for anyone that doesn't spot it. the first line states x < 0, so x is negative, at step 4, the equation is being divided by x, a negative number, so that sign has to change direction to 2x/x > x/x, which then leads to the right result of 2 > 1

I remember one tricky question based on contradictions.

The question was, there's a person who who always speaks the truth. There's another who always lies. A stranger comes to the place where there are these two men. The question is, with only one question to ask any of the two, he should be able to tell who's the person who speaks the truth always and who's the liar.

The solution is a bit tricky. The stranger ought to ask a question 'What would the liar say if  asked him who's the liar'. If he's the liar, he'd point to himself. That is because, in reality, the liar would have pointed to the truth speaker. But since he doesn't speak the truth, he'd point himself. If he's the one who always speaks the truth, he'd have pointed the liar. Since he does not lie, now he'd point to the liar. Either way, the index finger would be aimed at the liar. This way, the liar can easily be found.

In reality, we come across many contradictory statements. He's a regular late-comer.
There was  a law and order failure in the city.
The best I have come across is he is consistently inconsistent. A classic oxymoron!
I know him, he's unpredictable is another.

Last edited by JaneFairfax (2007-11-11 02:33:23)

Yep, it’s the same. That’s because in the the original equation, x is complex. Therefore, x should be a complex solution of x³ = −1, not x = −1. One must always be wary of such psychological pitfalls in mathematics.

Put it this way. You start with a quadratic equation, then you manipulate it and turn it into a cubic one. By doing so, you’ve introduced a stray solution to the equation – and it so happens that x = −1 is that stray solution. It’s like solving an equation involving surds (square roots): if you square both sides to get rid of the square roots, you introduce extra solutions into your equation; hence you need to substitute the solutions you’ve found back into the original equation to see which ones don’t fit.

Last edited by JaneFairfax (2007-11-19 23:49:47)