Proving Induction for All Integers

In summary: Because you used the fact, that ##\mathbb{Z}## is symmetric relative to the sign. You have proven the statement for all positive integers and zero, and then transformed this result to the negative integers by taking the power to ##-1##. In a way you used the same induction twice, or more precisely: you solved a portion of the statement and used another proof based on this result to do the rest.
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Mr Davis 97
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Homework Statement


Let ##\phi : G \to H## be a homomorphism. Prove that ##\phi (x^n) = \phi (x)^n## for all ##n \in \mathbb{Z}##

Homework Equations

The Attempt at a Solution


First, we note that ##\phi (x^0) = \phi(x)^0##. This is because ##1_G \cdot 1_G = 1_G \implies \phi (1_G 1_G) = \phi (1_G) \implies \phi (1_G)^2 = \phi (1_G) \implies \phi (1_G) = 1_H##.

Second, we show that ##\phi (x^n) = \phi (x)^n## where ##n \in \mathbb{Z}^+##. The base case is trivial. Now, suppose for some ##k## we have that ##\phi (x^k) = \phi (x)^k##. Then ##\phi (x^{k+1}) = \phi (x^k x) = \phi (x^k) \phi (x) = \phi (x)^k \phi (x) = \phi (x)^{k+1}##. This proves the result for positive integers.

Third, we prove the result for negative integers. First, we show that ##\phi (x^{-1}) = \phi (x)^{-1}##. ##xx^{-1} = 1_G \implies \phi (xx^{-1}) = 1_H \implies \phi(x) \phi(x^{-1}) = 1_H \implies \phi (x^{-1}) = \phi(x)^{-1}##. So, since the result ##\phi (x^n) = \phi (x)^n## is true for all positive integers, the result ##\phi (x^{-n}) = = \phi ((x^{-1})^n) = \phi (x^{-1})^n = ( \phi (x)^{-1} )^n = \phi (x)^{-n}## must also be true for all positive integers.

Is this argument fine?
 
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  • #2
Mr Davis 97 said:
Is this argument fine?
Apart from a double "=", yes. Maybe you should have mentioned ##x^0=1_G## and ##\phi(x)^0=1_H## but this is nit-picking. Your proof is fine.
 
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fresh_42 said:
Apart from a double "=", yes. Maybe you should have mentioned ##x^0=1_G## and ##\phi(x)^0=1_H## but this is nit-picking. Your proof is fine.
Great, but I have a couple of questions. So, I proved that ##\phi (x^n) = \phi (x)^n##. Why was it valid for me to substitute ##x## with ##x^{-1}## to get the result ##\phi (x^{-n}) = \phi (x)^{-n}##? Why didn't I have to do an induction a second time for the negative case?
 
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Mr Davis 97 said:
Great, but I have a couple of questions. So, I proved that ##\phi (x^n) = \phi (x)^n##. Why was it valid for me to substitute ##x## with ##x^{-1}## to get the result ##\phi (x^{-n}) = \phi (x)^{-n}##? Why didn't I have to do an induction a second time for the negative case?
Because you used the fact, that ##\mathbb{Z}## is symmetric relative to the sign. You have proven the statement for all positive integers and zero, and then transformed this result to the negative integers by taking the power to ##-1##. In a way you used the same induction twice, or more precisely: you solved a portion of the statement and used another proof based on this result to do the rest.
 
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Related to Proving Induction for All Integers

1. What is induction for all integers?

Induction for all integers is a mathematical method used to prove statements or propositions for an infinite set of numbers, specifically the set of all positive integers. It involves proving a base case, also known as the starting point, and then showing that if the statement is true for any given integer, it is also true for the next consecutive integer. This process is repeated until the statement is proven to be true for all integers.

2. How is induction for all integers different from regular mathematical induction?

Regular mathematical induction is used to prove statements for a finite set of numbers, while induction for all integers is used to prove statements for an infinite set of numbers. In regular mathematical induction, the base case is usually the number 1, while in induction for all integers, the base case is usually the number 0.

3. What are the steps involved in induction for all integers?

The steps involved in induction for all integers are:

  • Step 1: Prove the statement for the base case, usually the number 0.
  • Step 2: Assume the statement is true for some arbitrary integer, usually denoted as k.
  • Step 3: Use this assumption to prove that the statement is also true for the next consecutive integer, k+1.
  • Step 4: Repeat step 3 until the statement is proven to be true for all integers.

4. What are some examples of statements that can be proven using induction for all integers?

Examples of statements that can be proven using induction for all integers include:

  • The sum of the first n positive integers is n(n+1)/2.
  • The product of the first n positive integers is n! (n factorial).
  • The sum of the first n odd positive integers is n^2.

5. What are the limitations of induction for all integers?

Induction for all integers can only be used to prove statements for an infinite set of numbers. It cannot be used to prove statements for a finite set of numbers or for real numbers. Additionally, it requires a clear and specific base case, which can sometimes be difficult to determine. It also assumes that the next consecutive integer always exists, which may not always be true in certain mathematical scenarios.

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