Chapter 1, Problem 1.3P

To determine

To find:

The result produced from the given expressions and verify them using MATLAB.

Answer to Problem 1.3P

Solution:

The required result is verified.

Explanation of Solution

The given expressions are,

3==5+2

'b'<'a'+1

10>5+2

(10>5)+2

'c'=='d'-1&&2<4

'c'=='d'-1||2>4

xor('c'=='d'-1, 2>4)

xor('c'=='d'-1, 2<4)

10>5>2

There are only two possibilities that can be concluded from the expression. If the expression is true, then it will have the logical value $1$. If the expression is false, then it will have the logical value $0$. The given expression “3==5+2” is false. Hence, the output will be $0$.

Consider the value of a is $0$ and the value of b is $1$. The output will be,

$\begin{array}{c}b<a+1\\ 1<0+1\end{array}$

The obtained output is false. Hence, it will have the logical value $0$.

The given expression “10>5+2” is false. Hence, the output will be $1$.

In the expression “(10>5)+2”, first “(10>5)” will be evaluated. This expression is true. Hence, it will have the logical value $1$. Add “2” to this logical value. The final output is $3$.

Consider the value of c is $0$ and the value of d is $1$ in the expression “'c'=='d'-1&&2<4”. The expression becomes,

$\begin{array}{l}c==d-1\&\&2<4\\ 0==1-1\&\&2<4\end{array}$

In this expression, the order of preference is “$-,<,==,\&\&$”. Hence,

$\begin{array}{l}0==\left(1-1\right)\&\&2<4\\ 0==0\&\&\left(2<4\right)\\ \left(0==0\right)\&\&1\\ 1\&\&1\\ 1\end{array}$

Consider the value of c is $0$ and the value of d is $1$ in the expression “'c'=='d'-1||2>4”. The expression becomes,

$\begin{array}{l}c==d-1\left|\right|2>4\\ 0==1-1\left|\right|2>4\end{array}$

In this expression, the order of preference is “$-,<,==,\left|\right|$”. Hence,

$\begin{array}{l}0==\left(1-1\right)\left|\right|2>4\\ 0==0\left|\right|\left(2>4\right)\\ \left(0==0\right)\left|\right|0\\ 1\left|\right|0\\ 1\end{array}$

Consider the value of c is $0$ and the value of d is $1$ in the expression “xor('c'=='d'-1, 2>4)”. The expression becomes,

$\begin{array}{l}\text{xor}\left(c==d-1,2>4\right)\\ \text{xor}\left(0==1-1,2>4\right)\end{array}$

In this expression, the order of preference is “$-,<,==,\text{xor}$”. Hence,

$\begin{array}{l}\text{xor}\left(0==\left(1-1\right),2>4\right)\\ \text{xor}\left(0==0,\left(2>4\right)\right)\\ \text{xor}\left(\left(0==0\right),0\right)\\ \text{xor}\left(1,0\right)\\ 1\end{array}$

Consider the value of c is $0$ and the value of d is $1$ in the expression “xor('c'=='d'-1&&2<4)”. The expression becomes,

$\begin{array}{l}\text{xor}\left(c==d-1,2<4\right)\\ \text{xor}\left(0==1-1,2<4\right)\end{array}$

In this expression, the order of preference is “$-,<,==,\text{xor}$”. Hence,

$\begin{array}{l}\text{xor}\left(0==\left(1-1\right),2<4\right)\\ \text{xor}\left(0==0,\left(2<4\right)\right)\\ \text{xor}\left(\left(0==0\right),1\right)\\ \text{xor}\left(1,1\right)\\ 0\end{array}$

In the expression “10>5>2”, first “(10>5)” will be evaluated. This expression is true. Hence, it will have the logical value $1$. Now “(1>2)” is false. Hence, it will have the logical value $0$.

MATLAB Code:

3==5+2

'b'<'a'+1

10>5+2

(10>5)+2

'c'=='d'-1&&2<4

'c'=='d'-1||2>4

xor('c'=='d'-1, 2>4)

xor('c'=='d'-1, 2<4)

10>5>2

Save the MATLAB script with name, chapter1_54793_1_3P.m in the current folder. Execute the script by typing the script name at the command window to generate output.

Result:

Therefore, the required variable is created.