Inequality - Proof that √(a^2)<√(b^2) does not imply a<b

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    Inequality Proof
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The discussion centers around the mathematical statement that √(a^2) < √(b^2) does not imply a < b, illustrated by the counterexample of a = 1 and b = -2. Participants clarify that the book's example serves as a disproof rather than a formal proof. The distinction between proving and disproving statements is emphasized, noting that providing a single counterexample suffices to disprove an "if-then" statement. The original poster expresses a desire for a deeper algebraic understanding and seeks guidance on relevant topics to study. Overall, the conversation highlights the importance of understanding proofs and disproofs in mathematics.
Akitirija
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Hi everyone!

First of all thank you for a great forum! I downloaded the app and find it ingenious!

The problem stated above is from "3000 Solved Problems in Calculus".

The book solves this problem simply by stating: "No. Let a=1 and b=-2".

However, I am curious to know if it is possible to provide a more algebraic proof, or of this is the only way to prove it. As I really cannot provide any attempts of my own, I will just ask if anyone know what topic I have to study in order to find the answer. If I fail after that, I will return to you again.

(Maybe it should be mentioned that I am interested in doing Calculus 1,but I want the fundamentals in order first.) Yours sincerely,
Aki
 
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I am puzzled by this. Are you saying that this proof is too simple and you want a harder proof?

The problem is to show that "if \sqrt{a^2}&lt; \sqrt{b^2} then a< b" is NOT true. A standard way to show that an "if-then" statement is not true is to give a "counter example". While no number of "examples" will show that such a statement is true one example in which it does not work is enough to show that it is NOT true.
 
Ah, cool! I see your point. I'm not really used to proofs at all. Thank you very much for the quick reply!
 
Akitirija said:
Ah, cool! I see your point. I'm not really used to proofs at all. Thank you very much for the quick reply!

Just to be clear: what the book gave was not a "proof", it was a "disproof". Even if you do not know how to do proofs you may be able to do disproofs, as they are very different concepts.
 
I see. Thank you for the clarification!
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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