# Comparison Proof via axioms, almost done need hints for finish and proof read

• silvermane
In summary, the problem statement is to prove that 0 is less than or equal to 2x^2 - 3xy + 2y^2, using the given axioms. By using axiom 03), we are able to manipulate the expression and eventually come to four cases: when x and y are both positive, when x is positive and y is negative, when y is positive and x is negative, and when both x and y are negative. In each case, we are able to show that the expression is greater than or equal to 0, thereby proving the problem statement. Additionally, if the distributive property is allowed, the problem can be solved more easily by completing the square.
silvermane
Gold Member
1. The problem statement:

Prove that
$$0\leq2x^2 - 3xy + 2y^2$$​

2. These are the axioms we are permitted to use:

01) Exactly one of these hold: a<b, a=b, or b<a
02) If a<b, and b<c, then a<c
03) If a<b, then a+c < b+c for every c
04) If a<b and 0<c, then ac<bc.

## The Attempt at a Solution

By axiom 03) we are permitted to add $$+-y^2$$ to both sides:
$$0 = -y^2 + y^2 \leq 2x^2 -3xy+2y^2-y^2+y^2$$
$$0 \leq 2x^2 -3xy +y^2 + y^2$$

Again, by axiom 03) we add $$+-x^2$$ to both sides:
$$0 = x^2 - x^2 \leq 2x^2 - 3xy +y^2 +y^2 + x^2 - x^2 = x^2 - 3xy + y^2 +y^2 + x^2$$

By axiom 03) again, we add +-xy to both sides:
$$0 = xy - xy \leq 2x^2 - 3xy +y^2 +y^2 + x^2 - x^2 +xy - xy = x^2 - 3xy + y^2 +y^2 +xy +x^2$$

Case 1: If x=y=0, then our expression is zero

Case 2: If x doesn't = y, and y=0, then we have $$0<x^2+x^2=2x^2$$

Case 3: If x doesn't = y, and x=0, then we have $$0<y^2$$

Case 4: If x doesn't = y, and $$y\neq0$$,
$$0 < (x-y)^2 < (x-y)^2 +x^2 +xy +y^2$$ ??
then I'm stuck and need help

If anyone can help me figure out the last part, it would be greatly appreciated. I've worked very hard trying to figure it out, but that xy we would get could possibly be negative and I just need to prove that last part.

Thank you in advance for any hints/tips!

Case 1

$$x,y >0$$ This is trivial since everything is positive.

Case 2
$$x >0 ,y <0 \Rightarrow x>y$$

$$-x^{2}< xy \Rightarrow 0< xy + x^{2}$$

$$(x-y)^{2} + y^{2}< (x-y)^{2} +x^{2} + xy +y^{2}$$

Case 3

This is similar to case 2.

How come you can't use the distributive properties ?

Without them what does xy mean ? And how does it differ from yx ?

Assuming you could use axioms of multiplication and the distributive property you could divide by 2 and complete the square and the result falls into your hands.

$$x^2 - \frac{3}{2}xy + 2y^2 = (x- \frac{3}{2}y)^{2} + \frac{7}{16}y^{2}$$

## What is a comparison proof?

A comparison proof is a method of proving a mathematical statement by comparing it to a known or accepted fact or principle.

## What are axioms in a comparison proof?

Axioms are statements or principles that are accepted as true without needing to be proven. In a comparison proof, they serve as the basis for comparison with the statement being proved.

## How do you know when a comparison proof is almost done?

Typically, a comparison proof is almost done when you have successfully compared the statement being proved to the axioms and have shown that they are equivalent or that the statement follows logically from the axioms.

## What are some hints for finishing a comparison proof?

Some hints for finishing a comparison proof include checking your assumptions and logic, making sure all steps are clearly explained and justified, and double-checking your calculations. It can also be helpful to get feedback from others or to take a break and come back to the proof with fresh eyes.

## Why is proofreading important in a comparison proof?

Proofreading is important in a comparison proof to ensure that there are no errors or mistakes in the logic or calculations. It also helps to make the proof more clear and concise, making it easier for others to understand and verify.

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