NS13983
I get super confused in solving the below kind of inequalities.
Can some expert help?
Are the below working's correct?
1/| n | >n
• Case 1: n > 0… n 2 < 1… - 1 < n < 1
• Case 2: n < 0… - n 2 < 1… n 2 > -1…
x/| x | < x
• Case 1: x > 0… x^2 > x… as x > 0, divide by x without changing signs… x > 1
• Case 2: x < 0… - x^2 > x… as x < 0, divide and only flip sign… - x < 1… x > -1
1. \(\frac{1}{| n |}>n\) --> multiply by \(|n|\) (we can safely do that since |n|>0): \(n*|n| < 1\).
If \(n>0\), then we'll have \(n^2<1\) --> \(-1<n<1\). Since we consider the range when \(n>0\), then for this range we'll have \(0<n<1\).
If \(n<0\), then we'll have \(-n^2<1\) --> \(n^2>-1\). Which is true for any n from the range we consider. So, \(n*|n| < 1\) holds true for any negative value of n.
Thus \(\frac{1}{| n |}>n\) holds true if \(n<0\) and \(0<n<1\).
2. \(\frac{x}{|x|} < x\).
If \(x>0\), then we'll have \(\frac{x}{x} < x\) --> \(1<x\).
If \(x<0\), then we'll have \(\frac{x}{-x} < x\) --> \(-1<x\) and as we considering the range for which \(x<0\) then completer range would be: \(-1<x<0\).
Thus \(\frac{x}{|x|} < x\) holds true if \(-1<x<0\) or \(x>1\).
Hope it helps.