# Thread: describe an instance when an inequality is impossible to solve..

1. ## describe an instance when an inequality is impossible to solve..

Describe an instance when an inequality is impossible to solve..

not sure how to answer this...

2. Originally Posted by bigwave
Describe an instance when an inequality is impossible to solve..

not sure how to answer this...
x^2 < 0 is a simple example that comes to mind.

(I assume the question wants you to think of an inequality that has no solution).

3. I don't see whats impossible about x^2<0, but this one may be:

Continuum hypothesis - Wikipedia, the free encyclopedia

4. Originally Posted by ark600
I don't see whats impossible about x^2<0, but this one may be:

Continuum hypothesis - Wikipedia, the free encyclopedia
Because complex numbers can't be ordered...

5. Originally Posted by ark600
I don't see whats impossible about x^2<0, but this one may be:

Continuum hypothesis - Wikipedia, the free encyclopedia
The point is to provide help that is within the scope of what the OP is studying. Having reviewed the posting history of the OP, I concluded that the context of the question was real numbers - the impossibility of the inequality is obvious.

If the context is complex numbers, then something like |z - 1| + |z + 1| < 2 will work.

I very much doubt that the OP is at the level of understanding the content of the given link, which defeats the point of helping with the question.

6. Originally Posted by Prove It
Because complex numbers can't be ordered...
And so, the solution is that there is no solution.

(Sorry MrFantastic, I didn't read your bit at the end in brackets- I assumed it was part of your signature- I ignore them.)

7. Originally Posted by Prove It
Because complex numbers can't be ordered...
Thanks for your reply to ark600. However, although what you say is true, it's not relevant to the example I gave because x = i, for example, satisfies x^2 < 0.

But usage of the pronumeral x (rather than z) implies real numbers, so I'd have thought that the impossibility of x^2 < 0 would be obvious. The OP seems to get it.

8. ## devation

yes this question was in the algebra 1 level however it does have a section on imaginary numbers and I did also consider that as part of answer...

but here is another question (I have a limit of 2 questions)
Describe and draw the graph of the solution of all points that have a deviation less than 3 from the point (2, -1)

I assume that word "deviation" would imply a circle 3 units in radius whose center is pt (2, -1) the "less than" would imply up to the circle but not including it.

sorry but too temped to ask if a set of imaginary points can be restricted to the inside of circle. (this is not part of textbook question) I don't think it is but maybe it is.

9. Originally Posted by bigwave
yes this question was in the algebra 1 level however it does have a section on imaginary numbers and I did also consider that as part of answer...

but here is another question (I have a limit of 2 questions)
Describe and draw the graph of the solution of all points that have a deviation less than 3 from the point (2, -1)

I assume that word "deviation" would imply a circle 3 units in radius whose center is pt (2, -1) the "less than" would imply up to the circle but not including it.

sorry but too temped to ask if a set of imaginary points can be restricted to the inside of circle. (this is not part of textbook question) I don't think it is but maybe it is.
I agree with you that it is a circle of radius 3, centred at (2, -1), filled in and with the circle as a broken curve.

And yes, a set of imaginary points can be restricted to the inside of a circle. You would say that it is all $\displaystyle \displaystyle z \in C$ such that $\displaystyle \displaystyle |z| < r$.