Elementary math that professors cant solve

[plover]

Or for those whose tastes run to notation sadism:

[tex]- \lim_{0\rightarrow0} 0![/tex]​

 

[futb0l]

Strictly speaking [itex]0^0[/itex] is not defined. As:

[tex]x^0 = \left( x^1 \right) \left( x^{-1} \right)[/tex]

Therefore:

[tex]x^0 = \frac{x}{x}[/tex]

Which means [itex]x^0 = 1[/itex] when [itex]x \neq 0[/itex]

mmm.. if you do 0^1 in google, it will come up as 1.
and ... http://mathforum.org/dr.math/faq/faq.0.to.0.power.html

so i don't think
[itex]x^0 = 1[/itex] when [itex]x \neq 0[/itex]
is true.

 

[futb0l]

there should be a rule that says when any number is to the power of 0 it will be equal to 1.

 

[ExecNight]

Most of em are true solutions..

Now the funny thing here is we are getting something from noting..

How come we can get 1 from 0 by using only 0? That always makes my head iching...

 

[futb0l]

read that: http://mathforum.org/dr.math/faq/faq.0.to.0.power.html

 

[NoNose]

Hi
First: Sorry if my english isn´t correct or couldn´t be understand, but i´m trying to.

Without to resume the discussion, if factorial and bases are elemental math, i´m thinking i´ve found a solution for the problem down this text for all positive and negative real numbers and the 0:
[tex]\left( \left( x^2 \right) ^0 + \left( x^2 \right) ^0 + \left( x^2 \right) ^0 \right) ! =6[/tex]

There's another fun variation on this theme where you line up all the numbers from one to nine in threes and are supposed to make them add up to six by adding only plus, minus, division, multiplication, root and power signs (whole powers and roots, no logs!). You can also use ( and ) (forgot what they're called).

Like this:

1   1   1 = 6
2   2   2 = 6
3   3   3 = 6
4   4   4 = 6
5   5   5 = 6
6   6   6 = 6
7   7   7 = 6
8   8   8 = 6
9   9   9 = 6

For example (I hope I'm not ruining anything for anyone here ):
6 + 6 - 6 = 6

Have fun.

 

[Zurtex]

Hi
First: Sorry if my english isn´t correct or couldn´t be understand, but i´m trying to.

Without to resume the discussion, if factorial and bases are elemental math, i´m thinking i´ve found a solution for the problem down this text for all positive and negative real numbers and the 0:
[tex]\left( \left( x^2 \right) ^0 + \left( x^2 \right) ^0 + \left( x^2 \right) ^0 \right) ! =6[/tex]

I like that , but using the square function is kind of using a 2 really (where as the square root actually has a symbol). So perhaps before anyone complains about this it could be easily fixed as:

[tex]\left( |x|^0 + |x|^0 + |x|^0 \right) ! = 6[/tex]

For [itex]x \neq 0[/itex]

 

[Bartholomew]

10 (44-4)/4

I had to use one "44". Is there a way to get 10 without resorting to this?

Njorl

4 * 4 - 4! / 4 works

 

[physicsuser]

[tex]9+\frac{9}{9}=20_{(base 5)}[/tex]

I rarely do math for fun so my so I don't know much math indepth. From what I know base 2 or binary numbers are like this

01 this means that 0*(2^1)+1*(2^0)=1

So in base 5 it would be

0,1,2,3,4 = 0*(5^4)+1*(5^3)+2*(5^2)+3*(5^1)+4*(5^0)=194

how do you get 9 in base 5? Is it a different base system or something?


edit--------------

Oh I think I know

9 =14
14/14=1
14+1=20
right?