Question

Find a counterexample for the statement.
For every real number N > 0, there is some real number x such that
Nx > x.

Answer 1

I tried N= -1 and it was wrong

Answer 2

Anything negative. will be an answer.

Answer 3

bs it flips the </> sign(s)

Answer 4

we can rewrite it as $$x(N-1)>0$$ if \(N=1\), the expression is zero for all \(x\)

Answer 5

Agreed. :)

Answer 6

so no, negative \(N\) does not work, as we could just as easily pick negative \(x\) so that \(Nx\) is positive and therefore greater than \(x\)

Answer 7

Aren't we finding a 'counter example'? So if N is negative, then 0 is grater than any negative value right?
ex:
-10 < 0
-1 < 0
etc....?

Answer 8

it would equal itself in the zero case

Answer 9

So, here is how I'd approach this: what are they claiming? any positive number times a real will be greater than that real. Alright, does it seem reasonable? Not really. So we pick a number, x, to try. You tried x=-1 where we find N(-1)<-1not greater as required, so yes you found your counterexample.

Answer 10

however, they are just saying that there is some real number, so can we find one that does work? yea, any x>0. SO we come to the conclusion that in order for this statement to be true for all x, \(x> 0\)

Answer 11

This thing is saying that all real numbers greater than 0 can be multiplied by at least one value, x, with a product greater than x. This is not true for \[1>N>0\] There real numbers that violate the greater than x stuff.