Question

Suppose \([a],[b] \in \mathbb{Z}_{6}\) and \([a] \cdot[b]=[0]\). Is it necessarily true that either \([a]=[0]\) or \([b]=[0] ?\) What if \([a],[b] \in \mathbb{Z}_{7} ?\)

Short answer

No, it is not necessarily true that if \([a] \cdot [b] = [0]\) in \(\mathbb{Z}_{6}\), then either \([a] = [0]\) or \([b] = [0]\). Yes, it is true that if \([a] \cdot [b] = [0]\) in \(\mathbb{Z}_{7}\), then either \([a] = [0]\) or \([b] = [0]\).

Step-by-step solution

Understanding multiplication in \(\mathbb{Z}_{6}\)

In \(\mathbb{Z}_{6}\), \([a] \cdot [b] = [ab]\), where \(ab\) denotes regular multiplication, then taking the remainder when divided by 6. This is because \(\mathbb{Z}_{6}\) is the set of equivalence classes under the equivalence relation 'is congruent to modulo 6'. So if \([a] \cdot [b] = [0]\), that means \(ab = 6x\) for some integer \(x\), i.e., \(ab\) is divisible by 6.

Checking the condition in \(\mathbb{Z}_{6}\)

Now consider numbers in \(\mathbb{Z}_{6}\). The multiples of 6 that can be formed by multiplying two numbers between 0 and 6 (inclusive) come from not just \([0] = [0] \cdot [a]\) or \([0] = [a] \cdot [0]\), but also from \([2] = [3] \cdot [2]\) and \([2] = [2] \cdot [3]\). So, in \(\mathbb{Z}_{6}\), it is not necessarily true that if \([a] \cdot [b] = [0]\), either \([a] = [0]\) or \([b] = [0]\).

Understanding multiplication in \(\mathbb{Z}_{7}\)

Similar to \(\mathbb{Z}_{6}\), in \(\mathbb{Z}_{7}\), \([a] \cdot [b] = [ab]\), where \(ab\) denotes regular multiplication, then taking the remainder when divided by 7. This is because \(\mathbb{Z}_{7}\) is the set of equivalence classes under the equivalence relation 'is congruent to modulo 7'. If \([a] \cdot [b] = [0]\), then \(ab = 7x\) for some integer \(x\), i.e., \(ab\) is divisible by 7.

Checking the condition in \(\mathbb{Z}_{7}\)

Now consider numbers in \(\mathbb{Z}_{7}\). The multiples of 7 that can be formed by multiplying two numbers between 0 and 7 (inclusive) only come from \([0] = [0] \cdot [a]\) or \([0] = [a] \cdot [0]\). There are no other pairs of numbers that will result in a product divisible by 7. Therefore, in \(\mathbb{Z}_{7}\), if \([a] \cdot [b] = [0]\), it is necessarily true that either \([a] = [0]\) or \([b] = [0]\).