Question 962471
Your answers don't make sense, since none of them are irrational. What you need is a general proof. Suppose that x is rational, and is not zero. Also suppose that y is irrational. Assume, for the sake of contradiction, that xy is rational. Then
{{{xy=z}}}
where z ∈ ℚ. However, if this is true, we would be able to write that
{{{y=z/x}}}
or in other words, that y ∈ ℚ. But we assumed that y was irrational, contradicting ourselves in the process. We must conclude that a rational number times an irrational number is irrational.