## 0759: "3x9"

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phlip
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### Re: "3x9" Discussion (#759)

Sereue wrote:
am3930 wrote:Call it a property of two.

Or you could call them the definitions of exponentiation and multiplication:
Exponentiation is Repeated Multiplication --> a^b = a*...*a, b times;
Multiplication is Repeated Addition --> a*b = a+...+a, b times.

Right, which means it is a property of 2: namely, being the number of operands of a binary operator.

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`enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};void ┻━┻︵​╰(ಠ_ಠ ⚠) {exit((int)⚠);}`
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DavidRoss
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### Re: "3x9" Discussion (#759)

ringobob wrote:
What's so hard about that? Just think of a rectangle divided into either its rows or its columns.
Nothing's hard about that. It really only took me probably 45 minutes... in 2-3 minute chunks spread out over 5+ years. I always just accepted it, wondered on it briefly every now and then, and then finally visualized it similarly to what you said.

OK, now that that is behind you, spend the next 5 years attempting to visualize why the product of two negative numbers is positive. I have tried to find a simple way to explain the truth of that to someone who holds an ancient Greek view of numbers - negative numbers are only tolerated in the context of temporarily venturing below zero. Of course, by assuming that (6-4) * (3-1) = 4 and assuming that multiplication is distributive, i.e., that the equation can be converted to 6*3 + (- 4*3) + (6* -1) + (-4 * -1), and further accepting that debts can be negative assets and multiply a debt makes a larger debt, the equation simplifies to 18 - 12 - 6 + (-4 * -1) = 0 + (-4 * -1) = 4, so -a * -b = +ab. But that is bootstrapping, since the distributive nature of multiplication is not necessarily something that can be easily shown physically (to a skeptic).

phlip
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### Re: "3x9" Discussion (#759)

The distributive property of multiplication is easy enough to do physically: a rectangle (a+b) long and c wide can be split into two rectangles, one a by c and the other b by c.

For negative*negative=positive, I recommend drawing a full multiplication table, with positives and negatives, filling in three quadrants but leaving the negative*negative one blank. Now, you can take any given row or column in the parts that are filled in, and see that they're linear... eg, for the row that corresponds to the number 3, moving to the right adds 3 to the value, and moving to the left subtracts 3... and this happens everywhere on the row, regardless of whether it's 3*positive or 3*negative. So do the same thing in the negatives... and on the row for -3, you can see that everywhere on the row that's filled in so far, moving to the right subtracts 3, and moving to the left adds 3. So you can continue that pattern into the unfilled area, by continually adding 3. You can do the same by columns, and see that you get the same number in each cell. And you end up with (-a)(-b)=ab.

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`enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};void ┻━┻︵​╰(ಠ_ಠ ⚠) {exit((int)⚠);}`
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StNowhere
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### Re: "3x9" Discussion (#759)

Maybe I missed it, but I'm surprised no one has mentioned my favorite, the textbook trick.

On a test once, I was asked to prove a result that I could do fairly easily given a separate, somewhat nontrivial result that I was afraid I would have to prove in order to be allowed to use it. For some reason, I couldn't remember that proof, so when the time came for it, I used the line students always dread reading in textbooks.

"The proof is left to the reader as an exercise."

F-13
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### Re: "3x9" Discussion (#759)

Arakun wrote:What, no one posted this yet!?
Abbott And Costello 13 × 7 is 28

And this means I'm the second person who registered today just to say that this reminded me of the Abbott and Costello sketch.

And by the way, this is the best comic in a long, LONG time!!!!

Cathy
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### Re: "3x9" Discussion (#759)

I read this at 9:30 am, 30 minutes before my Cal 1 test and it made me feel better...

Now that I've TAKEN the test, and *sort of* used this... I feel a LOT better!

<3
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yurell wrote:We need fewer homoeopaths, that way they'll be more potent!

Louis XIV
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### Re: "3x9" Discussion (#759)

alemfi wrote:It's simply making use of the similarity in appearance of the square root symbol to the way one does long division.

