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### Re: Math proof

Posted: Sun Apr 27, 2008 6:00 pm
If anyone has any problems they would like to present to see if someone can prove them that would be a good exercise to. I'll put this one out there.( as a general statement, don't put your homework problems up here though ).

See if you can come up with a proof for DeMorgan's law, (A union B)^c = A^c intersection B^c. I'm using ^c to denote compliment.

Here's another proof to go through also. There are a whole lot of things wrong with this proof see if you can find them all.

Show that a + t^2 is a subspace of the polynomials given that a is a scalar and t is a variable.

take two vectors of the form a + t^2 and show that vector addition holds. Let a and b be arbitrary scalars, (a + t^2) + ( b + t^2 )= (a+b) + 2(t^2)= divide through by 2 = (a+b)/2 + t^2, so it's of the form a scalar added to a variable, so it's closed under addition.

Now show that it's closed under scalar multiplication. Let c be an arbitrary scalar, then c(a+t^2) = ca + ct^2 = ca/c + t^2 which is of the right form, so it's a subspace.

The correction to the composition of functions proof was correct

### Re: Math proof

Posted: Wed Apr 30, 2008 6:30 pm
What do you guys make of this?

_
.9=x
_
10x=9.9
_ _
10x - x = 9.9 - .9

9x = 9

x = 1
_
.9 = 1

Can you spot the mistake?
Or is there one?

### Re: Math proof

Posted: Wed Apr 30, 2008 9:57 pm
The problem is that you are not being consistent with what you're doing with the equation. If you subtract x from the left side you have to subtract x from the right side. so the equation would have to look like 10x - x = 9.9 - x, or 10x - .9 = 9 - .9. Even though x = .9 you can't interchange the two when manipulating the equation, you have to use the same thing for each side for an operation. That line is what leads to the incorrect solution.

It was a nice idea unfortunately incorrect

### Re: Math proof

Posted: Wed Apr 30, 2008 10:37 pm
Heh - I used to make these up daily in boring math professors classes.

Here is a recent one: I used wikipedia code for it cause its hard to type out
Code: Select all
$\sum_{n=5}^\infty \frac{n}{1} = \infty$<br />$\sum_{n=6}^\infty \frac{n}{1} = \infty$<br />$5 + \sum_{n=6}^\infty \frac{n}{1} = \sum_{n=5}^\infty \frac{n}{1}$<br />$5 + \sum_{n=6}^\infty \frac{n}{1} - \infty = \sum_{n=5}^\infty \frac{n}{1} - \infty$<br />$5 = 0$<br />

### Re: Math proof

Posted: Wed Apr 30, 2008 10:38 pm
Oh and no one add it to the page - I want to add it myself tommrow

### Re: Math proof

Posted: Thu May 01, 2008 12:03 pm
Heh, the _'s didn't space properly, I meant .9 to be .9 repeating infinitely, as in
.9999999999999999999999999999999999999999999999999999999999
ect

### Re: Math proof

Posted: Thu May 01, 2008 9:10 pm

### Re: Math proof

Posted: Fri May 02, 2008 12:54 am
Interesting, I haven't ever run into that before, but I don't know that much about real analysis or series.

Was the last one I posted too much?

### Re: Math proof

Posted: Fri May 02, 2008 5:33 am
*EDIT*
just forget it........

### Re: Math proof

Posted: Fri May 02, 2008 12:00 pm
krauser2288, please remove that or I will report you...