Why do negative numbers behave differently when multiplied?

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Negative numbers behave differently when multiplied due to the rules of arithmetic that govern signed numbers, where multiplying two negatives results in a positive, and a negative multiplied by a positive remains negative. A real-world example of this concept is modulation, where two signals are multiplied to shift a data waveform to radio frequencies, demonstrating the non-linear nature of multiplication. For educational purposes, it is suggested to use simple visual aids like colored sine waves to help 9th graders understand these concepts. Additionally, the idea of signed distances can illustrate that moving -3 meters to the right is equivalent to moving +3 meters to the left. Understanding these principles can enhance comprehension of both mathematical rules and their applications in real-world scenarios.
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I'm looking for a good/simple explanation for why the rules for multiplying signed numbers are the way they are.

i.e. why does (-)*(-)= (+); (-)*(+)=(-); etc.

Also, I'm looking for some good real world examples to where these situations apply.

Thanks for you help.
 
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One good real world example is modulation. Modulation is multiplying two signals together for some purpose, like shifting a data waveform up to RF waveform frequencies for transmission.

So the modulated waveform looks like M(t) = A sin(w1 * t) * B sin(w2 * t)

Draw out the waveforms and see how the sign convention represents reality in the modulation process. Does that help?
 
Sorry, but not really.

I should clarify. I'm looking for an example that I can share with my 9th grade students. :smile:
 
dfollett76 said:
Sorry, but not really.

I should clarify. I'm looking for an example that I can share with my 9th grade students. :smile:
Doh! :smile: Well, bright 9th graders would probably understand modulation -- just make some pretty slides up with colored sine waves...

Multiplication is a non-linear process, so your examples are going to have to involve non-linear phenomena (like modulation). Let's see, what else is an example of non-linear stuff...

Boy, that's a toughie. I'd stick with modulation and try to introduce it gently to them. :blushing:
 
0 * (-1) = 0
(1 + (-1)) * (-1) = 0
1 * (-1) + (-1) * (-1) = 0
(-1) + (-1) * (-1) = 0
1 + ((-1) + (-1) * (-1)) = 0 + 1
(1 + (-1)) + (-1) * (-1) = 1
0 + (-1) * (-1) = 1
(-1) * (-1) = 1

There's a proof in excruciating detail. :smile: Well, it takes a little bit more to prove (-1) * x = (-x), but this might be enough for them.
 
You might be able to concoct a "real world" example from the notion of a "signed distance". I.e. that (-3) meters to the right is the same thing as (+3) meters to the left.
 
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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