Gaussian Binomial Coefficient of 1

Theorem

$\dbinom 1 m_q = \delta_{0 m} + \delta_{1 m}$

That is:

$\dbinom 1 m_q = \begin{cases} 1 & : m = 0 \text { or } m = 1 \\ 0 & : \text{otherwise} \end{cases}$

where $\dbinom 1 m_q$ denotes a Gaussian binomial coefficient.

Proof

By definition of Gaussian binomial coefficient:

$\dbinom 1 m_q = \displaystyle \prod_{k \mathop = 0}^{m - 1} \dfrac {1 - q^{1 - k} } {1 - q^{m + 1} }$

When $m = 0$ the product on the right hand side is vacuous, and so:

$\dbinom 1 0_q = 0$

Let $m > 0$.

Then:

 $\displaystyle \dbinom 1 m_q$ $=$ $\displaystyle \prod_{k \mathop = 0}^{m - 1} \dfrac {1 - q^{1 - k} } {1 - q^{m + 1} }$ Definition of Gaussian Binomial Coefficient $\displaystyle$ $=$ $\displaystyle \left({\dfrac {1 - q^{1 - 0} } {1 - q^{0 + 1} } }\right) \left({\dfrac {1 - q^{1 - 1} } {1 - q^{1 + 1} } }\right) \left({\dfrac {1 - q^{1 - 2} } {1 - q^{2 + 1} } }\right) \cdots \left({\dfrac {1 - q^{1 - \left({m - 1}\right)} } {1 - q^{\left({m - 1}\right) + 1} } }\right)$ $\displaystyle$ $=$ $\displaystyle \left({\dfrac {1 - q} {1 - q} }\right) \left({\dfrac {1 - q^0} {1 - q^2} }\right) \left({\dfrac {1 - q^{-1} } {1 - q^3} }\right) \cdots \left({\dfrac {1 - q^{2 - m} } {1 - q^m} }\right)$

When $m > 0$ there exists a factor $1 - q^0 = 0$ in the numerator of the right hand side.

Hence when $m > 0$ we have that $\dbinom 1 m_q = 0$.

We are left with:

 $\displaystyle \dbinom 1 1_q$ $=$ $\displaystyle \prod_{k \mathop = 0}^0 \dfrac {1 - q^{1 - k} } {1 - q^{m + 1} }$ Definition of Gaussian Binomial Coefficient $\displaystyle$ $=$ $\displaystyle \dfrac {1 - q^{1 - 0} } {1 - q^{0 + 1} }$ $\displaystyle$ $=$ $\displaystyle \dfrac {1 - q} {1 - q}$ $\displaystyle$ $=$ $\displaystyle 1$

$\blacksquare$