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Topic: **statments****Question:**
Is this statement right? A proof by induction allows you to prove a statement about an arbitrary value by proving the statement is true when the value is equal to one.
Seems true to me since I usually use: assume balabala...

July 19, 2019 / By Edweena

It is not true. 1. "proving the statement is true when the value is equal to one" Basis can be any natural number, so not just 1. In addition, the statement may be true for 1 but the induction basis being a higher number, 7 for example. 2. A proof by induction allows you to prove a statement about an arbitrary value Not arbitrary but a statement for every natural number starting from basis. 3. Furthermore, it must be shown that IF the statement is supposed to be true for n Then it is also true for n+1.

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The principle of mathematical induction asserts the following. Given a statement (like 6 + 12 + ... + 6n = 3n(n+1) ) if: 1.) The statement is true for n = 1. 2.) And, whenever the statement is true for a positive integer k, then it is also true for k+1. Then we can conclude the statement must be true for all integers. This principle seems a bit wordy, but it makes intuitive sense. It is like climbing a ladder. You can "climb" an infinite ladder if you can: 1.) reach the first rung 2.) and whenever you can reach a rung, you can reach the next rung. Placed in this context the principle of mathematical induction makes "good" sense. Proving a statement by mathematical induction often includes two steps. STEP 1: Show the statement is true for n = 1. In plain English, we must show that the "sum" of the first 1 multiple of 6 is equal to 3(1)(1 + 1). In the case where n = 1, the statement simplifies to: 6 = 3(1)(1 + 1). And this is clearly a true statement. So the statement holds for n = 1. STEP 2: Show if it holds for k, then it holds for k + 1. This is often called the "inductive step". Assume that the statement is true for a positive integer k. That is, assume, 6 + 12 + ... + 6k = 3k(k + 1). This assumption is called the "induction hypothesis". We will now proceed to show that the statement must be true for the positive integer k + 1. We need to show that the sum of the first k + 1 multiples of 6 is equal to 3(k+1)( (k + 1) + 1). And we begin: 6 + 12 + ... + 6k + 6(k+1) = [6 + 12 + ... + 6k] + 6(k + 1) = 3k(k + 1) + 6(k + 1) (this is true by the induction hypothesis) = 3(k + 1) [ k + 2 ] (we factored out 3(k + 1) ) = 3(k + 1)[(k + 1) + 1] (rewritten to match the correct conclusion). And we have shown that the statement is true for the integer k + 1. We can conclude, by the Principle of Mathematical Induction that the statement must be true for all positive integers.

one million)At n=one million on the left we get we get 4(6)=24 on the main suitable 4(5)(15)/6=50 one million is fake 2)At n=one million on the left we get one million on the reight we get (6-3-one million)/2=one million So for 2 we've proved the preliminary term we ought to now instruct it for the n+one million term (3(n-one million)-2)^2 = (n+one million)(6(n+one million)^2 -3(n+one million) -one million)/2 (3n-5)^2 = (n+one million)(6(n^2+2n+one million)-3n+3-one million)/2 9n^2-30n+25 = (n+one million)(6n^2 +12n +6 -3n +2)/2 9n^2-30n+25 = 6n^3 +12n^2 -3n^2 +8 + 6n^2 +12n -3n +8 we now have an n^2 term on one area besides as an n^3 on the main suitable area ? those 2 are no longer equivalent.

prove true for base case, not necessarily 1 assume true for n then prove it true for n+1 hence proven by induction

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No. To prove by induction is a two step process. 1) prove that its true for 1. 2) prove that "true for N" implies "true for N+1" A proof by induction allows you to prove a statement about an arbitrary (natural number) value, by proving the statement about the truth at 1 AND by proving the inductive step.

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Ask them if they will accept a utility (gas, power, telephone) bill as proof of residency. I'd be leery of anyone asking for a copy of my bank statement.

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