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Soroban
 Post subject: A new world record!
PostPosted: Sun, 19 Jun 2011 00:02:20 UTC 
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Hello, everyone!

This problem appeared at another site, followed by my response.

Brace yourself . . . This is painful!

Quote:
\text{Simplify: }\:\sqrt{27x^3y^7} + \sqrt{48x^3y^7}

\text{I got: }\:9xy^4+16xy^4 \:=\:25x^2y^8 . . . . Wow!

I really stretched my imagination
. . . and still couldn't believe what I saw.

First of all, those square roots are probably cube roots: .\sqrt[3]{27x^3y^7} + \sqrt[3]{48x^3y^7}


\text{Then }\sqrt[3]{27x^3y^7}\text{ was simplified like this . . .}

\sqrt[3]{27x^3y^7} \;=\; \sqrt[3]{27}\cdot\sqrt[3]{x^3}\cdot\sqrt[3]{y^7}

. . \sqrt[3]{27} \:=\:27 \div 3 \:=\:9 ?

. . \sqrt[3]{x^3} \:=\:x^{3-3} \:=\:x ?

. . \sqrt[3]{y^7} \:=\:y^{7-3} \:=\:y^4 ?

\text{Hence: }\:\sqrt[3]{27x^3y^7} \;=\;9xy^4 ?


\text{And }\sqrt[3]{48x^3y^7}\text{ was simplified like this . . .}

\sqrt[3]{48x^3y^7} \;=\;\sqrt[3]{48}\cdot\sqrt[3]{x^3}\cdot\sqrt[3]{y^7}

. . \sqrt[3]{48} \:=\:48 \div 3 \:=\: 16 ?

. . \sqrt[3]{x^3} \:=\:x^{3-3} \:=\:x ?

. . \sqrt[3]{y^7} \:=\:y^{7-3} \:=\:y^4 ?

\text{Hence: }\:\sqrt[3]{48x^3y^7} \:=\:16xy^4 ?


\text{Then: }\:\sqrt[3]{27x^3y^7} + \sqrt[3]{48x^3y^7} \;=\;9xy^4 + 16xy^4 ?


\text{And evidently: }\:9xy^4 + 16xy^4 \;=\;(9+16)(x+x)(y^4+y^4) \;=\;25x^2y^8 ??


This should be in the Guinness Book
. . for "The Most Errors in One Solution".


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


And did you see what he did with the second problem?

Quote:
$\text{I got: }\;\frac{3x^2 - 5x + 4}{x+3} \;=\;3x^2 - 4x + 7

Did he really do this?

$\frac{3x^2 - 5x + 4}{x+3} \;=\;\frac{3x^2}{0x^2} + \frac{-5x}{1x} + \frac{4}{3} \;=\;3x^2 - 4x + 7
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Denis
 Post subject:
PostPosted: Sun, 19 Jun 2011 17:31:02 UTC 
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Soroban, are you his teacher ? :lol:

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outermeasure
 Post subject: Re: A new world record!
PostPosted: Sun, 19 Jun 2011 17:36:18 UTC 
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Without the URL I could only guess. I'm not sure about all those mistakes you (Soroban) are imagining... did (s)he simplified \sqrt[3]{x^3}=x correctly instead of going through two wrong steps x^{3-3}?

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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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Shadow
 Post subject:
PostPosted: Sun, 19 Jun 2011 18:01:13 UTC 
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Denis wrote:
Soroban, are you his teacher ? :lol:


I believe Soroban is retired.

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Denis
 Post subject:
PostPosted: Sun, 19 Jun 2011 20:31:43 UTC 
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Twas a joke Shadow; bad student = bad teacher :idea:

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Matt
 Post subject: Re: A new world record!
PostPosted: Mon, 20 Jun 2011 00:35:09 UTC 
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outermeasure wrote:
I'm not sure about all those mistakes you (Soroban) are imagining... did (s)he simplified \sqrt[3]{x^3}=x correctly instead of going through two wrong steps x^{3-3}?

