ECE 275: Place value number system
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\[
\newcommand{\bx}{\bar{x}}
\newcommand{\by}{\bar{y}}
\newcommand{\bz}{\bar{z}}
\newcommand{\bA}{\bar{A}}
\newcommand{\bB}{\bar{B}}
\newcommand{\bC}{\bar{C}}
\newcommand{\bD}{\bar{D}}
\newcommand{\bE}{\bar{E}}
\]
Announcments

Please wear masks properly.

Midterm is on Oct 6th, 9 AM in class.
Functional decomposition Ex4
  \(\bA \) 
  \(\bB\)  \(B \) 
  \(\bC\)  \(C \)  \(\bC\) 
\(\bD\)  \(\bE\) 
0  1  1  1 
\(E\) 
1  0  0  0 
\(D\) 
0  1  1  1 
\(\bE\) 
1  0  0  0 
  \(\bA \) 
  \(\bB\)  \(B \) 
  \(\bC\)  \(C \)  \(\bC\) 
\(\bD\)  \(\bE\) 
1  0  0  0 
\(E\) 
0  1  0  0 
\(D\) 
1  0  0  0 
\(\bE\) 
0  1  0  0 
Press b to see blackboard
Place value number system

Decimal: Base 10

Binary: Base 2
Decimal  Binary 
0  0 
1  1 
2  10 
3  11 
4  100 
5  101 
6  110 
7  111 
8  1000 
9  1001 
10  1010 
Value of a number

A number system with base \( b \) has at least \( b \) unique symbols with special.

A number system is has two operators \( + \) and \( \cdot \).
The number system must be closed under the two operators.

There is at least one symbol \( 0 \) such that \( x + 0 = x \)

There is at least one symbol \( 1 \) such that \( x \cdot 1 = x \)

A number in base \( b \) written as \( d_n \dots d_1 d_0 \)
has the value \[ d_n \dots d_1 d_0 = d_n b^n + \dots + d_1 b^1 + d_0 b^0 \]
Hexadecimal
Decimal  Hexadecimal 
0  0 
\(\vdots \)  \(\vdots\) 
9  9 
10  A 
11  B 
12  C 
13  D 
14  E 
15  F 
Arbitrarily new number system
Define two symbols \( 0, 1 \) as required.
Define rest of the symbols in increments of \( ♦ \)
The value of ♠♦♥ = \( 3\times 4^2 + 1 \times 4^1 + 0 \times 4^0\)