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Algebra (all content)
Course: Algebra (all content) > Unit 18
Lesson 9: Deductive and inductive reasoningUsing inductive reasoning (example 2)
Sal uses inductive reasoning in order to find the 50th element of a pattern toothpick shapes. Created by Sal Khan.
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It has to be a personal mental "block" but the only way i can intuitively reach the solution of this problem is
6n - (n-1). There's something wrong?(18 votes)
There's nothing wrong. You still get the same answer with your thinking.(2 votes)
What if n is negative? Would I be correct to say that the equation could work, but you just couldn't express it using toothpick diagrams?(4 votes)- The equation is intended to represent the pattern that is found in this real-life problem. One of the major keys to understand inductive reasoning is to know its boundaries. In this case, we start with the basic house shape and keep adding additions to it, so the formula only works for n=1. After this point, Sal found a way to make sense of the case where n=0, so the single "house wall" toothpick becomes the base case.
One of the fun things that happens in mathematics is that someone starts with a pattern like the one above for a restricted set (n>=1), and then figures out a way to extend the pattern into an area that was not originally envisioned. For example, the definition of the factorial where n! = n * (n-1) * ... * 1 started with only n>=1, since the terms were counting down to 1 and so would not make sense starting below 1. When we got to combinations and permutations, however, we saw another pattern where nPr = n!/(n-r)! and nCr = n!/[r!(n-r)!]. For this pattern, we needed to extend the definition of a factorial to include 0! = 1 so that nPn = n! and nCn = 1.
My point is that extending the equation into the realm of n<0 would require a different way of looking at the pattern that was not just adding a toothpick extension to a toothpick house.(7 votes)
- I don't understand how this is inductive reasoning. The question said it was a sequence, so it must have a pattern, and this is the only pattern that works. Didn't Sal deduct using logic the pattern rule, then apply it to the 50th element?(6 votes)
I understand your thought process. It's because he didn't logically reason/prove all the way to the 50th figure, but it is 10000% going to turn out to be 251 toothpicks. It's just not seen.
It's just like how you know...Notebook A is made of paper, and since it's made of paper, Notebook B that's just a different color but has the same feel and look is made of paper too. You can prove it but you haven't. There's better examples of this concept.(3 votes)
- im confused with this question: mike must belong to the bartenders and beverage union local 165, since almost every Los Vegas bartender does. thank you(3 votes)
- @patrickb20:
Short answer: No.
Long answer: This isn't covered in the video or the videos before it but I can try to explain it.
Assuming Mike is a Los Vegas bartender. If it's "almost every Los Vegas bartender", that doesn't mean that he's part of that group. He could be that group of Los Vegas bartenders that aren't a part of the Bartenders and Beverage Union Local 165..if I'm seeing it correctly.(2 votes)
- in the second example of inductive reasoning, is that characteristic of deductive reasoning being a principle part of inductive reasoning? i'm confused on the mechanics of inductive reasoning... i understand the purpose of it, but not the act of it... it just seems like when you've run out of road but can infer somewhat plausibly beyond it, but it's still a matter of deductive reasoning to reach those conclusions, is that not correct?
i don't mind the math i can appreciate it as an effective method, but could a more rhetorical example be if i'm not sure of a road to go down where it is split on the highway i can look at clues around me to hope i'm about to choose the right direction? i think there's something missing in that analogy as to the end-orientation of inductive reasoning... i thought that inductive reasoning meant that i knew my end expectation (ie x=12) and from that i can inductively reason the question may have been 10+2 or how many letters are in 'catastrophes'... this being the sort of reasoning productive in developing creative solutions to unanswerable questions(3 votes) - the title of this video is "inductive patterns" but the only reference to inductive or deductive is when Sal said "if we did our math or deduced correctly..." at5:58. is this inductive b/c we're generalizing from this sample? or is it deductive b/c this is a formula that's easily demonstrated to be fact?(2 votes)
- What if the blocks were separated, and that they don't share a common wall. So, the 50th term will just be the multiple of 6( 6n ), right?(1 vote)
Yes, because they wouldn't share a common wall, it would be a multiple of 6. We know that each "house" requires 6 toothpicks, all we have to do is multiply the the number of houses, which I will assign a variable of n, by the number of toothpicks needed for each house, which is 6. So the formula would be 6n = the number of toothpicks.
I hope this helped : )(2 votes)
The way I see it this is deductive reasoning if it is explicitly stated in the problem that this pattern is going to be consistent throughout until the fiftieth toothpick is reached. Otherwise it is inductive reasoning. Can someone clarify if this I am correct on this?(1 vote)- Inductive reasoning allows for the possibility that the conclusion is false, even if all of the premises are true.
A classical example of an incorrect inductive argument was presented by John Vickers:
All of the swans we have seen are white.
Therefore, all swans are white.
We could apply that to this saying each group of toothpicks has five more toothpicks than the previous term. (All of the swans we have seen are white.)
Therefore all the toothpicks will follow this pattern. (Therefore, all swans are white.)
Good question! This is inductive reasoning. There is no guarantee that, for example, the toothpicks are stacked so the shape changes or the number of toothpicks added is changed, etc.
source:
http://en.wikipedia.org/wiki/Inductive_reasoning#Inductive_vs._deductive_reasoning(2 votes)
- How do I complete this conjecture: The number of segments determined by n points is ...(1 vote)
I think you are talking about line segments that can be determined by n points? Are these supposed to be collinear points or do they have to be non-collinear points?
