-2 points
*

This only ever got handed down to us as gospel. Is there a compelling reason why we should accept that (-3) × (-3) = 9?

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5 points

Same reason that a double negative makes a positive.

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31 points

You can look at multiplication as a shorthand for repeated addition, so, for example:

3x3=0 + 3 + 3 + 3 = 9

In other words we have three lots of three. The zero will be handy later…

Next consider:

-3x3 = 0 + -3 + -3 + -3 = -9

Here we have three lots of minus three. So what happens if we instead have minus three lots of three? Instead of adding the threes, we subtract them:

3x-3 = 0 - 3 - 3 - 3 = -9

Finally, what if we want minus three lots of minus three? Subtracting a negative number is the equivalent of adding the positive value:

-3x-3 = 0 - -3 - -3 - -3 = 0 + 3 + 3 + 3 = 9

Do let me know if some of that isn’t clear.

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4 points

i think this is a really clean explanation of why (-3) * (-3) should equal 9. i wanted to point out that with a little more work, it’s possible to see why (-3) * (-3) must equal 9. and this is basically a consequence of the distributive law:

0  = 0 * (-3)
   = (3 + -3) * (-3)
   = 3 * (-3) + (-3) * (-3)
   = -9 + (-3) * (-3).

the first equality uses 0 * anything = 0. the second equality uses (3 + -3) = 0. the third equality uses the distribute law, and the fourth equality uses 3 * (-3) = -9, which was shown in the previous comment.

so, by adding 9 to both sides, we get:

9 = 9 - 9 + (-3) * (-3).

in other words, 9 = (-3) * (-3). this basically says that if we want the distribute law to be true, then we need to have (-3) * (-3) = 9.

it’s also worth mentioning that this is a specific instance of a proof that shows (-a) * (-b) = a * b is true for arbitrary rings. (a ring is basically a fancy name for a structure with addition and distribute multiplication.) so, any time you want to have any kind of multiplication that satisfies the distribute law, you need (-a) * (-b) = a * b.

in particular, (-A) * (-B) = A * B is also true when A and B are matrices. and you can prove this using the same argument that was used above.

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13 points
*

This was very clear. Now that I see it, I realize it’s the same reasoning why x^(-3) is 1/(x^3):

 2 × -3 = -6
 1 × -3 = -3
 0 × -3 =  0
-1 × -3 =  3

Thank you!

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5 points
*

Here’s another example:

A) -3 × (-3 + 3) = ?

You can solve this by figuring out the brackets first. -3 × 0 = 0

You can also solve this using the distributive property of multiplication, rewriting the equation as

A) -3 × (-3 + 3) = 0
(-3 × -3) + (-3 × 3) = 0
(-3 × -3) - 9 = 0
(-3 × -3) = 9

If (-3 × -3) didn’t equal 9 then you’d get different answers to equation A depending on what method you used to solve it.

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113 points

-3 feels like cheating.

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72 points

Eh, not really. It’s been a while, but I’m pretty sure the rule in algebra when solving for a squared variable like this is to use ± for exactly that reason.

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23 points
*

just wait for n-th roots of imaginary numbers :)

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11 points

Oh, I know this one! It’s pi!

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12 points

What, no. It’s… Eh close enough.

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4 points

TAU IS BETTER

/obligatory

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1 point

Help how do i take factorial of pi

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13 points

My teacher explained as sqrt(poop^2) = abs(poop). Yes, he wrote poop on the blackboard.

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3 points

He should have drawn a pile of poop instead 💩 (preferably without a face)

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20 points

Also math teacher…

“Show your work”

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