Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**JohnnyReinB****Member**- Registered: 2007-10-08
- Posts: 453

*Sometimes, It is the subtle mistakes that make the biggest mistakes*

Got any good contradicting equations? Maybe people could donate such, and others would try to find out what is wrong...

Let's start easy:

(this was in our exam, I thought it was pretty easy for us, until I asked my classmates what they answered...)

*Last edited by JohnnyReinB (2007-11-10 00:51:28)*

*"There is not a difference between an in-law and an outlaw, except maybe that an outlaw is wanted" *

Nisi Quam Primum, Nequequam

Offline

**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

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

The Beginning Of All Things To End.

The End Of All Things To Come.

Offline

**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

Heres an interesting one from wikipedia which cannot be done in the same way as yours to show its false.

for complex x:

*Last edited by luca-deltodesco (2007-11-10 00:42:05)*

The Beginning Of All Things To End.

The End Of All Things To Come.

Offline

**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 25,914

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.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Offline

**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

note quite johnny, its perfectly valid to only substitue into parts of the equation, the problem is something more ambiguous.

The Beginning Of All Things To End.

The End Of All Things To Come.

Offline

**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

*Last edited by JaneFairfax (2007-11-10 03:33:23)*

Offline

**JohnnyReinB****Member**- Registered: 2007-10-08
- Posts: 453

Ok, so it's like this one:

by dividing both sides by x

And, by substituting to the first equation, you get...

multiplying both sides by x...

finding the cube root...

So now, to check, substitute again...

Which is ,obviously, wrong...

Ok...right?

*Last edited by JohnnyReinB (2007-11-18 21:57:30)*

*"There is not a difference between an in-law and an outlaw, except maybe that an outlaw is wanted" *

Nisi Quam Primum, Nequequam

Offline

**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Yep, its the same. Thats because in the the original equation, *x* is complex. Therefore, *x* should be a *complex solution* of *x*[sup]3[/sup] = −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, youve introduced a stray solution to the equation and it so happens that *x* = −1 is that stray solution. Its 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 youve found back into the original equation to see which ones dont fit.

*Last edited by JaneFairfax (2007-11-19 00:49:47)*

Offline

**JohnnyReinB****Member**- Registered: 2007-10-08
- Posts: 453

So, how do you solve these equations?

*"There is not a difference between an in-law and an outlaw, except maybe that an outlaw is wanted" *

Nisi Quam Primum, Nequequam

Offline

Pages: **1**