How Can I Use Axioms to Prove x^2 ≥ 0?

[dreter98]

Homework Statement

I need to prove using the axioms ive added that X^2>=0 (x^2>0 or x^2=0). Im unsure what im doing ive attached what ive done so far

Relevant Equations

(A1) If x∈F and y∈F, then x+y∈F.
(A2) (commutativity of addition) x+y = y+x for all x,y∈F.
(A3) (associativity of addition) (x+y)+z = x+(y+z) for all x,y,z∈F.
(A4) There exists an element 0∈F such that 0+x = x for all x∈F.
(A5) For every element x∈F there exists an element−x∈F such that x+(−x) = 0.
(M1) If x∈F and y∈F, then xy∈F.
(M2) (commutativity of multiplication) xy = yx for all x,y∈F.
(M3) (associativity of multiplication) (xy)z = x(yz) for all x,y,z∈F.
(M4) There exists an element 1∈F (and 16= 0) such that 1x = x for all x∈F.
(D) (distributive law) x(y+z) = xy+xz for all x,y,z∈F.

(x)(x)>0 (D)
(x+(-x))(x+(-x)) >0 (A4)
x^2 + 2(-x)(x) + (-x)^2 >0 (D)
x^2 - 2x^2 + (-x)^2 >0
-x^2 + (-x)^2 >0

 

[Infrared]

None of the field axioms you listed say anything about the total order $<$. I suspect you should also have the following axioms to describe an ordered field.

For all $x,y,z\in F$,
1) If $x>y$, then $x+z>y+z$
2) If $x>0$ and $y>0$, then $xy>0$

Can you prove the assertion using these axioms? One approach is to use trichotomy and separately deal with the cases $x<0$ and $x=0$ and $x>0$.

 

[dreter98]

sorry yes I do have the order axioms, I have only just started learning this so I have no idea what the assertion is sorry

 

[Infrared]

By "the assertion" I just mean the problem statement that $x^2\geq 0$ for all field elements $x$. Try dividing into the cases $x>0$ and $x=0$ and $x<0$ like I suggested.

 

[dreter98]

How would I go about proving x=0 and then referencing into x^2 = 0, would it be (x)(x) = 0?

 

[Mark44]

Homework Statement:: I need to prove using the axioms I've added that X^2>=0 (x^2>0 or x^2=0). I am unsure what I am doing I've attached what I've done so far

(x)(x)>0 (D)

This isn't the distributive property. It says that a*(b + c) = a*b + a*c.

(x+(-x))(x+(-x)) >0 (A4)

No, this isn't true. By A5, x + (-x) = 0, so (x + (-x))(x + (-x)) can't be positive.

x^2 + 2(-x)(x) + (-x)^2 >0 (D)

Yes, this is distribution, but it takes a couple of steps to get here. Since the previous step is not true, that puts this step into question.

x^2 - 2x^2 + (-x)^2 >0
-x^2 + (-x)^2 >0

Again, not true. -x^2 + (-x)^2 = 0 -- it's not positive.

How would I go about proving x=0 and then referencing into x^2 = 0, would it be (x)(x) = 0?

Your goal isn't to prove that x = 0 -- it's to prove that for any arbitrary x in whatever field you're working with, that $x^2 > 0$ or $x^2 = 0$. You might start by looking at three separate cases: x < 0, x = 0, or x > 0, and using the axioms that @Infrared mentioned.