Prove that
(a) \[1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{1}{6} n (n+1)(2n+1)\]
(b) \[1^2 + 3^2 + 5^2 + \dots + (2n-1)^2 = \frac{1}{3} n (2n-1)(2n+1)\].
Is “I see what I eat” the same thing as “I eat what I see”?
To make it not so confusing let’s change the wording to make it more “mathematical”
“I see what I eat”=“If I eat it then I see it”
“I eat what I see”= “If I see it then I eat it”
Was the March Hare right? Is “I like what I get” the same thing as “I get what I like”?
Do you remember the example from the previous maths circle?
“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 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 \(a-b + (a+b) = b-a + (a+ b)\). It simplifies to \(2a = 2b\), or \(a=b\). Therefore, All NON-EQUAL NUMBERS ARE EQUAL! (This is gibberish, isn’t it?)”
Do you remember what the mistake was? In fact we have mixed up two things. It is indeed true “if \(x=y\), then \(x^2 = y^2\)”. But is not always true “if \(x^2 = y^2\), then \(x=y\).” For example, consider \(2^2 = (-2)^2\), but \(2 \neq (-2)!\) Therefore, from \((a-b)^2 = (b-a)^2\) we cannot conclude \(a-b = b-a\).
Was the Dormouse right? Is “I breathe when I sleep” the same thing as “I sleep when I breathe”?
Mary Ann yawns every time it rains. In fact Mary Ann is yawning right now. Is it raining at the moment?
The March Hare, the Hatter, and the Dormouse were accused of stealing some tarts. At the trial The March Hare said it was the Hatter who stole the tarts. The Dormouse and the Hatter made statements too, but nobody remembered what they said, and all the records were destroyed by Alice’s tears. As the trial proceeded, it became clear that the tarts were stolen by a person, who in fact was the only one who gave a truthful statement. Who stole the tarts?
Take \(x=-\frac{1}{2}\). It solves \[2x+1=0.\] Add \(x^2\) to both sides of the above equation \[x^2 + 2x +1 = x^2.\] Completing the square, we rewrite the equality as \[(x+1)^2 = x^2.\] Now, taking the square root from the both sides of the equality we get \[x+1=x.\] Subtracting \(x\) from both sides we get \[1=0\] Can you explain what went wrong in our reasoning? Why?
Now! – said the Hatter. – you might just as well say that “I see what I eat” is the same thing as “I don’t eat what I don’t see”!
– You might just as well say, – added the March Hare, –that “I like what I get” is the same thing as “I don’t get what I don’t like”!
– You might just as well say,– added the Dormouse, which seemed to be talking in its sleep, – that “I breathe when I sleep” is the same thing as “I don’t sleep when I don’t breathe”!
– It is the same thing indeed,– said the Hatter, and here the conversation dropped.
Do you agree with them? Why?
We already know that \(x=y\) does not follow from \(x^2=y^2\).
But is it true that if \(x^2 \neq y^2\) then \(x \neq y\)?