Given that it was originally a square root and not a cube root, it's the student who's full of crap.
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Shadow
 Post subject:
PostPosted: Mon, 20 Jun 2011 01:47:01 UTC 
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Denis wrote:
Twas a joke Shadow; bad student = bad teacher :idea:


*needs a sarcasm meter*

Though I do think bad teacher \Righarrow bad student, but perhaps not the other way around (necessarily).

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helmut
 Post subject:
PostPosted: Mon, 20 Jun 2011 02:05:27 UTC 
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Shadow wrote:
Denis wrote:
Twas a joke Shadow; bad student = bad teacher :idea:


*needs a sarcasm meter*

Though I do think bad teacher \Rightarrow bad student, but perhaps not the other way around (necessarily).

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Shadow
 Post subject:
PostPosted: Mon, 20 Jun 2011 06:59:59 UTC 
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helmut wrote:
Shadow wrote:
Denis wrote:
Twas a joke Shadow; bad student = bad teacher :idea:


*needs a sarcasm meter*

Though I do think bad teacher \Rightarrow bad student, but perhaps not the other way around (necessarily).


That's what I said all right. . .

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aswoods
 Post subject:
PostPosted: Mon, 20 Jun 2011 07:16:02 UTC 
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Shadow wrote:
That's what I said all right. . .


Actually what you said was "unparseable or potentially dangerous" ... presumably some sort of Lovecraftian oath put down in mathematical notation, and suppressed automatically by the server for the sake of readers' safety.
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Shadow
 Post subject:
PostPosted: Mon, 20 Jun 2011 07:31:12 UTC 
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aswoods wrote:
Shadow wrote:
That's what I said all right. . .


Actually what you said was "unparseable or potentially dangerous" ... presumably some sort of Lovecraftian oath put down in mathematical notation, and suppressed automatically by the server for the sake of readers' safety.


Well so I did. Thanks helmut! And you as well aswoods.

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outermeasure
 Post subject: Re: A new world record!
PostPosted: Mon, 20 Jun 2011 07:37:46 UTC 
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Matt wrote:
outermeasure wrote:
I'm not sure about all those mistakes you (Soroban) are imagining... did (s)he simplified \sqrt[3]{x^3}=x correctly instead of going through two wrong steps x^{3-3}?

Given that it was originally a square root and not a cube root, it's the student who's full of crap.


Sure, but the goal here is to find the minimum number of mistakes committed to get that "answer" (otherwise you can go on adding mistakes to make number of mistakes (or the density) committed arbitrarily high, even in the case when he/she/it did everything correctly).

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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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Soroban
 Post subject: Re: A new world record!
PostPosted: Mon, 20 Jun 2011 17:20:19 UTC 
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Hello, outermeasure!

Quote:
Without the URL I could only guess.
I'm not sure about all those mistakes you (Soroban) are imagining...
Did (s)he simplify \sqrt[3]{x^3}=x correctly
. . instead of going through two wrong steps x^{3-3}?

Good point . . .

I too thought he/she got one step correct.


\text{But if }\sqrt[3]{x^7} \:=\:x^{7-3} \:=\:x^4 . (the rule seems to be "subtract indices")

. . \text{then: }\,\sqrt[3]{x^3}\,\text{ must be }\,x^{3-3}
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outermeasure
 Post subject: Re: A new world record!
PostPosted: Mon, 20 Jun 2011 17:24:12 UTC 
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Soroban wrote:
Hello, outermeasure![/size]

Quote:
Without the URL I could only guess.
I'm not sure about all those mistakes you (Soroban) are imagining...
Did (s)he simplify \sqrt[3]{x^3}=x correctly
. . instead of going through two wrong steps x^{3-3}?

Good point . . .

I too thought he/she got one step correct.


\text{But if }\sqrt[3]{x^7} \:=\:x^{7-3} \:=\:x^4 . (the rule seems to be "subtract indices")

. . \text{then: }\,\sqrt[3]{x^3}\,\text{ must be }\,x^{3-3}


No, the rule could have been: "n-th root of x^n is x, otherwise subtract indices".

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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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math explorer
 Post subject: I wish I got $100 for every mistake I've found in math books
PostPosted: Tue, 21 Jun 2011 23:29:14 UTC 
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I could make a living at it.
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