One corner oriented wrong

[Feryll]

as im vindictive i would give them a maths problem for homework that was unsolvable........hehehehehehe that would teach them.

Good idea. What's the math problem?

x = x + 1

 

[SuperNerd]

as im vindictive i would give them a maths problem for homework that was unsolvable........hehehehehehe that would teach them.

Good idea. What's the math problem?

x = x + 1

Wow, you made my head aspload.

I'd give them √-1.

 

[Kenneth]

I'd give them √-1.

√-1 = √ [-1,0] = [0,1]

Or

√-1 = √ [-1,0] = [0,-1]

Both are correct, but also this:

√-1 = √ [-1,0,0] = [0,1,0]
√-1 = √ [-1,0,0] = [0,-1,0]
√-1 = √ [-1,0,0] = [0,0,1]
√-1 = √ [-1,0,0] = [0,0,-1]

And this:

√-1 = √ [-1,0,0] = [0,cos(45 degree),sin(45 degree)]

Or any point that is on a circle around the x-axis with radius = 1

Moving to 4D it is any point on a "sphear" around the x-axis with radius 1.

In 5D it is any point on a ???? (what is it??) around the x-axis and so on.

 

[~Phoenix Death~]

as im vindictive i would give them a maths problem for homework that was unsolvable........hehehehehehe that would teach them.

Good idea. What's the math problem?

x = x + 1

Hmm... the answer is very mysterious... No more numbers or the problem would probably be solvable. Awesomest math problem punishment. lol...

That problem is easily solvable, what's wrong with you?

 

[gibbleking]

sorry fella ...its just been solved by a hermit who wont accept the prize money..

 

[mrCage]

Hi all,

I'm a high school math teacher, and I use a cube whenever possible in discussing mathematics with my students. Last week I allowed some students to scramble my cube for me to solve, and came across a curious problem. There is one corner oriented wrong on my cube. It's not a matter that it just needs to be turned once or twice; it's backward. That is to say, going clockwise around it, this piece should be blue/orange/yellow, and instead it's blue/yellow/orange. The rest of the cube is correctly completed.

Obviously, I can't just pop the piece off and put it in correctly. So my questions are:
(1) What did my students do to my cube?
(2) What can I do to fix it?
(3) What should I do to the students who commit such a crime against mathematics?

Thanks for your help!

If you do not know anwer to 1 and 2 then dont be a math teacher

Per

 

[megaminxwin]

And now it's time to actually make a contribution in this thread.

1. Switch stickers.
2. Switch them back.

OR buy tiles so they can't switch the stickers

3. Teach them calculus:
'Why do we need to learn this, sir/miss?'
'Why did you switch the stickers on my cube?'
'But we -'
'Anyway...'

 

[canadiancuber]

as im vindictive i would give them a maths problem for homework that was unsolvable........hehehehehehe that would teach them.

Good idea. What's the math problem?

x = x + 1

Hmm... the answer is very mysterious... No more numbers or the problem would probably be solvable. Awesomest math problem punishment. lol...

That problem is easily solvable, what's wrong with you?

I'm too sleepy. Can't think.

what's the answer?


my guess

Spoiler

i don't have one......MONKEY

 

[ThatGuy]

x - x = x + 1 - x
0=1
All numbers are solutions
which is true because:
For notational purposes let a = b
a^2 = ab
a^2-2ab = ab-2ab = -ab
a^2 - 2ab + a^2 = -ab + a^2
2a^2 -2ab = -ab +a^2
2(a^2 - ab) = a^2 - ab
divide both sides by a^2 - ab
Yes I did divide by 0.
2 = 1
2 -1 = 1-1
1=0
transitive property
0=1
put anything discovered in spoilers. like why this works.

 

[adimare]

x - x = x + 1 - x
0=1
All numbers are solutions
which is true because:
For notational purposes let a = b
a^2 = ab
a^2-2ab = ab-2ab = -ab
a^2 - 2ab + a^2 = -ab + a^2
2a^2 -2ab = -ab +a^2
2(a^2 - ab) = a^2 - ab
divide both sides by a^2 - ab
Yes I did divide by 0.
2 = 1
2 -1 = 1-1
1=0
transitive property
0=1
put anything discovered in spoilers. like why this works.

Spoiler

I have a better one:

1 = 1
1 + 1 = 1 + 1
since 1 = √1
1 + 1 = 1 + √1
1 + 1 = 1 + √(-1 x -1)
since √ab = √a x √b
1 + 1 = 1 + √-1 x √-1
1 + 1 = 1 + i x i
1 + 1 = 1 + i^2
1 + 1 = 1 - 1
2 = 0

 

[miniGOINGS]

1 + 1 = 1 + √-1 x √-1
1 + 1 = 1 + i x i
1 + 1 = 1 + i^2
1 + 1 = 1 - 1
2 = 0

(-1)^2 =/= (-1)

 

[CubesOfTheWorld]

Peeled the stickers like the non-cubers they are.

 

[TrollingHard]

Easy:

It doesn't happen.
It's similar to proving that you can divide by zero: all math would be incorrect.

 

[adimare]

1 + 1 = 1 + √-1 x √-1
1 + 1 = 1 + i x i
1 + 1 = 1 + i^2
1 + 1 = 1 - 1
2 = 0

(-1)^2 =/= (-1)

I never wrote that (-1)^2 = -1
I wrote that i^2 = -1

 

[qqwref]

Even better: http://mzrg.com/math/0=1.shtml

 

[Forte]

since √ab = √a x √b

Not when a and b are both negative :rolleyes:

Even better: http://mzrg.com/math/0=1.shtml

Is it that you didn't factor in the constant from the integrals? (the +C)

 

[adimare]

since √ab = √a x √b

Not when a and b are both negative :rolleyes:

You got me

Even better: http://mzrg.com/math/0=1.shtml

Is it that you didn't factor in the constant from the integrals? (the +C)

You got him. If you're going to abuse integration by parts, 1/x works just as well.

\( \int{\frac{1}{x}dx} \)
\( u = \frac{1}{x} \) and \( dv = dx \)
\( du = -\frac{1}{x^2}dx \) and \( v = x \)

\( \int{\frac{1}{x}dx}=\frac{x}{x}-\int{-\frac{1}{x^2}xdx} \)

\( \int{\frac{1}{x}dx}=1+\int{\frac{1}{x}dx} \)

\( 0=1 \)

 

[Rinfiyks]

Let x = -1/(-1/(-1/(-1/(-1/(-1...
x = -1/x
x^2 = -1
x = ± i
± i = -1/(-1/(-1/(-1/(-1/(-1...

 

[Forte]

Let x = -1/(-1/(-1/(-1/(-1/(-1...

That can't exist because each time you do the -1 thing, you're flipping it from 1 to -1, so it's "divergent" (it never settles onto one number). I think that's what it is

 

[Rinfiyks]

Let x = -1/(-1/(-1/(-1/(-1/(-1...

That can't exist because each time you do the -1 thing, you're flipping it from 1 to -1, so it's "divergent" (it never settles onto one number). I think that's what it is

It converges from the "bottom" of the infinite fraction though
\(
\frac{-1}{0-\frac{-1}{0-\frac{-1}{0-\frac{-1}{0-\frac{-1}{...}}}}}
\)
(I put the zeroes in there so it's easier to see)