Question

Prove using the notion without loss of generality that \({\bf{min}}\left( {{\bf{x,y}}} \right){\bf{ = }}\frac{{\left( {{\bf{x + y}} - \left| {{\bf{x - y}}} \right|} \right)}}{{\bf{2}}}\) and \({\bf{max}}\left( {{\bf{x,y}}} \right){\bf{ = }}\frac{{\left( {{\bf{x + y + }}\left| {{\bf{x - y}}} \right|} \right)}}{{\bf{2}}}\) whenever x and y are real numbers.

Short answer

It is proved that \(\frac{{x + y - \left| {x - y} \right|}}{2} = y = \min \left( {x,y} \right)\) and \(\frac{{x + y + \left| {x - y} \right|}}{2} = x = \max \left( {x,y} \right)\)

Step-by-step solution

Without loss of generality

Assume that \({\bf{x}}\)is odd and \({\bf{y}}\) is even without loss of generality.

Without loss of generality is satisfied here as when\({\bf{x}}\)and\({\bf{y}}\)are interchanged in the given expression, the value of original expression would remain the same.

Find the two addends of the given expression 

Assume \(x = \max \left( {x,y} \right)\)

Since \(\left| {x - y} \right| = \left| {y - x} \right|\) . x and y can be interchanged to obtain the same result in the case that \(y = \max \left( {x,y} \right)\).

Now, \(y = \min \left( {x,y} \right)\)

Simplify the given fractions to obtain:

\(\begin{array}{c}\frac{{x + y - \left| {x - y} \right|}}{2} = \frac{{x + y - \left( {x - y} \right)}}{2}\\ = \frac{{x + y - x + y}}{2}\\ = \frac{{2y}}{2}\\ = y\end{array}\)

\(\frac{{x + y - \left| {x - y} \right|}}{2} = y = \min \left( {x,y} \right)\)

Also,

\(\begin{array}{c}\frac{{x + y + \left| {x - y} \right|}}{2} = \frac{{x + y + \left( {x - y} \right)}}{2}\\ = \frac{{x + y + x - y}}{2}\\ = \frac{{2x}}{2}\\ = x\end{array}\)

Thus,

\(\begin{array}{c}\frac{{x + y + \left| {x - y} \right|}}{2} = x\\ = \max \left( {x,y} \right)\end{array}\)

Thus, both identities are proved without loss of generality assuming \(x \ge y\).