How do we know they are all triangular numbers?
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Module 3 Week 4 Day 13 Challenge Part 1 Explanation
How do we know that each case is actually a triangular number for every single case and that it's not a coincidence?
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@tidyboar This is one way to see it (the way I see it), but not the only way:
We're making cases based on a "ceiling" that we choose for our numbers. There is actually both a "ceiling" and a "floor." The ceiling is fixed for each case, and is equal to the value of \(c.\) For each fixed value of \(c,\) the floor can be raised up and up until it is the same height as the ceiling. For each level of the floor, we have some corresponding values of \((a, b).\) But as the floor is raised, we get fewer and fewer possible values. In fact, raising the floor by \(1\) will decrease the number of values of \((a, b)\) by one. This is how we get the "triangle."
Do you see that in the bottom triangle, the value of \(a\) is the floor? In the first row, the "floor" is equal to \(0,\) so there are many possibilities (3 of them). In the second row, the "floor" is raised up to \(1,\) so there are fewer possible values for \((a,b)\) that go up to our ceiling of \(c = 2.\) In the third row, the "floor" is equal to the "ceiling," so there is only one possible value: that value where both \(a\) and \(b\) are equal to the value of the floor/ceiling.
Sorry if that sounded confusing, and please let me know if you would like more explanation on this topic!
Thanks for asking and being curious! -
sorry i don't see why they're triangular numbers
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@tidyboar Triangular numbers are the sum of successive consecutive numbers starting from \(1,\) like \(\textcolor{red}{1, 3, 6,} \textcolor{red}{10}, \textcolor{red}{15} \text{ and } \textcolor{red}{21}.\)
$$\begin{aligned} 1 &= \textcolor{red}{1} \\ 1 + 2 &= \textcolor{red}{3} \\ 1 + 2 + 3 &= \textcolor{red}{6} \\ 1 + 2 + 3 + 4 &= \textcolor{red}{10} \\ 1 + 2 + 3 + 4 + 5 &= \textcolor{red}{15} \\ 1 + 2 + 3 + 4 + 5 + 6 &= \textcolor{red}{21} \\ \end{aligned} $$They are called triangular numbers because if you were to create a triangle out of rows of dots, where each successive row had one more dot than the previous row, then the arrangement would look like a triangle!
$$\begin{aligned} \bullet &= \textcolor{blue}{1} \\ \bullet \text{ } \text{ }\text{ }\bullet &= \textcolor{blue}{2} \\ \bullet \text{ }\text{ }\text{ } \bullet \text{ }\text{ }\text{ } \bullet &= \textcolor{blue}{3} \\ \bullet \text{ }\text{ }\text{ } \bullet \text{ }\text{ }\text{ } \bullet \text{ }\text{ }\text{ } \bullet &= \textcolor{blue}{4} \\ \bullet \text{ }\text{ }\text{ } \bullet \text{ }\text{ }\text{ } \bullet \text{ }\text{ }\text{ } \bullet \text{ }\text{ }\text{ } \bullet &= \textcolor{blue}{5} \\ \bullet \text{ }\text{ }\text{ } \bullet \text{ }\text{ }\text{ } \bullet \text{ }\text{ }\text{ } \bullet \text{ }\text{ }\text{ } \bullet \text{ }\text{ }\text{ } \bullet &= \textcolor{blue}{6} \\ \\ \text{ total } = \textcolor{blue}{1 + 2 + 3 + 4 + 5 + 6 =} \textcolor{red}{ 21 } \\ \end{aligned} $$The above illustrates the sixth triangular number, which is \(\textcolor{red}{21}.\)
In Module 4 Day 5, Prof. Loh goes over how to add up the sum of numbers in such a sequence, where each term increases by the same amount. Such a sequence is called an arithmetic sequence.
Going back to the video, we see that there were \(3\) ways to assign values to \(a\) and \(b\) for Case 2, where \(c = 1.\) Oh, \(\textcolor{red}{3}\) is a triangular number, since \( 1 + 2 = \textcolor{red}{3}!\)
There were \(6\) ways to assign values to \(a\) and \(b\) for Case 3, where \(c = 2.\) Wow, \(\textcolor{red}{6}\) is also a triangular number since it equals \( 1 + 2 + 3 = \textcolor{red}{6}!\)
Why does the number of entries take the form \(1 + 2 + 3 + \ldots \) anyway...?
As I mentioned before, you can imagine that the value of \(b\) for the \(c = 2\) case must be less than or equal to \(2.\) The number \(2\) is like a "ceiling." Neither \(a\) nor \(b\) can exceed the value of this ceiling. They can be equal to it, but not greater than it. Likewise, the value of \(a\) is like a "floor," because it determines all the values that are in the middle that are possible. If the value of \(a\) is smaller, we have more possible values of \(b\) that work. The result of this is that we get the pattern \(1 + 2 + 3 + \dots \).
Please let me know if this is still unclear, and I'll be happy to try to elucidate upon it!
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Thank you I understand now!
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