Help prove false: For any real number x, If x2>=1, then x>=1

[ANJ]

The problem says: Show that the conjuncture is false by finding a counterexample.
For any real number x, If x²>=1, then x>=1

 

[JeffM]

The problem says: Show that the conjuncture is false by finding a counterexample.
For any real number x, If x²>=1, then x>=1

And what have you thought about?

 

[ANJ]

I have thought about using a fraction. Would that work?

 

[JeffM]

I have thought about using a fraction. Would that work?

OK. At least you are thinking. That's good.

\(\displaystyle -1 < x < 1 \implies 0 \le x^2 < 1.\) But that is not a counter-example to the proposition given. Why not?

 

[stapel]

The problem says: Show that the conjuncture is false by finding a counterexample.
For any real number x, If x²>=1, then x>=1

You might find it helpful to look at a graph of y = x^2; specifically, look at where y > 1, and look at ALL the x-values (numbers on the x-axis) that lead upward to where y > 1.