Let \(n\) be some positive number. It is obvious that \[2n-1<2n.\] Take another positive number \(a\), and multiply both sides of the inequality by \((-a)\) \[-2na +a< -2na.\] Now, subtracting \((-2na)\) from both sides of the inequality we get \(a<0\).
Thus, ALL POSITIVE NUMBERS ARE NEGATIVE!
Suppose \(a \neq b\). We can write \[-a = b - (a+b)\] and \[-b = a - (a+b)\] Since \((-a)b = a(-b)\), then \[( b - (a+b))b = a(a - (a+b))\] Removing the brackets, we have \[b^2 - (a+b)b = a^2 - a(a+b)\] Adding \(\left(\frac{a+b}{2}\right)^2\) to each member of the equality we may complete the square of the differences of two numbers \[\left(b - \frac{a+b}{2}\right)^2 = \left(a - \frac{a+b}{2}\right)^2\] From the equality of the squares we conclude the equality of the bases \[b - \frac{a+b}{2} = a - \frac{a+b}{2}.\] Adding \(\frac{a+b}{2}\) to both sides of equality we get \(a=b\). Therefore, WE HAVE SHOWED THAT FROM \(a\neq b\) IT FOLLOWS \(a=b\).
Let \(x\) be equal to 1. Then we can write \(x^2=1\), or putting it differently \(x^2 -1 =0\). By using the difference of two squares formula we get \[(x+1)(x-1)=0\] Dividing both sides of the equality by \(x-1\) we obtain \[x+1=0,\] in other words \(x=-1\). But earlier we assumed \(x=1\). THUS \[-1=1\ !\]
In every right-angled triangle the arm is greater than the hypotenuse. Consider a triangle \(ABC\) with right angle at \(C\).
The difference of the squares of the hypothenuse and one of the arms is \(AB^2 -BC^2\). This expression can be represented in the form of a product \[AB^2 -BC^2 = (AB - BC)(AB+BC)\] or \[AB^2 -BC^2 = -(BC - AB)(AB+BC)\] Dividing the right hand sides by the product \(-(AB-BC)(AB+BC)\), we obtain the proportion \[\frac{AB+BC}{-(AB+BC)} = \frac{BC-AB}{AB-BC}.\] Since the positive quantity is greater than the negative one we have \(AB+BC>-(AB+BC)\). But then also \(BC-AB>AB-BC\), and therefore \(2BC>2AB\), or \(BC>AB\), i.e. THE ARM IS GREATER THAN THE HYPOTENUSE!
This academic year Harry decided not only to attend Maths Circles, but also to join his local Chess Club. Harry’s chess set was very old and some pieces were missing so he ordered a new one. When it arrived, he found out to his surprise that the set consisted of 32 knights of different colours. He was a bit upset but he decided to spend some time on solving the problem he heard on the last Saturday’s Maths Circle session. The task was to find out if it is possible to put more than 30 knights on a chessboard in such a way that they do not attack each other. Do you think it is possible or not?
After listening to Harry’s complaints the delivery service promised him to deliver a very expensive chess set together with some books on chess strategies and puzzles. This week one of the tasks was to put 14 bishops on a chessboard so that they do not attack each other. Harry solved this problem and smiled hoping he is not getting 32 identical bishops this time. Can you solve it?
On the way back from his weekly maths circle Harry created the following puzzle:
Put 48 rooks on a \(10\times10\) board so that each rook attacks only 2 or 4 empty cells.
When he showed this problem to the teachers next Saturday they were very impressed and decided to include it in the next problem set. Try to find a suitable placement of rooks.
A boy is playing on a \(4\times10\) board. He is trying to put 8 bishops on the board so that each cell is attacked by one of the bishops. Finally he manages to solve this problem.
(a) Can you show a possible solution?
(b) Can you do the same thing with 7 bishops?
More problems about chessboard and chess pieces:
(a) Can it be true that there are only 8 knights on a \(4\times12\) board and each empty cell is attacked by at least one of the knights?
(b) Put some number of knights on a chessboard in such a way that each knight attacks exactly three other knights.
Draw 16 diagonals inside some cells of a \(5\times5\) square in such a way that no two of these diagonals share any points.