Thanks for the explanation. I didn't get it, mainly because in Germany we do long divisions like this:

fakepants
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### Re: "3x9" Discussion (#759)

I recall a student asking a prof if he could do a problem a certain way, as it produced the correct answer. But the method was rubbish. The prof, after showing him he'd simply gotten lucky, said something like, "If you make one mistake, you almost always get the wrong answer--but at least you know it. If you make lots of mistakes, sometimes you get lucky, and don't even realize you made one mistake."

squareroot
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### Re: "3x9" Discussion (#759)

fakepants wrote:I recall a student asking a prof if he could do a problem a certain way, as it produced the correct answer. But the method was rubbish. The prof, after showing him he'd simply gotten lucky, said something like, "If you make one mistake, you almost always get the wrong answer--but at least you know it. If you make lots of mistakes, sometimes you get lucky, and don't even realize you made one mistake."

Math is the onlz place where two wrongs make a right: -(-x)=x.
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internetcommenter
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### Re: "3x9" Discussion (#759)

squareroot wrote:Math is the onlz place where two wrongs make a right: -(-x)=x.

Or as my Linear Algebra professor use to say:
Any even number of wrongs make a right.

fakepants
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### Re: "3x9" Discussion (#759)

internetcommenter wrote:
squareroot wrote:Math is the onlz place where two wrongs make a right: -(-x)=x.

Or as my Linear Algebra professor use to say:
Any even number of wrongs make a right.

Haha! Precisely.

kg333
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### Re: "3x9" Discussion (#759)

RogueCynic wrote:Also, I heard a rumor someone in Somerville Ma. used a similar complex formula to fill in the dollar amount on a check to pay an electric bill. He was protesting the utility's complex billing system. I can't remember the person's name though.

The one I'm familiar with was actually done by the author...Randall wrote a check to Verizon doing exactly that .

Here's the link...Have mercy, mods, I know I don't have the posts yet; hopefully it's ok since it's still on-site?
http://xkcd.com/verizon/

scarletmanuka
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### Re: "3x9" Discussion (#759)

When I was an honours / postgrad mathematics student, I used to run some tutorials for first-year students. I always checked line by line and made sure they didn't try anything like this. And yes, I checked carefully across page boundaries too. If there seemed to be an unsupported jump in the logic from one line to the next, they'd lose one or more marks (depending on how big the gap was). Some students were honest about it, some tried to hide it. I always appreciated the honest ones. And for every error, I tried to explain what was wrong, why it was wrong and what they should have done instead (if that wasn't obvious, that is; "you missed a minus sign here" doesn't need any further elaboration).

But I know many of the other tutors just checked key points in the working out. They got through their assignment loads far faster than I did, but I hope I provided a better service to my students.

I had a couple of cases where students used a non-obvious methodology, very different to the one on the answer sheet, but managed nevertheless to correctly derive the answer. They got full marks for that, of course. The answer sheets were often buggy, too - this helped me to catch cheaters more than once.

I think the funniest one I got was for one question where the answer sheet was fairly wrong. Two students in my group (whom I'd previously noted as copying occasional questions from each other) had copied the answer sheet exactly. Identical diagram, same choice of variable names, every error in the logic faithfully reproduced. Their reward (as sanctioned by the official policy at the time) was a mark of 0 for the whole assignment.

sr123
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### Re: "3x9" Discussion (#759)

Is there a name for this class of problems?

Another one: a^b = a/\b, somehow. That is, a raised to the power of b is misread as a XOR b in a binary base. Or maybe it would be better done as a set construct, except for finding a well-defined power notation?

Ugh, bitwise exponentials - does anyone know of an algorithm to calculate them? I'll try to derive one myself based on the bitwise multiplication http://en.wikipedia.org/wiki/Bitwise_operation#Applications bitshift-addition technique.

phlip
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### Re: "3x9" Discussion (#759)

sr123 wrote:Another one: a^b = a/\b, somehow. That is, a raised to the power of b is misread as a XOR b in a binary base.

I think the only way that can happen is 10 = 1⊕0 = 1 (using not-caret notation for both operators for clarity).  No, wait, also: -1-2 = -1⊕-2 = 1, assuming 2's complement for the negative numbers in the bitwise XOR.

Code: Select all

`enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};void ┻━┻︵​╰(ಠ_ಠ ⚠) {exit((int)⚠);}`
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Trung2695
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### Re: 0759: "3x9"

I don't know if anyone posted this, but this is from a 2010 examiners report of the STEP exam (the exam you take in some universities in the UK!):
On some questions, some candidates tried to work forwards from the given question and
backwards from the answer, hoping that they would meet somewhere in the middle. While
this worked on occasion, it often required fudging, and did bring to mind a recent web
comic: http://xkcd.com/759.

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