How do you solve this inequality (0<1/x<2) I'm in calculus, but I forgot how to do this. Thanks

How do you solve this inequality (0<1/x<2) I'm in calculus, but I forgot how to do this. Thanks
How about inveting both side
wgat you get is 0<x/1<1/2
therefore x is
0<x<1/2
that is incorrect, in inequalities you cannot invert both sidesHow about inveting both side
wgat you get is 0<x/1<1/2
therefore x is
0<x<1/2
3>2
1/3 is not > 1/2
Here is how to do it:
0<1/x<2
Multiply by x:
0 < 1 < 2x
1/2 < x
therefore this inequality remains true for all values greater then 1/2.
thanks for the answers, but I found out how to do it veli had the right concept but you also have to flip the signs. so if the inequality was
2<1/x<3 the answer will be 1/2>x>1/3. you have to flip the signs and take the reciprocal of every thing.
Hmm, I will try not to be offended but my answer is correct....
If you flip the sign of Veli inequality you get x > 1/2
I believe that is the same as my answer 1/2 < x.
Both methods are fine...
Still it is correct to flip the inequality when inverting. You also flip the inequality if you multiply by  1.
Hmmm... no one has mentioned that the left hand side is incorrect as well. If you include the limit called infinity in your number set then the correct operation would be....Originally Posted by veli
0<1/x<2
to
infinity > x > 1/2
which brings up the intriguing question of how you would handle
2<1/x<2
well you can seperate this into
2<1/x<0 OR 0<1/x<2
then our operation gives us
1/2 > x > infinity OR infinity > x > 1/2
Notice the use of the limit 0 to avoid the obviously wrong conclusion
1/2 > x > infinity OR infinity > x > 1/2 which makes no sense.
I have also left out the possibility that 1/x = 0 under the assumption that a only real solutions were desired.
Of course multiplying by x is more sensible using a more elementary level of mathematics but it is not as interesting.
« mathematics of the mind  Circles enclosed in regular polygons » 