Someone newly asked me why an adverse $ imes$ a negative is positive, and also why an adverse $ imes$ a positive is negative, etc.

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I went ahead and offered them a proof by contradiction favor so:

Assume $(-x) cdot (-y) = -xy$

Then divide both sides by $(-x)$ and you gain $(-y) = y$

Due to the fact that we have actually a contradiction, then our initially presumption need to be incorrect.

I"m guessing I did somepoint wrong here. Because the conclusion of $(-x) cdot (-y) = (xy)$ is difficult to derive from what I composed.

Is tbelow a better way to define this? Is my proof incorrect? Also, what would certainly be an intuitive method to describe the negation principle, if tright here is one?


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edited Dec 20 "16 at 2:24
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BLAZE
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This is pretty soft, however I experienced an analogy digital to define this once.

If you film a guy running forwards ($+$) and also then play the film forward ($+$) he is still running forward ($+$). If you play the film backward ($-$) he shows up to be running backwards ($-$) so the outcome of multiplying a positive and an unfavorable is negative. Same goes for if you film a male running backwards ($-$) and play it usually ($+$) he shows up to be still running backwards ($-$). Now, if you film a male running backwards ($-$) and play it backwards ($-$) he appears to be running forward ($+$). The level to which you rate up the rewind does not matter ($-3x$ or $-4x$) these results hold true. $$ extbackward imes extbackward = extforward$$$$ extnegative imes extnegative = extpositive$$It"s not perfect, yet it introduces the concept of the number line having directions at least.


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edited Sep 15 "16 at 18:25
answered Nov 8 "15 at 21:08
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miradulomiradulo
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Informal justification of $ extpositive imes extnegative = extnegative$

Continue the pattern:

$$eginarrayr2 & imes & 3 & = & 6\2 & imes & 2 & = & 4\2 & imes & 1 & = & 2\2 & imes & 0 & = & 0\2 & imes & -1 & = & ? & ( extAnswer = -2 )\2 & imes & -2 & = & ? & ( extAnswer = -4 )\2 & imes & -3 & = & ? & ( extAnswer = -6 )\endarray$$

The number on the right-hand also side keeps decreasing by 2.

Informal justification of $ extnegative imes extnegative = extpositive$

Continue the pattern:

$$eginarrayr2 & imes & -3 & = & -6\1 & imes & -3 & = & -3\0 & imes & -3 & = & 0\-1 & imes & -3 & = & ? & ( extAnswer = 3 )\-2 & imes & -3 & = & ? & ( extAnswer = 6 )\-3 & imes & -3 & = & ? & ( extAnswer = 9 )\endarray$$

The number on the right-hand also side keeps increasing by 3.


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Jordan Gray
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Dan ChristensenDan Christensen
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Well if I were to define this in an intuitive means to someone (or at least try), I would certainly prefer to think of an analogy via walking over the real line, by agreeing that walking left will certainly be walking in the negative direction and walking appropriate in the positive direction.

Then I will try to convey the concept that if you are multiplying two numbers (let"s suppose they are integers to make points much easier to picture) then a product as $2*3$ would certainly just intend that you have to walk appropriate (in the positive direction) a distance of $2$ (say miles for instance) 3 times, that is, first you walk $2$ miles, then one more $2$ miles and also lastly an additional $2$ miles to the appropriate.

Now you image wbelow you"re at? Well, you"re at the ideal of the origin so you are in the positive area. But in the exact same means you can play this principle via an adverse times a positive.

With the very same instance in mind, what would $-2*3$ mean? First, expect that the $-2$ simply mentions that you will certainly have to walk left a distance of $2$ miles. Then how many kind of times you will certainly walk that distance? Just as prior to $3$ times and also in the end you"ll be $6$ miles to the left of the origin so you"ll be in the negative section.

Finally, you"ll need to attempt to picture what might $(-2)*(-3)$ expect. Maybe you might think of the negative sign in the second aspect to indicate that you readjust direction, that is, it provides you revolve roughly and also begin walking the stated distance. So in this situation the $-2$ tells you to walk left a distance of $2$ miles yet the $-3$ tells you to first turn roughly, and also then walk $3$ times the $2$ miles in the other direction, so you"ll end up walking ideal and also finish in the allude that is $6$ miles to the appropriate of the beginning, so you"ll be in the positive area, and also $(-2)*(-3) = 6$.

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I do not recognize if this will certainly help, but it"s the only method I can think of this in some intuitive sense.