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

N=1

Answer 2

for any number \[0<N<1\]

Answer 3

yes in fact\[0<N\le1\]

Answer 4

I enjoyed this problem immensely :)

Answer 5

also you might want to first write the negation of the statement before finding the counterexample!!!!

Answer 6

Statement: For every real number N > 0, there is some real number x such that Nx > x.
Negation : There exists a real number N>0 such that for all real numbers x it is true that Nx <= x

Answer 7

and any N in (0,1] works

Answer 8

the negation *is* a counterexample, so proving the negation is finding at least one N :)

Answer 9

I have a feeling @malia667 didnt write the equation exactly asked.
It is missig some absolute value symbols!!!

Answer 10

electro, can you show me where you would need an absolute value?

Answer 11

thanks guys :))
n=1 was correct

Answer 12

more to come

Answer 13

\[|Nx|>|x|\]
since x is any real number, it could be negative too

Answer 14

it doesnt matter if x is negative

Answer 15

\[5(-3)<-3\]

Answer 16

N=5
and
x=-3
the inequality flips

Answer 17

.yes but the counterexample addresses that

Answer 18

it may but there are infinite answers.
the question would have no meaning to it

Answer 19

it is true that for all x, Nx > x for N inside (0,1]

Answer 20

the given inequality is false for each and every N

Answer 21

i need to ask another question please :))

Answer 22

it you put an absolute sign, then it'd make sense

Answer 23

yes so the above claim goes wrong too when x<0

Answer 24

no, because it says 'there exists an x '

Answer 25

it didnt say for all x in the original statement