv, STRL, DMU, [email protected] – p. 3
Octal and Hexadecimal
With octal we have that the base is 8 and the digitsare from 0 to 8 − 1 = 7. Example:
N = 75361203(8) = 7756356413220130 = 7×87+5×86+
3×85+6×84+1×83+2×82+0×81+3×80 = 16114307.
Hexadecimal is base 16 and the digits are from 0 to15. However, with digits larger then 9, e.g. 13, we arerisking confusion between the digit 13 and the twodigit number 1316 = 1910. So, we use the notation,10 = A, 11 = B, 12 = C, 13 = D, 14 = E, 15 = F .
N = A3B6(16) = A332B160 =
10 × 163 + 3 × 162 + 11 × 161 + 6 × 160 = 41910.
J.Dimitorv, STRL, DMU, [email protected] – p. 4
From base 10 to 2, 8 and 16
We saw how we convert from Binary, Octal and Hexinto Decimal. Let us look into the dual problem.
We go back to
N = dn−1× bn−1 +dn−2× bn−2 + ...+d1× b1 +d0× b0
and convert this into
N = b(dn−1 × bn−2 + dn−2 × bn−3 + ... + d1 × b0)+d0
and further
N = b × N1 + b0.
J.Dimitorv, STRL, DMU, [email protected] – p. 5
From base 10 to 2, 8 and 16
N = b(dn−1 × bn−2 + dn−2 × bn−3 + ... + d1 × b0)+d0
N = b × N1 + b0.
This tells us that the rightmost digit b0 in N ’srepresentation in base b is the remainder of thedevision of N with b. Again we can do the samewith N1 and find the second digit b1, etc.
J.Dimitorv, STRL, DMU, [email protected] – p. 6
From base 10 to 2, 8 and 16
Example: Let us convert 25(10) into binary.
25
J.Dimitorv, STRL, DMU, [email protected] – p. 7
From base 10 to 2, 8 and 16
Example: Let us convert 25(10) into binary.
2 2512 r1 = d0
J.Dimitorv, STRL, DMU, [email protected] – p. 7
From base 10 to 2, 8 and 16
Example: Let us convert 25(10) into binary.
2 252 12 r1 = d0
6 r0 = d1
J.Dimitorv, STRL, DMU, [email protected] – p. 7
From base 10 to 2, 8 and 16
Example: Let us convert 25(10) into binary.
2 252 12 r1 = d0
2 6 r0 = d1
3 r0 = d2
J.Dimitorv, STRL, DMU, [email protected] – p. 7
From base 10 to 2, 8 and 16
Example: Let us convert 25(10) into binary.
2 252 12 r1 = d0
2 6 r0 = d1
2 3 r0 = d2
1 r1 = d3
J.Dimitorv, STRL, DMU, [email protected] – p. 7
From base 10 to 2, 8 and 16
Example: Let us convert 25(10) into binary.
2 252 12 r1 = d0
2 6 r0 = d1
2 3 r0 = d2
2 1 r1 = d3
0 r1 = d4
J.Dimitorv, STRL, DMU, [email protected] – p. 7
From base 10 to 2, 8 and 16
Example: Let us convert 25(10) into binary.
2 252 12 r1 = d0
2 6 r0 = d1
2 3 r0 = d2
2 1 r1 = d3
0 r1 = d4
Therefore our number 25(10) = 1d41d30d20d11d0
(2).
