Question

Use the method of proof by contradiction to prove the following statements. (In each case, you should also think about how a direct or contrapositive proof would work. You will find in most cases that proof by contradiction is easier.) If \(b \in \mathbb{Z}\) and \(b \nmid k\) for every \(k \in \mathbb{N},\) then \(b=0\).

Short answer

The initial statement 'If \(b\) is an integer and \(b\) doesn't divide any natural number \(k\), then \(b\) equals to zero.' is successfully proven by contradiction. If \(b\) didn't equal zero, that would mean it could divide itself, contradicting the assertion that \(b\) doesn't divide any natural number. Hence, \(b\) must be zero.

Step-by-step solution

Assume the Contradictory

Begin the proof by contradiction by assuming the opposite of what you want to prove. Here, we want to prove that if \(b\) doesn't divide any natural number \(k\), then \(b\) equals to zero, so let's assume \(b \neq 0\).

Find a Contradiction

According to the division property, if \(b \neq 0\), \(b\) should divide itself. This leads to a contradiction because we initially stated that \(b\) doesn't divide any natural number \(k\), implied that \(b\) doesn't divide itself as well.

Conclusion

Since our assumption led to a contradiction, the assumption must be false, hence, \(b\) must be equal to zero, proving the initial statement.