Question

If x, y belongs to real number and xy=0 ,then prove that x=0 or y=0 or both x=0, y=0.​

Answer 1

Step-by-step explanation:

The goal is to prove "if xy=0, then x=0 or y=0." As you said, this means we assume xy=0 then prove the statement "x=0 or y=0."

When they break it into cases, they are considering either x=0 or x!=0. Since x is some real number, and every real number is either 0 or not 0, these two cases give all possibilities.

When broken into two cases, we get to include the hypotheses of each case, that is, in one case we assume "xy=0" and "x=0." In the other case, we assume "xy=0" and "x!=0". Since the two cases cover every possibility, we just prove the two cases.

In the first case, we have assumed "xy=0" and "x=0." Since "x=0" it follows that "x=0 or y=0". That is, we have the implication "if 'xy=0' and 'x=0' then 'x=0 or y=0'."

The second case follows as in their proof, so we have the implication "if 'xy=0' and 'x!=0' then 'x=0 or y=0'." These two implications give the full result.

As an aside, proving "if x=0 then xy=0" looks more like this: suppose x=0. Then, substituting and applying the distributive property, we have xy=0y=(1-1)y=y-y=0.

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