Necessary Condition for Existence of BIBD

From ProofWiki
Jump to: navigation, search

Theorem

Let there exist be a BIBD with parameters $v, b, r, k, \lambda$.

Then the following are true:

$(1): \quad b k = r v$
$(2): \quad \lambda \left({v-1}\right) = r \left({k-1}\right)$
$(3): \quad \displaystyle b \binom k 2 = \lambda \binom v 2$
$(4): \quad k < v$
$(5): \quad r > \lambda$

All of $v, b, r, k, \lambda$ are integers.


Some sources prefer to report $(3)$ as:

$\displaystyle b = \frac { {v \choose 2} }{ {k \choose 2} }\lambda$

which is less appealing visually, and typographically horrendous.


Proof

$(1)$: We have by definition that each point is in exactly $r$ blocks, and each block is of size $k$.

We have that $b k$ is the number of blocks times the size of each block.

We also have that $r v$ is the number of points times the number of blocks each point is in.

The two must clearly be equal.


$(2)$: Comparing the left and right hand sides of the equation we can see that:

LHS: An arbitrary point must be paired with $v-1$ other points.

If $\lambda>1$ then every point is paired $\lambda \left({v-1}\right)$ times.


RHS: An arbitrary point is paired with $k-1$ other points for each of the $r$ blocks it is in.

Therefore it is paired $r \left({k-1}\right)$ times.

Both values give the number of times an arbitrary point is paired, therefore LHS = RHS.


$(3)$: From equation $(1)$, we have that $r = \dfrac {bk} v$

From $(2)$ we have that $r = \dfrac {v-1} {k-1} \lambda$.

So:

\(\displaystyle \frac {bk} v\) \(=\) \(\displaystyle \frac {v-1} {k-1} \lambda\) $\quad$ substituting for $r$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle b k \left({k-1}\right)\) \(=\) \(\displaystyle \lambda v \left({v-1}\right)\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle b \binom k 2\) \(=\) \(\displaystyle \lambda \binom v 2\) $\quad$ from Binomial Coefficient with Two $\quad$

$\blacksquare$