make byte plural consistent

1 files changed, 6 insertions, 6 deletions

diff --git a/tex-stuff/math.tex b/tex-stuff/math.tex
index 929d658..8df2064 100644
--- a/tex-stuff/math.tex
+++ b/tex-stuff/math.tex
@@ -112,7 +112,7 @@ the price $p_{b_a}$ bidder $a$ is willing to pay. $\forall j: p_j < p_{j+1}$.
 
 The message has $k$ parts, each consisting of $10$ Points plus an additional $3$
 Points for the last proof. Therefore the message is $10k*32 + 3*32 = 320k + 96$
-bytes large.
+byte large.
 
 \begin{enumerate}
 	\item $\forall j:$ Set $B_{aj}=\begin{cases}G & \mathrm{if}\quad j=b_a\\0 & \mathrm{else}\end{cases}$ and publish $\alpha_{aj}=B_{aj}+r_{aj}Y$ and $\beta_{aj}=r_{aj}G$.
@@ -125,7 +125,7 @@ bytes large.
 \subsubsection{Round 2: Compute outcome}
 
 The message has $nk$ parts, each consisting of $5$ Points. Therefore the message
-is $5nk*32 = 160nk$ bytes large.
+is $5nk*32 = 160nk$ byte large.
 
 $\forall i,j:$ Compute and publish \\[2.0ex]
 $\gamma_{ij}^{\times a} = m_{ij}^{+a}\displaystyle\left(\left(\sum_{h=1}^n\sum_{d=j+1}^k\alpha_{hd}\right)+\left(\sum_{d=1}^{j-1}\alpha_{id}\right)+\left(\sum_{h=1}^{i-1}\alpha_{hj}\right)\right)$ and \\[2.0ex]
@@ -157,7 +157,7 @@ $\delta_j^{\times a} = m_j^{+a}\displaystyle\left(\sum_{h=1}^n\sum_{d=j+1}^k \be
 with a corresponding Proof 2 for $\displaystyle ECDL\left(m_j^{+a}\left(\sum_{h=1}^n\sum_{d=j+1}^k\alpha_{hd}\right)\right) = ECDL\left(m_j^{+a}\left(\sum_{h=1}^n\sum_{d=j+1}^k \beta_{hd}\right)\right)$. \\[2.0ex]
 
 The message has $k$ parts, each consisting of $5$ Points. Therefore the message
-is $5k*32 = 160k$ bytes large. Note that compared to auctions with private
+is $5k*32 = 160k$ byte large. Note that compared to auctions with private
 outcome the message size is reduced by a factor of $n$ because we don't need to
 compute different outcome functions for each bidder when the outcome should be
 public. Therefore we don't need $nk$ blinding factors $m_{ij}^{+a}$ in this
@@ -170,7 +170,7 @@ x_{+a}\left(\sum_{h=1}^n\delta_j^{\times h}\right)$ with a Proof 2 showing
 $ECDL(\varphi_j^{\times a}) = ECDL(Y_{\times a})$ \\[2.0ex]
 
 This message has $k$ parts, each consisting of $4$ Points. Therefore the message
-is $4k*32 = 128k$ bytes large.
+is $4k*32 = 128k$ byte large.
 
 \subsubsection{Epilogue: Outcome determination}
 
@@ -219,12 +219,12 @@ M+1st highest bid.
 The bidders also have to use Proof 2 to show that $\displaystyle ECDL_Y\left(\left(\sum_{j=1}^{k/n}\alpha_{a,jn+a}\right) - G\right) = ECDL_G\left(\sum_{j=1}^{k/n}\beta_{a,jn+a}\right)$. \\[2.0ex]
 This is to ensure bidders have only chosen valid bids for their bid index, since
 in M+1st price auctions the amount of possible prices is multiplied by $n$ to
-prevent ties. This increases the message size by $96$ bytes.
+prevent ties. This increases the message size by $96$ byte.
 
 \subsubsection{Round 2: Compute outcome}
 
 The message has $nk$ parts, each consisting of $5$ Points. Therefore the message
-is $5nk*32 = 160nk$ bytes large.
+is $5nk*32 = 160nk$ byte large.
 
 $\forall i,j:$ Compute and publish \\[2.0ex]
 $\gamma_{ij}^{\times a} = m_{ij}^{+a}\displaystyle\left(\sum_{h=1}^n\left(\sum_{d=j}^k\alpha_{hd}+\sum_{d=j+1}^k\alpha_{hd}\right)+\left(2M+2\right)\sum_{d=1}^{j}\alpha_{id} - \left(2M+1\right)G \right)$ and \\[2.0ex]