Today is thanksgiving day in our jurisdiction; and, being that it falls on the th — a number that is “HTMLy” (read as extremely) significant to us — we’ve chosen to mark the occasion with math. While this choice may seem unconventional, it doesn’t rule out the fact that our genuine gratitude to The Generous Giver is unconditional.
Now to the math itself. Consider the prime factorizations of and :
and an enumeration of the first four prime numbers:
Clearly, we have that ; this was what motivated the theorems in this post. By the way, our use of the word theorem is a deliberate exaggeration, but since we’re not dealing with advanced math, we hope that we’ll be offered exoneration, despite our overt exaggeration.
Let be an arithmetic sequence. Then .
If are as given, we can suppose that the common difference of the arithmetic sequence is , and then write . Consider :
Since the difference , we conclude that .
Let be integers that form an arithmetic sequence. Then can never be a perfect square.
If are integers that form an arithmetic sequence, then the common difference must be an integer as well (note that we exclude the case when ). From our previous calculation, we saw that . Since is not a perfect square when is an integer (except for ), so is .
Let be consecutive, positive even numbers. Then .
This should have been better described as a corollary (to the first “theorem”), but we don’t want to introduce another jargon here. We’ll ask you to prove it in the exercises at the end; for now, let’s see a few instances:
- Consider the first four positive even numbers: . We have that ;
- Consider the numbers . We have that ;
- Consider the numbers . We have that ;
- And so on. You feel that?
Let be consecutive, positive odd numbers. Then .
Actually, the “positive” requirement in the two preceding theorems can be dropped, what’s essential is that none of the numbers should be zero. Few examples below:
- Consider the first four positive odd numbers: . We have that ;
- Consider the numbers . We have ;
- Consider the numbers . We have .
Let be consecutive integers, none of which is zero. Then, .
Consecutive integers form an arithmetic sequence with a common difference of . So, in view of our first theorem, we have that .
The next result is well-known, and is in a class of its own:
Let be four consecutive terms of a Fibonacci sequence. Then .
Usually, the terms in a Fibonacci sequence are denoted by ; we’ve used for convenience. Also, in order to keep things simple, we’ve omitted the proof of the above result, but we invite our readers to try it out — this can be accomplished, for example, by using the fact that the th term is given by
and then expanding and simplifying .
Let be a geometric sequence. Then .
Let be as given. Suppose that the common ratio of the geometric sequence is , then we can write . Now:
which is what we want.
Linear systems with special coefficients
Do our “theorems” have any where we can apply them? Well, somewhat — think linear systems. If we have a linear system in two variables and , like so:
then the solution can be given by (Cramer’s rule ):
As can be seen from the above explicit formula, the quantity is akin to what we’ve been playing around with. That makes us to ask: “What if the coefficients in a linear system are so arranged that they’re consecutive terms of any of the number patterns we’ve considered in this post”? This is certainly an obscure possibility, but we’ve included it out of curiosity.
Let be consecutive integers, none of which is zero. Then the linear system
has integer solutions, if and are even numbers.
The proof relies on the fact that the quantity (called the determinant) appears in the denominator when the above linear system is solved. Recall that the solution is:
Since are consecutive non-zero integers, we know, from Theorem 5, that , or . Thus the solution may be re-written as:
Since both and are even numbers, we know that the numerators above, namely and , are both even. Therefore, the expressions for and are divisible by , resulting in integer quotients.
has integer solutions if and are integers.
Notice that we have if are consecutive terms of a Fibonacci sequence (Theorem 6 above), and so the general solution
simplifies to , which are integers.
has no solution so long as (and and are not both zero).
A proof of our present theorem relies on the fact that when form a geometric sequence (Theorem 7), together with the fact that the solution to the linear system is:
Since division by zero is not allowed, we’ll be unable to determine and from the above expressions, thus no solution. But wait. How about the assumption that and are not both zero? If they are both zero, then we’ll have a trivial solution: . What of the assumption that ? Suppose that . Then, we can divide equation (8) by , obtaining
However, since form a geometric sequence with common ratio , it follows that and . This would then mean that equation (9) is the same as:
which is exactly equation (7). This is saying that if , then equations (7) and (8) are the same, meaning that there’s an infinite number of solutions to the original linear system. So we want for our theorem to hold.
If four numbers are arranged in ascending order, then the quantity always takes on constant values, under certain circumstances. In the case of even/odd numbers, it is ; in the case of Fibonacci numbers, it is ; in the case of geometric sequences, it is .
- Construct an example to show that does not always imply that .
- Without requiring that form a geometric sequence, prove that it is still possible to have , where are non-zero integers that are not all equal and have been sorted in ascending order.
- Let be numbers. Under what conditions, other than the numbers forming an arithmetic sequence, can we always have ?
- Let be consecutive positive even numbers. Prove that .
- Let be consecutive positive odd numbers. Prove that .
In the spirit of today’s occasion, the poster finds it appropriate to appreciate God for a certain Thursday, June 14, 2018.
Thursday, June 14, 2018. Beautiful, beautiful day. Thursday, June 14, 2018. Beautiful, beautiful day. The poster can’t refrain from singing this refrain again and again. He’s got chance today; he just can’t refrain; he just chants that refrain.
No doubt we should keep our personal things to ourselves, but the poster couldn’t help it this time, especially considering the coincidence with today’s date, today’s event, and “that Thursday”. The magnitude of what happened makes the poster to offer tribute and gratitude to The One whose gracious attribute makes Him to distribute good things generously to all men.
So, what happened exactly? Refer to a strange word the poster used at the beginning of this post, namely “HTMLy”. Read it as “each-team-ly” (imagine how a 2 year old might pronounce it ). “HTMLy” was coined to capture two words: HTML and extremely. “That Thursday” is extremely significant to the poster because that was the day the poster was introduced to you guessed it HTML. That singular and spectacular Thursday transformed the poster’s way of thinking, the poster’s way of teaching, and — of course — the poster’s way of “thanking”.
Did you get the point the poster is trying to make? Make it a point of duty to be grateful, even for things that are little.