Prove, by PMI, that sum of any 3 consecutive integers is divisible by 3?
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Mathematical induction is a method of mathematical proof typically used to establish that a given statement is true of all natural numbers (non-negative integers). It is done by proving that the first statement in the infinite sequence of statements is true, and then proving that if any one statement in the infinite sequence of statements is true, then so is the next one.
So, lets start with the equation where we have
3[int(x)] = y > 100
Where y is greater than 100, and x can only exist as an integer.
Therefore, we can use the PMI, which states that if any one statement in an infinite sequence of statements is true, then so is the next one.
To make the above equation / half inequality work, we have to plug in a number that is 34 or greater. so let's use 34, the first number in the sequence. (which includes 3 digit numbers)
3[int(34)] = ? = 102
Now, let's turn it around and say, 102, the first three digit number, divisible by three has digits that give a sum of 3 (1+0+2). We can even do it again, with 35... 105, the second three digit number divisible by three has digits that add to 6 (1+0+5), proves that there is a pattern that satisfies the PMI.
So, lets start with the equation where we have
3[int(x)] = y > 100
Where y is greater than 100, and x can only exist as an integer.
Therefore, we can use the PMI, which states that if any one statement in an infinite sequence of statements is true, then so is the next one.
To make the above equation / half inequality work, we have to plug in a number that is 34 or greater. so let's use 34, the first number in the sequence. (which includes 3 digit numbers)
3[int(34)] = ? = 102
Now, let's turn it around and say, 102, the first three digit number, divisible by three has digits that give a sum of 3 (1+0+2). We can even do it again, with 35... 105, the second three digit number divisible by three has digits that add to 6 (1+0+5), proves that there is a pattern that satisfies the PMI.
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In the base case, let n=1 be the smallest of the three consecutive integers. Then 1+2+3 = 6 is divisible by 3.
In the recursive case, suppose three consecutive integers starting at n are divisible by 3. Then n+(n+1)+(n+2) = 3m for some integer m. Adding 1 to each of these n's and simplifying, this becomes
(n+1) + (n+2) + (n+3) = 3m+3 = 3(m+1),
completing the recursive case. Thus the sum of any three consecutive positive integers is divisible by 3.
You need to do a second case for the negative integers, and special cases for those that cross the line between positive and negative, depending on precisely what the question actually wants. To be honest, this is a very poor example of an inductive proof. The usual way one would do it, n + (n+1) + (n+2) = 3n+3 = 3(n+1), is much more direct and uses the "for all elimination" first order rule rather than the inductive axiom scheme, making it both simpler and logically more desirable. However, I should be fair and say that the associative and commutative laws that I implicitly made use of are proven, from Peano arithmetic at least, via heavy use of induction.
In the recursive case, suppose three consecutive integers starting at n are divisible by 3. Then n+(n+1)+(n+2) = 3m for some integer m. Adding 1 to each of these n's and simplifying, this becomes
(n+1) + (n+2) + (n+3) = 3m+3 = 3(m+1),
completing the recursive case. Thus the sum of any three consecutive positive integers is divisible by 3.
You need to do a second case for the negative integers, and special cases for those that cross the line between positive and negative, depending on precisely what the question actually wants. To be honest, this is a very poor example of an inductive proof. The usual way one would do it, n + (n+1) + (n+2) = 3n+3 = 3(n+1), is much more direct and uses the "for all elimination" first order rule rather than the inductive axiom scheme, making it both simpler and logically more desirable. However, I should be fair and say that the associative and commutative laws that I implicitly made use of are proven, from Peano arithmetic at least, via heavy use of induction.