Let a be a real number and let m, n be natural numbers such that m < n. Prove that if 0 < a < 1 --> a^n < a^m < 1.
So, what I thought to start was: since 0 < a
0 < a^n Also a < 1, a^n < 1. By the property of real #s where if a < b there is a real number t such that a < t < b, since a^n < 1, a^n < a^m < 1. But I don't think that is enough to justify the problem. HELP!
well, a^n = a^(m)*a^(n-m)
n-m > 0, so a^(n-m) < 1
To formally prove the statement "if 0 < a < 1, then a^n < a^m < 1" for real number a and natural numbers m, n such that m < n, we can use the properties of exponents.
Here's a step-by-step proof:
1. Given: 0 < a < 1 and m < n.
2. We want to prove: a^n < a^m < 1.
Proof:
Step 3: Start by raising a to the power of n.
- Since 0 < a < 1, multiplying a by itself (raising to a power) makes it smaller.
- Thus, a^n < a, as a^n is smaller than a.
Step 4: Now, let's raise a to the power of m.
- Again, multiplying a by itself makes it even smaller.
- Since m < n, a raised to the power of m (a^m) is smaller than a^n.
Step 5: Combining the results from steps 3 and 4, we have a^n > a^m.
Step 6: Next, we need to prove that a^m < 1:
- Since 0 < a < 1, raising a to a power makes it even smaller.
- Therefore, a^m < a, as a^m is smaller than a.
- If a is smaller than 1, then a^m (which is smaller than a) is also smaller than 1.
Step 7: Combining the results from steps 5 and 6, we have a^n > a^m > 1.
Hence, we have proven that if 0 < a < 1 and m < n, then a^n < a^m < 1.
Note: It is important to note that mathematical proofs should be rigorous and make use of logical steps. You may need to provide more detailed reasoning and justification based on the properties of exponents if required.