Prove 2ab<= a^2+b^2 using order axioms

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In summary, the conversation discusses how to prove the inequality 2ab <= a^2+b^2 using order axioms and other basic axioms for real numbers. The main approach discussed is to use the fact that (a-b)^2 is greater or equal to zero, but there is also discussion of constructing a^2+b^2-2ab using the axioms. The conversation also touches on proving that a^2 is always greater or equal to zero, and briefly mentions proving that (-x)(-y)=-(xy).
  • #1
whyme1010
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Homework Statement



prove 2ab<= a^2+b^2 using order axioms
Of course use of other basic axioms for real numbers are also okay.

Homework Equations


axioms for set of real numbers.


The Attempt at a Solution


The easy way to do this would be just subtract 2ab from both sides, factor, and see that (a-b)^2 is greater or equal to zero.

But we have to use the basic axioms. So I tried constructing a^2+b^2-2ab by multiplying (a-b)(a-b) using the axioms. I'm not sure I was rigorous enough. Also I'm not sure whether I can just say a square of a real number is greater than zero. Though it is easy to prove.
 
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  • #2
You can use only the order axioms and the field axioms. I don't think you can use squared numbers are greater than zero.
 
  • #3
well then I can just prove that a^2 is always greater or equal to zero.

a>0
Then a(a)>(a)0
a^3>0

a<0
a+(-a)<0+(-a)
0<-a
-a(0)<(-a)(-a)
0<(-1)(-1)(a)(a)
0<a

a=0
a(a)=0(a)
a^2=0

So a^2 >=0
 
  • #4
whyme1010 said:
well then I can just prove that a^2 is always greater or equal to zero.

a>0
Then a(a)>(a)0
a^3>0

a<0
a+(-a)<0+(-a)
0<-a
-a(0)<(-a)(-a)
0<(-1)(-1)(a)(a)
0<a

a=0
a(a)=0(a)
a^2=0

So a^2 >=0

Hmmm I guess so. did you prove that (-x)(-y)=-(xy)?
 
  • #5
yeah. I also solved this by using three lemmas. =) Don't know how to close the thread though.
 
  • #6
happysauce said:
Hmmm I guess so. did you prove that (-x)(-y)=-(xy)?

I sure hope he didn't.
 
  • #7
Mentallic said:
I sure hope he didn't.

woops I mean (x)(-y) = -(xy)
 

Related to Prove 2ab<= a^2+b^2 using order axioms

1. How do you use order axioms to prove 2ab <= a^2+b^2?

To prove 2ab <= a^2+b^2 using order axioms, we must first understand the order axioms. These are mathematical rules that define the properties of real numbers, such as transitivity, trichotomy, and the distributive property. Using these axioms, we can manipulate the expressions 2ab and a^2+b^2 to show that 2ab is less than or equal to a^2+b^2.

2. What is the importance of proving 2ab <= a^2+b^2 using order axioms?

Proving 2ab <= a^2+b^2 using order axioms allows us to establish the relationship between these two expressions in a rigorous and logical manner. This proof is important because it demonstrates the validity of the order axioms and serves as a foundation for further mathematical reasoning.

3. Can you provide an example of using order axioms to prove 2ab <= a^2+b^2?

Sure, let's say we have the expressions 2(3)(4) and (3)^2+(4)^2. Using the distributive property, we can rewrite 2(3)(4) as 2(3)+2(4), which is equal to 6+8=14. Similarly, (3)^2+(4)^2 can be expanded to 9+16=25. Now, we can use the transitive property to show that 14 is less than or equal to 25, thus proving 2(3)(4) <= (3)^2+(4)^2.

4. Are there any alternative methods for proving 2ab <= a^2+b^2?

Yes, there are other methods for proving 2ab <= a^2+b^2, such as using algebraic manipulation or geometric proofs. However, using order axioms is a straightforward and reliable approach that relies on fundamental principles of mathematics.

5. How could understanding the proof of 2ab <= a^2+b^2 using order axioms benefit me as a scientist?

Understanding this proof can benefit you as a scientist by developing your critical thinking and logical reasoning skills. It also demonstrates the importance of using axioms and proof in mathematics, which is the foundation of many scientific theories and experiments.

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