J.Dimitorv, STRL, DMU, [email protected] – p. 7
Binary arithmetic
Let us start with decimal arithmetic.1 1
7 5+ 3 8
1 1 3In binary we have exactly the same process but weremember that 0 + 0 = 0, 0 + 1 = 1 + 0 = 1 and1 + 1 = 10, i.e. 0 and 1 carry.Example:
1 11 1 = 3
+ 1 1 = 31 1 0 = 6
J.Dimitorv, STRL, DMU, [email protected] – p. 8
Octal and Hexadecimal arithmetic
Similarly, in the case of octal we have1 1
5 7 = 47+ 2 3 = 19
1 0 2 = 66and hexadecimal
1 15 A = 90
+ D B = 2191 3 5 = 309
J.Dimitorv, STRL, DMU, [email protected] – p. 9
Negative numbers
Negative numbers come from the need to solveequations like
N + x = 0
where x is the unknown. In the arithmetic HomoSapiens uses, we denote x = −N . With binary wewill do the same:
0 1 0 0 1 1 0 1+
0 0 0 0 0 0 0 0
J.Dimitorv, STRL, DMU, [email protected] – p. 10
Negative numbers
Negative numbers come from the need to solveequations like
N + x = 0
where x is the unknown. In the arithmetic HomoSapiens uses, we denote x = −N . With binary wewill do the same:
10 1 0 0 1 1 0 1
+ 10 0 0 0 0 0 0 0
J.Dimitorv, STRL, DMU, [email protected] – p. 10
Negative numbers
Negative numbers come from the need to solveequations like
N + x = 0
where x is the unknown. In the arithmetic HomoSapiens uses, we denote x = −N . With binary wewill do the same:
1 10 1 0 0 1 1 0 1
+ 1 10 0 0 0 0 0 0 0
J.Dimitorv, STRL, DMU, [email protected] – p. 10
Negative numbers
Negative numbers come from the need to solveequations like
N + x = 0
where x is the unknown. In the arithmetic HomoSapiens uses, we denote x = −N . With binary wewill do the same:
1 1 10 1 0 0 1 1 0 1
+ 0 1 10 0 0 0 0 0 0 0
J.Dimitorv, STRL, DMU, [email protected] – p. 10
Negative numbers
Negative numbers come from the need to solveequations like
N + x = 0
where x is the unknown. In the arithmetic HomoSapiens uses, we denote x = −N . With binary wewill do the same:
1 1 1 10 1 0 0 1 1 0 1
+ 0 0 1 10 0 0 0 0 0 0 0
J.Dimitorv, STRL, DMU, [email protected] – p. 10
Negative numbers
Negative numbers come from the need to solveequations like
N + x = 0
where x is the unknown. In the arithmetic HomoSapiens uses, we denote x = −N . With binary wewill do the same:
1 1 1 1 10 1 0 0 1 1 0 1
+ 1 0 0 1 10 0 0 0 0 0 0 0
J.Dimitorv, STRL, DMU, [email protected] – p. 10
Negative numbers
Negative numbers come from the need to solveequations like
N + x = 0
where x is the unknown. In the arithmetic HomoSapiens uses, we denote x = −N . With binary wewill do the same:
1 1 1 1 1 10 1 0 0 1 1 0 1
+ 1 1 0 0 1 10 0 0 0 0 0 0 0
J.Dimitorv, STRL, DMU, [email protected] – p. 10
Negative numbers
Negative numbers come from the need to solveequations like
N + x = 0
where x is the unknown. In the arithmetic HomoSapiens uses, we denote x = −N . With binary wewill do the same:
1 1 1 1 1 1 10 1 0 0 1 1 0 1
+ 0 1 1 0 0 1 10 0 0 0 0 0 0 0
J.Dimitorv, STRL, DMU, [email protected] – p. 10
Negative numbers
Negative numbers come from the need to solveequations like
N + x = 0
where x is the unknown. In the arithmetic HomoSapiens uses, we denote x = −N . With binary wewill do the same:1 1 1 1 1 1 1 1
0 1 0 0 1 1 0 1+ 1 0 1 1 0 0 1 1
0 0 0 0 0 0 0 0
J.Dimitorv, STRL, DMU, [email protected] – p. 10
Negative numbers
Negative numbers come from the need to solveequations like
N + x = 0
where x is the unknown. In the arithmetic HomoSapiens uses, we denote x = −N . With binary wewill do the same:1 1 1 1 1 1 1 1
0 1 0 0 1 1 0 1+ 1 0 1 1 0 0 1 1
0 0 0 0 0 0 0 0and we will ignore the last carry 1.
J.Dimitorv, STRL, DMU, [email protected] – p. 10
2’s complement in base 2
In our previous example we saw that the negativefor 01001101 is 10110011. We can also find thenegative if we do a simple procedure.
0 1 0 0 1 1 0 1
+
J.Dimitorv, STRL, DMU, [email protected] – p. 11
2’s complement in base 2
In our previous example we saw that the negativefor 01001101 is 10110011. We can also find thenegative if we do a simple procedure.
0 1 0 0 1 1 0 10
+
J.Dimitorv, STRL, DMU, [email protected] – p. 11
2’s complement in base 2
In our previous example we saw that the negativefor 01001101 is 10110011. We can also find thenegative if we do a simple procedure.
0 1 0 0 1 1 0 11 0
+
J.Dimitorv, STRL, DMU, [email protected] – p. 11
2’s complement in base 2
In our previous example we saw that the negativefor 01001101 is 10110011. We can also find thenegative if we do a simple procedure.
0 1 0 0 1 1 0 10 1 0
+
J.Dimitorv, STRL, DMU, [email protected] – p. 11
2’s complement in base 2
In our previous example we saw that the negativefor 01001101 is 10110011. We can also find thenegative if we do a simple procedure.
0 1 0 0 1 1 0 10 0 1 0
+
J.Dimitorv, STRL, DMU, [email protected] – p. 11
2’s complement in base 2
In our previous example we saw that the negativefor 01001101 is 10110011. We can also find thenegative if we do a simple procedure.
0 1 0 0 1 1 0 11 0 0 1 0
+
J.Dimitorv, STRL, DMU, [email protected] – p. 11
2’s complement in base 2
In our previous example we saw that the negativefor 01001101 is 10110011. We can also find thenegative if we do a simple procedure.
