Algebra (all content)
Sal analyzes a solution of a mathematical problem to determine whether it uses inductive reasoning. Created by Sal Khan and Monterey Institute for Technology and Education.
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- So would this be deductive ?(4 votes)
- If there is no conjecture in the proof, it is deductive.
If there is conjecture, for example, assume that A is true for all n, then it is inductive.(4 votes)
- At first it says Luis, but later in the question it says Carlos...
Did anybody else catch that?(3 votes)
- Can anyone exploain to me how to do thes problems? I am trying to wrap my mind and still cannot get it...... :( I really am bad with inductive reasoning.....
i know the pattern but i do not know how to write a conjecture
2,4,16...... (squaring the previous answer)
1/2,1/4,1/8... ( as each number progresses.... 2^previous+1)
-3,6,-9,12... (- or + depending..... (previous number + or - [previous subtraction or additon -or+ 3])(1 vote)
- The question is old but I will answer it anyway. Just for fun (:
The first sequence 2, 4, 16, 256 , 65536, … can be written as:
2^1, 2^2, 2^4, 2^8, 2^16, …
If we only take care of the exponent, they go in this manner:
1, 2, 4, 8, 16, which is the same as:
2^0, 2^1, 2^2, 2^3, 2^4, …
As it can be noted, the 1st of these is raised to 1 – 1 = 0. The 2nd one to 2 – 1 = 0. Therefore, the nth terms in this sequence (not the original), will be a 2 raised to the n – 1 power:
Since 2^(n-1) is the exponent of the nth term in the original sequence, we can place it in that one, but being careful of putting it between parenthesis, otherwise the rule of exponent will take 2^(n-1) as if it were 2*(n-1):
In the second sequence: 1/2, 1/4, 1/8, ... if we just look at the denominators, the 1st is 2^1, the 2nd is 2^2 and so on. And so the nth term would be 2^n. But how do we get the 1 as numerator? By rule of exponent, we know that a^(-b) = (1/a)^b. Thus the nth term in this sequence is:
In the third sequence: -3, 6, -9, 12, ... it looks like the 2nd term is the 1st one multiplied by 2, the 3rd terms is the 1st again times 3, and so on. This part can be assumed to be 3*n for the nth term of the sequence. Now, the first term looks like if it were multiplied by -1, the 2nd by 1, the 3rd by -1 and so on. We can conjecture that it’s a positive 1 when even and a negative 1 when odd. -1 raised to an even/odd power gives us this same pattern. So the nth term of the sequence is:
(-1)^n * 3n(5 votes)
- Isn't it inductive reasoning because Luis conjectured the answer he found for this problem to be to for ALL x and y? Not just the x and y in this problem. It seems he is making a inference about what x and y represent in general.(1 vote)
- I didn't think of that. Not sure why you were downvoted (i've since rebalanced it) maybe because somebody thought you were wrong? I think you could be right. I wish someone who knows more would answer.(1 vote)
- I'm learning a lot about inductive vs deductive reasoning. I would like to practice this concept. The test I need to take involves both reasonings. Is there a place I can go to practice this concept?(2 votes)
- Around1:00Carlos said that I conjecture that the expression=x^3-y^3. Then Sal said that it wasn't a conjecture because a conjecture is making a generalization. However couldn't he be making a generalization for all numbers, as he didn't prove the result for imaginary numbers?(1 vote)
- Yes and no. First, Sal is correct, it's not a conjecture.
Think of it this way, a conjecture is basically you saying, based on what we have, I think that this will be true, but I'm not sure.
So, if at the begining he said, I conjecture that (stuff) = x^3 - y^3, but didn't prove it, then that is a conjecture.
Now since he went on and made a proof using algebraic manipulation, it is no longer a conjecture about something that is unknown, but it is now a proof.
A conjecture can be proved or disproved because it is unknown at the time, but a proof, if done correctly, is 100% fact and nothing you say or I say will change the fact that it is correct.
To answer why it is general... Yes, it is the general case for all x and y that the function will be of the form x^3 - y^3, but that doesn't make it a conjecture. It is a generalization based on a proof.(0 votes)
- For this example I understand that inductive reasoning was used, however why was the letter "n" used in the example for Inductive reasoning? Could we have used "x", or "y" or does the "n" reflect the word "number"?(0 votes)
- Compare the three inductive reasoning?(0 votes)
- Hi boyratana,
It would be easier to answer your question if you were more specific. Also there is only one inductive reasoning...
- Sam :|(0 votes)
Luis looked at the expression x minus y times x squared plus xy plus y squared, and wrote the following. He said, using the distributive property-- so he took x minus y and he distributed this expression onto the x and this expression onto the negative y. So that's why we have an x times this entire expression over here. x times the entire expression minus y times this entire expression. Then he says, using the distributive property again, this is equal to-- so then he distributed this x into that and got all of the first three terms here. He got these first three terms. And then he distributed the y. He distributed the y over here and he got these three terms. And then he saw, looks like that this term x squared y cancels out with a minus x squared y, and the xy squared cancels out with the negative xy squared. Then he saw that it equals x to the third minus y to the third. And he wrote down, I conjecture that x minus y times x squared plus xy plus y squared is equal to x to the third minus y to the third for all x and y. Did Carlos use inductive reasoning? Explain. Well, inductive reasoning is looking at a sample of things-- looking at some set of data, if you will-- and then making a generalization based on that. You're not 100% sure, but based on what you've seen so far, you think that the pattern would continue. Or you think it might be true for all things that have that type of property or whatever. Now in this situation, he didn't look at some type of a sample. He actually just did a proof. He multiplied this out algebraically. In fact, it's incorrect for him to say I conjecture here. A conjecture is a statement or proposition that is unproven, but it's probably going to be true. It's unproven but it seems reasonable, or it seems likely that it's true. This isn't a conjecture. This is proven. He proved that x minus y times x squared plus xy plus y squared is equal to x to the third minus y third. He should have written-- and this is a much stronger thing to say-- he should have said, I proved that this is true for all x and y. So to answer the question, did he use inductive reasoning? No. I would say that he made an outright proof. No, he made a proof. Inductive reasoning would have been, if he would have saw, if you would have given him 5 minus 2 is equal to-- or 5 minus 2 times 5 squared plus 5 times 2 plus 2 squared, and you saw that that was equal to the same thing as 5 to the third minus 2 to the third. And then let's say he did it for, I don't know, one in seven in a couple of examples. And it kept holding for all the examples that it was the first number cubed minus the second number cubed. Then it would have been inductive reasoning to say that that is true for all numbers x and y. But here it's not an induction. He didn't use induction. Or I shouldn't say induction. He didn't use inductive reasoning. He outright proved that this statement is true for all x and y.