Well, the first thing you should probably do if you don't have this filed away as geometric/mathematic trivia is to sketch out some tests to figure out how many segments you can draw if the number of points is 1, 2, 3, 4, 5, and maybe a point or so more, to see if you can see a pattern. A little table would help. It is kind of interesting that you cannot draw a line segment within one point, so the first result is zero. Two points give you one segment between them. Three give you three! What happens with 4 dots? Yikes, it jumps up to 6. Now what about 5 points?
1 0
2 1
3 3
4 6
5 ?
6 ?
Once you have figured out how the numbers of line segments are increasing, then you want to try to state the number of segments in terms of n, where n= the item number of the porcupines you are drawing, which happens to also be the number of points.
Next, see if your conjecture (your guess about how the number of segments is related to n the number of points) works for each of the examples you already have. This pattern isn't simple in the sense of number of sides = n/2
Hint, this is a famous factorial exercise.(2 votes)
is anyone else watching during a class for a teacher and then you can't hear because you don't have ear-phones and then you are confused about everything that is going on.(1 vote)
I am not even watching it so i'm just as confused as you are.(2 votes)
5:58Video transcript
We're asked how many toothpicks
will be needed to form the 50th figure
in this sequence? And they've drawn the first four
figures in the sequence. This is the first, the
second, the third. Now, let's see what's
happening here. So in this first figure, we have
this nice little house that they've made out of
toothpicks, and it has one, two, three, four, five,
six toothpicks. So the first figure in our
sequence has six toothpicks. Now, what happens in the
second figure here? Well, it looks just like
our first figure. We have our one, two, three,
four, five, six toothpicks, but then they add
some toothpicks. It looks like they added one,
two, three, four, five, toothpicks. So it's 6 plus 5 or
11 toothpicks. So let me write this down. It is the 6 toothpicks plus 5
more, which is equal to 11 toothpicks. And then when we go to the third
figure, it looks just like the second figure, so it
has the 11 toothpicks, so the 11 toothpicks are one, two,
three, four, five, six, seven, eight, nine, ten, eleven, just
like the second figure. And then they add some more. They add one, two, three,
four, five. So it looks like every time,
when you draw part of a another house, because it can
share a wall with the previous house or share a toothpick
with the previous set of houses, you add five
toothpicks. So for the third term in our
sequence, we had 11 and when we had our second term,
or our two houses, and then we add five more. So to our 50th term, so 11
plus 5 is going to be 16. And then for the fourth term,
it's going to be 16 plus 5, which is equal to 21. Now, there's a couple of
ways to think about it. You can think about however
many more you are than 1. So let's say you are in the
nth term of the sequence. If you are in the nth term,
however more you are than 1, so if you're the nth term,
you're going to be n minus 1 more than 1. If that is confusing, we'll do
it with real numbers, so it gets a little bit
more tangible. So the nth term, you are n minus
1 more or greater than-- I'll just say more-- than 1. For example, if n is 2,
you are 1 more than 1. If n is 3, you are 3 minus 1,
which is 2 more than 1. So however much more
you are than 1, you multiply that by 5. We're 2 more than 1, so we add
10 to the number of toothpicks we have here. We're 3 more than 1 here, so
we add 15 to the number of toothpicks we have there. So you could say that for the
nth term, the number of toothpicks is equal to the
number you're more than 1, < n minus 1 times 5. That's how much you're going
to add above and beyond the amount of toothpicks
in just the first sequence, so times 5. n minus 1 times 5 plus the
number of toothpicks that you would just have in the
first sequence or just this one house. And we already counted
that: plus 6. So that's one way to
think about it. And if this looks complicated,
you just say, well, look, if I put a 4 here, I'm 3 more than
1, so 4 minus 1 is 3, so that'll be 3 times
5, which is 15. And then you add the number
of toothpicks in 1 and then you get 6. Now another way, and many of you
all might find this easier to think about, is even in 1,
you could imagine a term here-- let me do it this way. You can imagine a term here
that they didn't draw. You can imagine a 0th term. Let me just draw it here. Imagine a 0th term, and the 0th
term would just be kind of a left wall of the house, or
in this case, the left toothpick of a house. And then the first one, you're
adding 5 toothpicks to that. In the second one, you're adding
5 toothpicks to that. And when you think about it this
way, it actually becomes a little simpler to think
in terms of n. Here, you could say, well, the
nth term-- let me do this in a different color. You could say that the nth term
is going to have-- or maybe we should say number of
toothpicks in nth figure is going to be equal to 1. So in the 0th figure, which I
just made up, you have at least 1 toothpick, and then
whatever term in the sequence you are, you multiply that times
5 and you add that to the number of toothpicks in the
0th figure, so it's 1 plus 5 times n, which is actually a
simpler way to think about. The first figure is going to
be that 1 in the 0th figure plus 5: 6 toothpicks. Even in the 0th figure,
it works out. If you put a 0 here,
0 plus 1 is 1. If you put the fourth figure
here, 5 times 4 is 20 plus 1 is 21. Now let's answer
their question. We need to figure out the
toothpicks in the 50th figure in the sequence. Well, we just put
50 for n here. So for the 50th figure, we
could use either formula. We have 1 plus 5 times 50. 5 times 50 is 250 plus
1 is equal to 251. And now I said you could use
either formula because they should reduce to each other, if
we did our math right or if we deduced properly. So this one up here, let's just
verify this is the same thing as this over here. If we multiply 5 times n minus
1, we get 5n minus 5, right? We just distributed the 5. And then we have that plus 6. Negative 5 plus 6 is plus 1. So it's 5n plus 1
or 1 plus 5n. Either way, your 50th figure
is going to have 251.