0 1 0 0 1 1 0 11 1 0 0 1 0
+
J.Dimitorv, STRL, DMU, [email protected] – p. 11
2’s complement in base 2
In our previous example we saw that the negativefor 01001101 is 10110011. We can also find thenegative if we do a simple procedure.
0 1 0 0 1 1 0 10 1 1 0 0 1 0
+
J.Dimitorv, STRL, DMU, [email protected] – p. 11
2’s complement in base 2
In our previous example we saw that the negativefor 01001101 is 10110011. We can also find thenegative if we do a simple procedure.
0 1 0 0 1 1 0 11 0 1 1 0 0 1 0
+
J.Dimitorv, STRL, DMU, [email protected] – p. 11
2’s complement in base 2
In our previous example we saw that the negativefor 01001101 is 10110011. We can also find thenegative if we do a simple procedure.
0 1 0 0 1 1 0 11 0 1 1 0 0 1 0
+ 11 0 1 1 0 0 1 1
J.Dimitorv, STRL, DMU, [email protected] – p. 11
2’s complement in base 2
In our previous example we saw that the negativefor 01001101 is 10110011. We can also find thenegative if we do a simple procedure.
0 1 0 0 1 1 0 11 0 1 1 0 0 1 0
+ 11 0 1 1 0 0 1 1
What did we actually do? We will demonstrate itwith 2’s complement in Hex
J.Dimitorv, STRL, DMU, [email protected] – p. 11
2’s complement in base 16
Let us arrange the digits in base 16 as0 1 2 3 4 5 6 7 8 9 A B C D E F
How will we find the inverse for 75(16)?
7 5
J.Dimitorv, STRL, DMU, [email protected] – p. 12
2’s complement in base 16
Let us arrange the digits in base 16 as0 1 2 3 4 5 6 7 8 9 A B C D E F
↑
How will we find the inverse for 75(16)?
7 5
J.Dimitorv, STRL, DMU, [email protected] – p. 12
2’s complement in base 16
Let us arrange the digits in base 16 as0 1 2 3 4 5 6 7 8 9 A B C D E F
↑ ↑
How will we find the inverse for 75(16)?
7 5
J.Dimitorv, STRL, DMU, [email protected] – p. 12
2’s complement in base 16
Let us arrange the digits in base 16 as0 1 2 3 4 5 6 7 8 9 A B C D E F
↑ ↑
How will we find the inverse for 75(16)?
7 5A
J.Dimitorv, STRL, DMU, [email protected] – p. 12
2’s complement in base 16
Let us arrange the digits in base 16 as0 1 2 3 4 5 6 7 8 9 A B C D E F
↑ ↑
↑
How will we find the inverse for 75(16)?
7 5A
J.Dimitorv, STRL, DMU, [email protected] – p. 12
2’s complement in base 16
Let us arrange the digits in base 16 as0 1 2 3 4 5 6 7 8 9 A B C D E F
↑ ↑
↑ ↑
How will we find the inverse for 75(16)?
7 5A
J.Dimitorv, STRL, DMU, [email protected] – p. 12
2’s complement in base 16
Let us arrange the digits in base 16 as0 1 2 3 4 5 6 7 8 9 A B C D E F
↑ ↑
↑ ↑
How will we find the inverse for 75(16)?
7 58 A
J.Dimitorv, STRL, DMU, [email protected] – p. 12
2’s complement in base 16
Let us arrange the digits in base 16 as0 1 2 3 4 5 6 7 8 9 A B C D E F
↑ ↑
↑ ↑
How will we find the inverse for 75(16)?
7 58 A
+ 18 B
Therefore 8A is the inverse and 8B is the 2’scomplement in Hex. We can check this by adding75 + 8B
J.Dimitorv, STRL, DMU, [email protected] – p. 12
Real numbers
Do computers work with reals?
Answer: NO
J.Dimitorv, STRL, DMU, [email protected] – p. 13
Real numbers
Do computers work with reals?
Answer: NO
They work with rational numbers thatapproximate reals.
J.Dimitorv, STRL, DMU, [email protected] – p. 13
Real numbers
Do computers work with reals?
Answer: NO
They work with rational numbers thatapproximate reals.
We use mantissa M and exponent E torepresent numbers as
M × 10E ,
where M and E are signed numbers.
J.Dimitorv, STRL, DMU, [email protected] – p. 13
Real numbers
Do computers work with reals?
Answer: NO
They work with rational numbers thatapproximate reals.
We use mantissa M and exponent E torepresent numbers as
M × 10E ,
where M and E are signed numbers.
This representation is called floating pointbecause by changing the exponent we movethe decimal point.
J.Dimitorv, STRL, DMU, [email protected] – p. 13
