If x and y are integers, is ax > ay?

Sir, from statement 1 how can we conclude that a^2 is positive

The square of a number (more generally an even power of a number) is always non-negative, so 0 or positive. \(a^3*x > a^3*y\) also implies that \(a \neq 0\), becasue if it were, then we'd have \(a^3*x =0 = a^3*y\). Thus, in this a^2 is not only non-negative but positive.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

The first step of VA(Variable Approach) method is modifying the original condition and the question, and rechecking the number of variables and the number of equations. We can modify the original condition and question as follows.

\(ax > ay <=> a^3x > a^3y\) if \(a≠0\).
\(ax > ay <=> -ax < -ay\).

Condition 1)
Since \(a^3x > a^3y\), \(a\) is not zero.
By dividing both sides by \(a^2\), we have \(ax > ay\) since \(a^2 > 0\).
This is sufficient.

Conditin 2)
By multiplying \(-ax < -ay\) by \(-1\), we have \(ax > ay\).
This is sufficient.

Happy Studying !!!

What if a is negative? Will statement 2 still hold?

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