### 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

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