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intertubes wrote: Wait, why is he dividing into a square root symbol? /doesn't get that
iGeek wrote:On a physics exam, a friend of mine couldn't remember whether a particular equation had "squared" or "times two" in it, so he decided to pick a number and try it both ways and see which one looked right. Unfortunately, he realized later, the number he picked was 2.
2^2 = 2*2
I'm not convinced that that one doesn't have a more satisfying explanation. Took me forever to come up with a human satisfying explanation for the commutative property of multiplication.Initially this looks like another example of this, but it actually makes sense why this works
iGeek wrote:On a physics exam, a friend of mine couldn't remember whether a particular equation had "squared" or "times two" in it, so he decided to pick a number and try it both ways and see which one looked right. Unfortunately, he realized later, the number he picked was 2.
2^2 = 2*2
ringobob wrote:I'm not convinced that that one doesn't have a more satisfying explanation. Took me forever to come up with a human satisfying explanation for the commutative property of multiplication.
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.What's so hard about that? Just think of a rectangle divided into either its rows or its columns.
Turtle_ wrote:I did something similar one time. I was trying to remember whether the full law of sines said that sinA / a = R or a / sinA = R, so I tried an example and got sinA / a seemed like R. The real answer was a / sinA = 2R
Nereid wrote:Dimensional analysis is your friend.Turtle_ wrote:I did something similar one time. I was trying to remember whether the full law of sines said that sinA / a = R or a / sinA = R, so I tried an example and got sinA / a seemed like R. The real answer was a / sinA = 2R
Shay Guy wrote:iGeek wrote:On a physics exam, a friend of mine couldn't remember whether a particular equation had "squared" or "times two" in it, so he decided to pick a number and try it both ways and see which one looked right. Unfortunately, he realized later, the number he picked was 2.
2^2 = 2*2
You've heard of Graham's number, right? That absurdly huge number equivalent to 3↑↑↑...↑↑↑3, with a much smaller but still absurdly huge number of up arrows?
Change those 3s to 2s, and it reduces to 4.
Two and two is four, no matter how you do it.
happysteve wrote:Kinda reminds me of this fun math tidbit...
Solve for x:
x^2 - x - 20 = 10 ... hmm okay, factor the left hand side...
(x + 4)(x - 5) = 10
ah great, now it's just a matter of solving for (x + 4) = 10 and (x - 5) = 10
x is either 6 or -5
Check with the original statement:
6^2 - 6 - 20 = 10 ... yup, that works
(-5)^2 - (-5) - 20 = 10 ... yup, that works too.
yay, problem solved.
Shay Guy wrote:
You've heard of Graham's number, right? That absurdly huge number equivalent to 3↑↑↑...↑↑↑3, with a much smaller but still absurdly huge number of up arrows?
Change those 3s to 2s, and it reduces to 4.
Two and two is four, no matter how you do it.
Grumbleduke wrote:A quick bit of algebra shows that this will work for any two numbers where:\begin{eqnarray*}a \times b & = & a \times \sqrt{b^2} \\ & = & a \sqrt{b^2}\\ & = & \frac{b^2}{a} \\ \therefore a^2 & = & b \end{eqnarray*}
It's not all that exciting; try it with 2 and 4, or -3 and 9...
However, the alt text is very good advice.
Edit: Ah, assuming that a isn't 0 of course - although I think it still works algebraicly, if you use l'Hopital's rule on the fraction.
xkcd wrote:Alt-text:
Handy exam trick: when you know the answer but not the correct derivation, derive blindly forward from the givens and backward from the answer, and join the chains once the equations start looking similar. Sometimes the graders don't notice the seam.
Shivari wrote:... I missed that part where that was interesting and not just what you always do in that scenario.
I wonder in how many cases this works.masamune55 wrote:*MIND BLOWN*
Shivari wrote:happysteve wrote:Kinda reminds me of this fun math tidbit...
Solve for x:
x^2 - x - 20 = 10 ... hmm okay, factor the left hand side...
(x + 4)(x - 5) = 10
ah great, now it's just a matter of solving for (x + 4) = 10 and (x - 5) = 10
x is either 6 or -5
Check with the original statement:
6^2 - 6 - 20 = 10 ... yup, that works
(-5)^2 - (-5) - 20 = 10 ... yup, that works too.
yay, problem solved.
... I missed that part where that was interesting and not just what you always do in that scenario.
happysteve wrote:Kinda reminds me of this fun math tidbit...
Solve for x:
x^2 - x - 20 = 10 ... hmm okay, factor the left hand side...
(x + 4)(x - 5) = 10
ah great, now it's just a matter of solving for (x + 4) = 10 and (x - 5) = 10
x is either 6 or -5
Check with the original statement:
6^2 - 6 - 20 = 10 ... yup, that works
(-5)^2 - (-5) - 20 = 10 ... yup, that works too.
yay, problem solved.
airshowfan wrote:Well, I do usually manage to close up the seam. And when I don't, the gaping hole is pretty obvious (and I have to resist the temptation to write "Then a miracle happens" into the gap)
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