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 \(1=0\).
Let’s prove that \(1=2\). Take a number \(a\) and suppose \(b=a\). After multiplying both sides we have \(a^2=ab\). Subtract \(b^2\) from both sides to get \(a^2-b^2=ab-b^2\). The left hand side is a difference of two squares so \((a-b)(a+b)=b(a-b)\). We can cancel out \(a-b\) and obtain that \(a+b=b\). But remember from the start that \(a=b\), so substituting \(a\) for \(b\) we see that \(2b=b\), dividing by \(b\) we see that \(2=1\).
John drunk a \(\frac16\) of a full cup of black coffee and then filled the cup back up with milk. Then he drunk a third of what he had in the cup. Then, he refilled it back to full with milk again, and after that, drunk a half of the cup. Finally, he once again refilled the cup with milk and drunk everything he had. What did he drink more of - coffee or milk?
A round necklace contains \(45\) beads of two different colours: red and blue. Show that it is possible to find two beads of the same colour next to each other.
A broken calculator can only do several operations: multiply by \(2\), divide by \(2\), multiply by \(3\), divide by \(3\), multiply by \(5\), and divide by \(5\). Using this calculator any number of times, could you start with the number \(12\) and end up with \(49\)?
The numbers \(1\) through \(12\) are written on a board. You can erase any two of these numbers (call them \(a\) and \(b\)) and replace them with the number \(a+b-1\). Notice that in doing so, you remove one number from the total, so after \(11\) such operations, there will be just one number left. What could this number be?
If a magician puts \(1\) dove into his hat, he pulls out \(2\) rabbits and \(2\) flowers from it. If the magician puts \(1\) rabbit in, he pulls out \(2\) flowers and \(2\) doves. If he puts \(1\) flower in, he pulls out \(1\) rabbit and \(3\) doves. The magician starts with \(1\) rabbit. Could he end up with the same number of rabbits, doves, and flowers after performing his hat trick several times?