Chapter 14.1, Problem 14WE

(a)

To determine

To find: The mapping is one-to-one mapping.

Answer to Problem 14WE

The mapping is one-to-one mapping.

Explanation of Solution

Given information:

The given diagram is shown below.

  

Calculation:

It is shown that every point of the red square has only one image on the blue square therefore, it is a one-to-one mapping

Therefore, the mapping is one-to-one mapping.

(b)

To determine

To find: The diagram and locate a point $X$ that is its own image.

Answer to Problem 14WE

The only point the center $O$ that maps its image to itself.

Explanation of Solution

Given information:

The given diagram is shown below.

  

Calculation:

There is only one point, the center $O$ that maps its image to itself.

  

Therefore, the only point the center $O$ that maps its image to itself.

(c)

To determine

To find: Two points $R\text{and}S$ on the red square and their images $R\text{'}\text{and}S\text{'}$ on the blue square that have their property that $RS\ne R\text{'}S\text{'}$ .

Answer to Problem 14WE

There is no segment lie on the square.

Explanation of Solution

Given information:

The given diagram is shown below.

  

Calculation:

As it is one-to-one mapping, therefore there is no segment lie on the square which differs in

length with its image.

  

  $RS\ne R\text{'}S\text{'}$ .

Therefore, there is no segment lie on the square.

(d)

To determine

To find: The mapping preserve distance.

Answer to Problem 14WE

The mapping preserve distance.

Explanation of Solution

Given information:

The given diagram is shown below.

  

Calculation:

As it is one-to-one mapping, therefore there is no segment lie on the square which differs in

length with its image.

Therefore, the mapping preserve distance.

(e)

To determine

To find: A mapping from the red square on to the blue square that does preserve distance.

Answer to Problem 14WE

The mapping preserve distance.

Explanation of Solution

Given information:

The given diagram is shown below.

  

Calculation:

Consider the length between two points on the diagonal vertices of the red square. The image of segment will have same length when it will lie on the blue square.

Therefore, the mapping preserve distance.