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#1 2013-09-11 18:58:46

Bob
Administrator
Registered: 2010-06-20
Posts: 10,052

proof that x² ≥ 0 for Equation_Kitty

I have moved this post for Equation_Kitty.

Hi all!! I'm super new to Math is fun forum. I stumbled onto here while researching how to prove x² ≥  0. I need help with this proof I know it's very basic. I need someone to show me the light!

could you simply say "for x² ≥ 0 based on the non-negative property of squares, a square of a number will always be greater than or equal to zero. Thus x² ≥ 0 is true for all integers...?

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#2 2013-09-11 19:09:11

Bob
Administrator
Registered: 2010-06-20
Posts: 10,052

Re: proof that x² ≥ 0 for Equation_Kitty

hi Equation_Kitty,

Welcome to the forum.

To avoid confusion between your posts and Derick Mixon's thread, I've moved your question here in it's own thread.

[If you click on the Help Me link you'll see you can start your own new topic by clicking the 'Post New Topic' link on the right.]

Now to think about an answer.  smile

Yes, what you have said is OK, as long as x ∈ {reals}

There's a larger set called {complex numbers} in which square roots of negative numbers exist.  As you specified {integers} your argument is OK.

There's probably a way to write this out rigorously if that is what is required.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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