The numbers \(x\) and \(y\) satisfy \(x+3 = y+5\). Prove that \(x>y\).
The numbers \(x\) and \(y\) satisfy \(x+7 \geq y+8\). Prove that \(x>y\).
Prove that there are infinitely many natural numbers \(\{1,2,3,4,...\}\).
Prove that there are infinitely many prime numbers \(\{2,3,5,7,11,13...\}\).
Is it possible to colour the cells of a \(3\times 3\) board red and yellow such that there are the same number of red cells and yellow cells?
There are \(24\) children in the class and some of them are friends with each other. The following rules apply:
If someone (say Alice) is a friend with someone else (say Bob), then the second student (Bob) is also a friend with the first (Alice).
If Alice is friend with Bob and Bob is friend with Claire, then Alice is also friend with Claire.
Find a misconception in the following statement: under the above conditions Alice is friend with herself.
Let’s prove that \(1\) is the
largest natural number.
Let \(n\) be the largest natural
number. Then, \(n^2\), being a natural
number, is less than or equal to \(n\).
Therefore \(n^2-n=n(n-1)\leq 0\).
Hence, \(0\leq n\leq 1\). Therefore
\(n=1\).
Theorem: If we mark \(n\) points on
a circle and connect each point to every other point by a straight line,
the lines divide the interior of the circle is into is \(2n-1\) regions.
"Proof": First, let’s have a look at the smallest natural numbers.
When \(n=1\) there is one region (the whole disc).
When \(n=2\) there are two regions (two half-discs).
When \(n=3\) there are \(4\) regions (three lune-like regions and one triangle in the middle).
When \(n=4\) there are \(8\) regions, and if you’re still not convinced then try \(n=5\) and you’ll find \(16\) regions if you count carefully.
Our proof in general will be by induction on \(n\). Assuming the theorem is true for \(n\) points, consider a circle with \(n+1\) points on it. Connecting \(n\) of them together in pairs produces \(2n-1\) regions in the disc, and then connecting the remaining point to all the others will divide the previous regions into two parts, thereby giving us \(2\times (2n-1)=2n\) regions.
Let’s "prove" that the number \(1\) is a multiple of \(3\). We will use the symbol \(\equiv\) to denote "congruent modulo \(3\)". Thus, what we need to prove is that \(1\equiv 0\) modulo \(3\). Let’s see: \(1\equiv 4\) modulo \(3\) means that \(2^1\equiv 2^4\) modulo \(3\), thus \(2\equiv 16\) modulo \(3\), however \(16\) gives the remainder \(1\) after division by \(3\), thus we get \(2\equiv 1\) modulo \(3\), next \(2-1\equiv 1-1\) modulo \(3\), and thus \(1\equiv 0\) modulo \(3\). Which means that \(1\) is divisible by \(3\).
Recall that \((n+1)^2=n^2+2n+1\) and
after expansion we get \((n+1)^2-(2n+1)=n^2\). Subtract \(n(2n+1)\) from both sides \((n+1)^2-(2n+1)-n(2n+1)=n^2-n(2n+1)\) and
rewrite it as \((n+1)^2-(n+1)(2n+1)=n^2-n(2n+1)\).
Now we add \(\frac{(2n+1)^2}{4}\) to
both sides: \((n+1)^2-(n+1)(2n+1)+\frac{(2n+1)^2}{4}=n^2-n(2n+1)+\frac{(2n+1)^2}{4}\).
Factor both sides into square: \(((n+1)-\frac{2n+1}{2})^2=(n-\frac{2n+1}{2})^2\).
Now take the square root: \((n+1)-\frac{2n+1}{2}=n-\frac{2n+1}{2}\).
Add \(\frac{2n+1}{2}\) to both sides
and we get \(n+1=n\) which is
equivalent to \(1=0\).