Chapter 1.4, Problem 17GP

To determine

Whether the given conjecture is true. If not, then give a counterexample.

Answer to Problem 17GP

  $\text{Not true}$

Explanation of Solution

Given information:

All multiples of 3 are odd.

Concept Used:

Commutative property:

  $a+b=b+a$

  $a\cdot b=b\cdot a$

Associative property:

  $\left(a+b\right)+c=a+\left(b+c\right)$

  $\left(a\cdot b\right)\cdot c=a\cdot \left(b\cdot c\right)$

Distributive property:

  $a\left(b+c\right)=ab+ac$

  $\left(b+c\right)a=ba+ca$

Additive identity:

  $a+0=0+a=a$

Multiplicative identity:

  $a\cdot 1=1\cdot a=a$

Zero property of Multiplication:

  $a\cdot 0=0\cdot a=0$

Calculation:

The given conjecture is not correct because all the multiples of 3 are not odd. Only the odd multiples of 3 are odd, while the even multiples of 3 are all even.

For example, the second multiple of 3 is 6, which is even. Similarly, the fourth multiple of 3 is 12, which is even as well. Similarly, all the even multiples of 3 are even numbers as well.

Thus, the conjecture that the all multiples of 3 are oddis not true.