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Mathematical induction

  1. Apr 2, 2004 #1
    I have a question about mathematical induction, I know that this stuff is really easy I just can't seem to get the point here. I lost alot of math alegbra more likely to say because of moving and now I am having trouble with algebra in my first semester of college.
    I don't really need anyone to solve it for me, I was just wondering if anyone had a clear explanation of the concept for mathematical induction? Like this problem here asks to prove each statement by mathematical induction.

    1/3+1/9+1/27+....+1/3^n= 1/2 ( 1-1/3^n)

    Basically I know that you have to first say that it will also work for (k+1) right? In this case you cannot add (k+1) you have to make it as a power when proving it which gets me confused.
    Thank you for you r help I appreciate it. Just a basic concept is what I want. :frown:
  2. jcsd
  3. Apr 2, 2004 #2

    matt grime

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    You need a set of statements indexed by the set of natural numbers. You need to show that if the statement is true for the one indexed by k that it is true for the one indexed by k+1, you also need to show that it is true for theone indexed by 1 (or some particular number) and then it is true for all n bigger than 1 (or whatever it was that you initially proved). Just read the definitions, it's rather trivial.l
  4. Apr 18, 2004 #3


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    I know what you mean. I remember when I first learned induction it seemed kinda weird. I had a really cool highschool algebra teacher that used a good analogy that really helped me understand the concept. If you think of a ladder with many rungs on it, proving something by inductuion is basically saying that you can climb as high as you want on the ladder, because there will always be a rung when you need it. You must show that there is a least one rung on the ladder first. This is done by showing the statement is true for some value n. (usually n=1, but it dosn't have to.) Once this is done, it means that I have a ladder and there is at least one rung on it. Then you must show that if you have any rung (could be the first, tenth, trillionth, dosn't matter which one... you just need one, and you proved that in the first part) you can take the next step to the next rung. This is known as the "induction hypothesis," the part where you assume it's true for k and show that this implies it's true for k+1. Once this is proven, you have shown that for any rung on the ladder, there is another rung above it. Since you first showed that there is a first rung (when you showed it was true for n), you now know that there are an infinite amount of rungs above it, and hence the statement is true for all natural numbers greater than your starting value n. I hope this helped you, cause it really helped me. Just keep thinking, and you'll figure it out.

  5. Apr 18, 2004 #4
    Here's a good analogy someone explained to me once:

    Let's say you need to cross a river but you can't swim. Fortunately, there are rocks in the river which you may be able to hop across. How do you know if you can cross the river? Two pieces of information is needed to prove you can: 1) Whether or not you can get to the first rock (the rock closest to your side of the shore) 2) Whether or not you can hop from one rock to the next. The same goes with mathematical induction. You need to prove that something is true for the first case. Then you must prove that if something is true for any given case, it is also true for the immediately following case.

    Take your example: First prove that equation is right for the case where n=1 (which is just plug-n-chug). Then prove that if it's true for n=k, it's also true for the case of n=k+1.
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