Why Does Multiplying Two Negative Numbers Yield a Positive?

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Multiplying two negative numbers results in a positive product, which can be proven using the definition of positivity. When both a and b are negative, their opposites (-a and -b) are positive. By applying the properties of positive numbers, one can show that the product ab is greater than zero. Understanding the axioms for positivity is crucial in this proof. The discussion emphasizes the need to work from the definitions provided in mathematical texts to reach a valid conclusion.
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Homework Statement


prove that if a is less than zero , and if b is less than zero then ab is greater than zero.

I have been having troubles with this problem.
thanks.


Homework Equations





The Attempt at a Solution

 
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What have you tried so far? Do you know what axioms for positivity you can use?
 
Office_Shredder said:
What have you tried so far? Do you know what axioms for positivity you can use?

Im reading the subject in the book of Spivak and frankly I don't understand what he says.
on page 12 he defined P to be a positive number thenhe said that for a number a only one of this three equalities is correct

a=0, a is a is part of P, and - a is part of P. I don't understand the last one since he defined P as the set of all the positive numbers maybe there might be a mistake in my book though.

is there any other way to prove this theorem?
 
For example, if a=-3, then -a is in P, not a.

You're going to have to use his definition of positivity to do the problem. You can't prove that something has a certain property without using its defining features!

As a starting point: We know that a<0 and b<0 here, so (-a)>0 and (-b)>0. Try to work from here
 
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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