# Converse of a proof

Let $$\pi$$ be the partition of $$n=a_1+a_2+...+a_r$$, where $$a_1\geq a_2 \geq ....\geq a_r\gt0$$ and $$\pi$$’ be the partition, $$n=b_1+b_2+....+b_s”.$$Prove that $$\pi$$’ is the conjugate of $$\pi$$ if and only if $$r=b_1$$ and $$s=a_1$$ and $$b_j$$ is the number of parts in the partition of $$\pi$$ that are $$\geq j$$ for $$j=1,2,....s$$

What i did so far:

If we consider the ferrers diagram of the partition $$n=a_1+a_2+...+a_r$$ then $$a_r$$ denotes the number of rows of the representation and r denotes the number of columns in the diagram. Similarly for the other representation $$b_s$$ denotes the number of rows of the representation while s denotes the number of columns of the representation.

Now if the conjugate of $$n=b_1+b_2+...b_s$$ were to be $$n=a_1+a_2+...+a_r$$ then the largest of all $$b_i$$’s i.e $$b_1$$ will be the number of columns, i.e $$b_1=r$$. Similarly the number of terms i.e s will be the number of rows, i.e $$s=a_1$$.

So, can anyone help me in proceeding towards the converse part ?, also is $$s=a_1$$ and $$r=b_1$$ enough to state that $$\pi$$ is the conjugate of $$\pi$$?