Why Does Mathematical Induction Prove Formulas 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.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top