Take any two non-equal numbers \(a\) and \(b\), then we can write \[a^2 - 2ab + b^2 = b^2 - 2ab + a^2\] Using the formula \((x-y)^2 = x^2 - 2xy + y^2\), we complete the squares and rewrite the equality above as \[(a-b)^2 = (b-a)^2.\] As we take a square root from the both sides of the equality, we get \[a-b = b-a.\] Finally, adding to both sides \(a+b\) we get \[\begin{aligned} a-b + (a+b) &= b-a + (a+ b)\\ 2a&= 2b\\ a&=b. \end{aligned}\] Therefore, All NON-EQUAL NUMBERS ARE EQUAL! (This is gibberish, isn’t it?)
Consider equation \[x-a=0\] Dividing both sides of this equation by \(x-a\), we get \[\frac{x-a}{x-a} = \frac{0}{x-a}.\] But \(\frac{x-a}{x-a}=1\) and \(\frac{0}{x-a}=0\). Therefore, we get \[1=0.\]
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!
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?