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= Number Laws = toc

Solutions to Extension Questions

Extension 1
How could you __prove__ that the set of rational numbers is closed for addition, subtraction, multiplication and division?

__**Solution:**__

A rational number is any number that can be written as a fraction with an integer as numerator and an integer as denominator.

Choose __any__ two rational numbers A and B. Since A and B are rational, they can be written as fractions.

math \\ \text{Let } A = \dfrac{a}{c} \qquad \textit{ where a and c are integers and } c \neq 0 \\ \\ \text{Let } B = \dfrac{b}{d} \qquad \textit{ where b and d are integers and } d \neq 0 math

To prove A and B are closed for addition:

math \\ A + B = \dfrac{a}{c} + \dfrac{b}{d} \qquad \qquad \textit{change to common denominator = cd} \\ \\ . \qquad \; = \dfrac{ad}{cd} + \dfrac{bc}{cd} \qquad \qquad \textit{now add}\\ \\ . \qquad \; = \dfrac{ad+bc}{cd} math

Since the set of integers is closed for multiplication and since c and d are integers, it follows that the value (cd) must be an integer.

Also, since the set of integers is closed for both addition and multiplication and since a,b,c,d are all integers, it follows that the value (ad + bc) must be an integer.

Hence we have found that for __any__ rational numbers A and B, A + B gives a fraction with an integer as numerator and an integer as denominator (ie a rational number).

Therefore, we have proved that the set of rational numbers is closed for addition.

We can perform similar calculations with A and B to prove that the set of rational numbers is closed for subtraction, multiplication and division.

Extension 2
How could you prove that the set of integer powers of 2{2 n where n is any integer} is closed for multiplication and division?

__**Solution**__

Let A and B be members of the set of integer powers of 2.

Then math \\ \text{Let } A = 2^a \qquad \textit{ where a is an integer} \\ \\ \text{Let } B = 2^b \qquad \textit{ where b is an integer} math

To prove that A and B are closed for multiplication

math \\ A \times B = 2^a \times 2^b \qquad \qquad \textit{use index laws} \\ \\ . \qquad \; \;= 2^{a+b} math

Since the set of integers is closed for addition and since a and b are integers, it follows that (a + b) is also an integer.

Hence we have found that for __any__ members of the set, A and B, it is true that A × B can also be written as 2 n and so is a member of the same set.

Therefore we have proved that the set of integer powers of 2 is closed for multiplication.

We can perform similar calculations to prove that A and B are closed for division as well.

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