From fe58d96e50e33b05f2a45f1493eca39ec9b3d030 Mon Sep 17 00:00:00 2001
From: Maarten <mzaanen@users.noreply.github.com>
Date: Sat, 5 Sep 2015 12:35:58 +0200
Subject: [PATCH] Fixed #25355 -- Made two tweaks to
 docs/topics/db/aggregation.txt.

---
 docs/topics/db/aggregation.txt | 4 ++--
 1 file changed, 2 insertions(+), 2 deletions(-)

diff --git a/docs/topics/db/aggregation.txt b/docs/topics/db/aggregation.txt
index 409f58aa4c..65a1eb7ee6 100644
--- a/docs/topics/db/aggregation.txt
+++ b/docs/topics/db/aggregation.txt
@@ -73,7 +73,7 @@ In a hurry? Here's how to do common aggregate queries, assuming the models above
     {'price_per_page': 0.4470664529184653}
 
     # All the following queries involve traversing the Book<->Publisher
-    # many-to-many relationship backward
+    # foreign key relationship backwards.
 
     # Each publisher, each with a count of books as a "num_books" attribute.
     >>> from django.db.models import Count
@@ -242,7 +242,7 @@ price field of the book model to produce a minimum and maximum value.
 
 The same rules apply to the ``aggregate()`` clause. If you wanted to
 know the lowest and highest price of any book that is available for sale
-in a store, you could use the aggregate::
+in any of the stores, you could use the aggregate::
 
     >>> Store.objects.aggregate(min_price=Min('books__price'), max_price=Max('books__price'))