Question

Show that if \(a|b{\rm{ and b|a}}\), where a and b are integers, then \(a = b{\rm{ or a = - b}}\)

Short answer

\(a = b\)or \(a = - b\).

Step-by-step solution

Step 1

DEFINITIONS

a divides b if there exists an integer c such that \(b = ac\)

Notation: \(a|b\).

Step 2

SOLUTION

Given: a and b are integers with \(a|b\) and \(b|a\)

To proof: \(a = b\)or \(a = - b\)

PROOF

Since \(a|b\), there exists an integer c such that:

\(b = ac\)

Since \(b|a\), there exists an integer d such that:

\(a = bd\)

Combining these two equations, we then obtain:

\(a = bd = (ac)d = a(cd)\)

By the identity property of multiplication, cd then has to be equal to \({\rm{1}}\).

\(cd = 1\)

Since c and d are both integers whose product is \(1\), c and d are either both \(1\) or both \( - 1\).

\(c = d = 1{\rm{ or c = d = - 1}}\)

Let us use the values for d on the equation \(a = bd\)

\(a = bd = b1 = b\)

\(a = bd = b( - 1) = - b\)

Thus we have then shown \(a = b\)or \(a = - b\).