# Algebra/Exponents

• March 9th 2010, 06:16 PM
Alan306090
Algebra/Exponents
If (a + b)^1/2 = (a - b )^-1/2 why must a^2 - b^2 = 1 must be true?

If Set X has x members and yet Y has y members, Set Z consists of all members that are in either set X or set Y with the exception of the k common members (k > 0), then why must x + y - 2k must equal the number of members in set Z?
• March 9th 2010, 06:23 PM
Prove It
Quote:

Originally Posted by Alan306090
If (a + b)^1/2 = (a - b )^-1/2 why must a^2 - b^2 = 1 must be true?

If Set X has x members and yet Y has y members, Set Z consists of all members that are in either set X or set Y with the exception of the k common members (k > 0), then why must x + y - 2k must equal the number of members in set Z?

$(a + b)^{\frac{1}{2}} = (a - b)^{-\frac{1}{2}}$

$(a + b)^{\frac{1}{2}} = \frac{1}{(a - b)^{\frac{1}{2}}}$

$(a + b)^{\frac{1}{2}}(a - b)^{\frac{1}{2}} = 1$

$[(a + b)(a - b)]^{\frac{1}{2}} = 1$

$(a^2 - b^2)^{\frac{1}{2}} = 1$

$a^2 - b^2 = 1^2$

$a^2 - b^2 = 1$.
• March 9th 2010, 06:25 PM
Prove It
Quote:

Originally Posted by Alan306090
If (a + b)^1/2 = (a - b )^-1/2 why must a^2 - b^2 = 1 must be true?

If Set X has x members and yet Y has y members, Set Z consists of all members that are in either set X or set Y with the exception of the k common members (k > 0), then why must x + y - 2k must equal the number of members in set Z?

You should know that

$n(X \cup Y) = n(X) + n(Y) - n(X \cap Y)$.

So $n(X \cup Y) = x + y - k$.

But you want to exclude the ones that are already in both.

So $n(Z) = n(X \cup Y) - n(X \cap Y)$

$= x + y - k - k$

$= x + y - 2k$.
• March 9th 2010, 06:28 PM
Alan306090
Quote:

Originally Posted by Prove It
You should know that

$n(X \cup Y) = n(X) + n(Y) - n(X \cap Y)$.

So $n(X \cup Y) = x + y - k$.

But you want to exclude the ones that are already in both.

So $n(Z) = n(X \cup Y) - n(X \cap Y)$

$= x + y - k - k$

$= x + y - 2k$.

Can you elaborate a bit more please?
• March 9th 2010, 06:30 PM
Prove It
Quote:

Originally Posted by Prove It
You should know that

$n(X \cup Y) = n(X) + n(Y) - n(X \cap Y)$.

This means that the number of elements in the union (in other words, the number of elements in X or Y or both) has to be the same as the number in X + the number in Y - the number in both (because you've counted those twice).

So $n(X \cup Y) = x + y - k$.

But you want to exclude the ones that are already in both.

So you're getting rid of the ones in the intersection.

So $n(Z) = n(X \cup Y) - n(X \cap Y)$

$= x + y - k - k$

$= x + y - 2k$.

...