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!
Having had experience with some faulty proofs above, can you now answer the following questions
(a) From the equality \((a-b)^2=(m-n)^2\) may one draw the conclusion that \(a-b=m-n\)?
(b) For what value of \(x\) do the following expressions lose their meaning?
(1) \(\frac{x^3-1}{x-1}\), (2) \(\frac{1}{x^2-1}\), (3) \(\frac{x+1}{1-2^x}\).
(c) If \(a>b\), can one conclude that \(ac>bc\) for any number \(c\)?
We prove by mathematical induction that all horses in the world are of the same colour.
Base case: There is a single horse. It has some coat colour. Because there are no other horses, all the horses have the same coat colour.
Induction step: We have \(n\) horses. We assume all of them have the same coat colour. Now we add an additional \((n+1)\)st horse. We don’t know what colour it has, but if we for now get rid of one horse from the group we had before, we suddenly have a group of \(n\) horses which includes the new one. Since we have our claim proven for \(n\), all of these horses have the same coat colour and therefore the new horse has the same coat colour as all the other ones. So every group of \(n+1\) horses has the same colour.
The third step: due to mathematical induction rule, all the horses in the world have the same coat colour. THUS WE HAVE PROVED THAT ALL HORSES IN THE WORLD ARE OF THE SAME COLOUR!
There are six natural numbers, all different, which sum up to 22. Can you find those numbers? Are they unique, or is there another bunch of such numbers?
In how many ways can you rearrange the numbers 1, 2, ..., 100 so the neighbouring numbers differ by not more than 1?
There are one hundred natural numbers, they are all different, and sum up to 5050. Can you find those numbers? Are they unique, or is there another bunch of such numbers?
The Hatter made 44 hats. Can he put his hats into 9 piles in such a way that the number of hats in each pile is different?
Alice finally decided to do some arithmetic. She took four different integer numbers, calculated their pairwise sums and products, and the results ( the pairwise sums and products) wrote down in her wonderful book. What could be the smallest number of different numbers Alice wrote in her book?
Alice wants to write down the numbers from 1 to 16 in such a way that the sum of two neighbouring numbers will be a square number. The Hatter tells Alice that he can write down the numbers with this property in a line, but he believes that it is absolutely impossible to write the numbers with this property in a circle. Show that he is right.