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• Freshmanʻs Dream | Proof

Freshmanʻs Dream | Proof

This article is all about freshmanʻs dream in the modern day. I will be mentioning it in the sense of the prime characteristic not regular binomial expansion. So, the setting we will be discussing is over the defined ring $R$. We define the commutative ring to be $R \equiv (S, +, \cdot)$ over a set of the real numbers, $S$. Further, we define the ring to be of characteristic $p$ so any multiple of $p$ evaluates to $0$. Thereʻs really no restriction so if we really want, $S$ could be $\mathbb{R}$, the entire set of reals. So, now for the conditions. We have the number $p$ which is always a prime number and $x,y \in R$ which means that $x$ and $y$ are inside the ring $R$. We are interested in the expression

in this setting of $(1)$. This isnʻt the same as regular algebra, really this is abstract algebra. Binomial expansion of $(1)$ gives the $n$th coefficient given as…

Now, each coefficient in the expansion is a multiple of $p$, so they will all evaluate to $0$ for some $n$. If we only look at the terms for $n$ when $n \in \lbrace1,…,p-1\rbrace$ we notice that these donʻt evaluate to $1/p$, the inverse of $p$. This is because we are choosing items from a set of $p$. They can never be fractional. These terms when $n \in \lbrace1,…,p-1\rbrace$ would all be $0$. The resulting binomial terms $x^{\alpha}y^{p-\alpha}$ would all be finite numbers but when multiplied with the coefficient in $(2)$ we get $0$. Hence, the only terms we would be left with would be when $n = 0$ and $n = p$ since they evaluate to 1 by direct inspection. In other words…

The terms that match these coefficients are $x^p$ and $y^p$ which leaves us with…

Which is what we sought to show. $\blacksquare$