Prove a^(m+n) = a^m + a^n with mathematical induction

In summary, the conversation discusses the use of mathematical induction to prove two equations: 1) a^(m+n) = a^m + a^n and 2) (a^m)^n = a^(mn). The conversation raises the issue that the first equation is not true, but the second equation can be proven using induction. The suggested approach is to use n=0 as the base case and prove the corrected thesis.
  • #1
gluttony
1
0
prove by induction that, 1) a^(m+n) = a^m + a^n

2) (a^m)^n=a^(mn)


2. Homework Equations -- mathematical induction
3. The Attempt at a Solution
the equations hold for a = 1.
let's say the equation hold for a = s; then s^(m+n) = s^m + s^n
now for a = s+1; (s+1)^(m+n) = (binomial expansion of s+1)

how can i prove that (s+1)^(m+n) = (s+1)^m * (s+1)^n
 
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  • #2
gluttony said:
prove by induction that, 1) a^(m+n) = a^m + a^n
You can't prove that because it's not true. 22+1=23=8, 22+21=4+2=6, and 8 is obviously not equal to 6.

What you can prove is that am^n = am*an.
3. The Attempt at a Solution
the equations hold for a = 1.
let's say the equation hold for a = s; then s^(m+n) = s^m + s^n
now for a = s+1; (s+1)^(m+n) = (binomial expansion of s+1)

how can i prove that (s+1)^(m+n) = (s+1)^m * (s+1)^n
Wrong approach. Try starting with n=0 as your base case rather than a=1. And of course you do want to prove the corrected thesis.
 

1. How is mathematical induction used to prove the equation a^(m+n) = a^m + a^n?

Mathematical induction is a method of proof that uses the principle of mathematical induction, which states that if a statement is true for a base case (usually the smallest possible value of the variable) and if it can be shown that the statement holds for any subsequent case by using the previous case, then the statement is true for all cases. In the case of proving a^(m+n) = a^m + a^n, mathematical induction is used to show that the equation holds for all values of m and n.

2. What is the base case and how is it used in the proof?

The base case is the starting point of the proof, usually the smallest possible value of the variable. In this case, the base case is typically m=1 and n=1. The base case is used to show that the equation holds true for a specific case, and then the subsequent cases are built upon this initial case using the inductive hypothesis.

3. What is the inductive hypothesis and how is it used in the proof?

The inductive hypothesis is the assumption that the statement is true for some value of n. In the proof of a^(m+n) = a^m + a^n, the inductive hypothesis is used to show that if the statement holds true for some value of n, then it also holds true for the next value of n, thus proving the statement for all values of n.

4. How is the inductive step used to complete the proof?

The inductive step is used to extend the proof from the base case to all subsequent cases. It involves using the inductive hypothesis to show that if the statement is true for some value of n, then it is also true for the next value of n. This completes the proof by showing that the statement holds true for all values of n, and therefore, for all values of m and n.

5. Why is mathematical induction a valid method of proof for equations like a^(m+n) = a^m + a^n?

Mathematical induction is a valid method of proof for equations like a^(m+n) = a^m + a^n because it relies on the principle of mathematical induction, which is a fundamental principle in mathematics. It has been proven to be a valid method of proof and is widely used in mathematical proofs to show that a statement holds true for all cases, even when there are infinitely many cases to consider.

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