Can Real Number Properties Simplify Complex Analysis Proofs?

[trojansc82]

Homework Statement

Let x, y, and z be real numbers. Prove the following:

1. If x * z = y * z, then x = y.

2. If x is not equal to 0, then x^2 > 0. (consider the two cases x > 0 and x < 0 ).

3. 0 < 1

4. For each n ∈ N, if 0 < x < y, then x^n < y^n

5. If x > 1, then x^2 > x.

6. If 0 < x < 1, then x^2 < 1

7. If 0 < x < y, then 0 < √x < √y

8. If x > 0, then 1/x > 0. If x < 0, then 1/x < 0.

9. If 0 < x < y, then 0 < 1/y < 1/x.

10. If xy > 0, then either (i) x > 0 and y > 0, or (ii) x < 0 and y < 0.

Homework Equations

Ordered Field Axioms

The Attempt at a Solution

Need help with these

 

[mighty2000]

problems.

I am happy to assist you with proving these statements. I will provide a brief explanation for each statement and its proof using the Ordered Field Axioms.

1. If x * z = y * z, then x = y.
Proof: We can use the axiom of cancellation, which states that if a * x = b * x, then a = b. Since x * z = y * z, we can substitute a = x, b = y, and x = y.

2. If x is not equal to 0, then x^2 > 0.
Proof: We can use the axiom of trichotomy, which states that for any real numbers a and b, exactly one of the following is true: a < b, a = b, or a > b. Since x is not equal to 0, we can consider two cases: x > 0 and x < 0.

- Case 1: x > 0
In this case, we can use the axiom of closure under multiplication, which states that if a and b are positive, then ab is also positive. Since x > 0, x * x = x^2 is also positive.

- Case 2: x < 0
In this case, we can use the axiom of closure under multiplication and the axiom of additive inverses, which states that for any real number a, there exists a unique real number -a such that a + (-a) = 0. Since x < 0, -x > 0 and (-x) * (-x) = x^2 is also positive.

3. 0 < 1
Proof: This follows directly from the axiom of ordering, which states that for any real numbers a and b, if a < b, then a + c < b + c for any real number c. In this case, we can let a = 0 and b = 1, and since 0 < 1, 0 + 1 < 1 + 1, or 1 < 2.

4. For each n ∈ N, if 0 < x < y, then x^n < y^n.
Proof: We can use the axiom of induction, which states that if a property is true for the base case and if it is true for n, then it is also true for n+1. In this case,