Why Does Mathematical Induction Prove Formulas True for All Positive Integers?

AI Thread Summary
Mathematical induction proves formulas true for all positive integers by establishing a base case and an inductive step. The base case confirms the formula is true for n=1. The inductive step shows that if the formula holds for an arbitrary positive integer k, it must also hold for k+1. This creates a chain reaction, confirming the formula for all subsequent integers. Thus, once the base case and inductive step are validated, the formula is proven true for all positive integers.
andrewkg
Messages
86
Reaction score
0
Hello I'm learning about proofs and in my book there's a sect. On mathematical induction. And I'm trying understand why this makes it true for all values.
1+3+5...2n-1=n^2
Suppose that the formula is known to be true for n=1, and suppose that as a result of assuming that it is true for n=k, where k is an arbitrary positive integer, we can prove that it is also true for n=k+1.
Then the formula is true for all k.

Why does this addition of 1 make it true for all k?
 
Mathematics news on Phys.org
You know it's true for n=1 and you know that for every n where it's true, it's also true for n+1. Since you proved it for 1, this implies it's true for 1+1 = 2. Now, since you know it's true for 2, it must be true for 2+1 = 3. Now since you know it's true for 3, it's also true for 3+1 = 4. And so on, so it's true for every positive integer.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top