Thread: inequality

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

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 sides

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.

ya you where right too .

Originally Posted by veli

How about inveting both side
wgat you get is 0<x/1<1/2
therefore x is
0<x<1/2

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....

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.