Confusion with induction problem.

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SUMMARY

The discussion centers on proving three fundamental laws of exponents for all integers, specifically for non-zero real numbers \( a \) and \( b \). The laws include \( a^n a^m = a^{n+m} \), \( (a^n)^m = a^{nm} \), and \( a^m b^m = (ab)^m \). The participant expresses confusion about extending the proof to negative integers and seeks clarification on the induction approach. The reference to Munkres' "Topology" suggests a structured method for proving these laws using induction.

PREREQUISITES
  • Understanding of exponentiation rules
  • Familiarity with mathematical induction
  • Knowledge of integer sets, specifically \( \mathbb{Z} \) and \( \mathbb{N} \)
  • Basic concepts from real analysis
NEXT STEPS
  • Study the principles of mathematical induction in depth
  • Review the properties of exponents in real numbers
  • Explore proofs involving negative integers in exponentiation
  • Examine Munkres' "Topology" for additional context on mathematical proofs
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Students of mathematics, particularly those studying algebra and real analysis, as well as educators looking for effective methods to teach exponentiation and induction.

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Homework Statement


Let a in R and a != 0. Define a^0=1 and for all n in the positive integers, a^-n = 1/a^n.

Show,

a^n a^m =a^(n+m)

(a^n)^m = a^nm

a^m b^m =(ab)^m

for all a,b != 0 and n,m in Z.


The Attempt at a Solution



Notice that last part where it says Z, not just N. I just proved these 3 laws using induction for n,m in N. I am confused now on how to approach the problem.

Do I fix n, and let m = -1 for the base case. Then assume the statement is true to m<0 and show it is also true for m-1 ?

Thank you for your time.

EDIT: P35 in Munkres, Topology
 
Last edited:
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If you have proved that anam[/sup]= an+m for positive integers then you should be able to prove, in the same way, that a-na-m= a-n-m for m and n any positive integers.
 

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