[ASK] Mathematical Induction: Prove 7^n-2^n is divisible by 5.

In summary, the conversation discusses how to prove by mathematical induction that 7^n-2^n is divisible by 5. The method of induction is used to show that for n = 1 and n = k, the expression is divisible by 5. However, when trying to prove for n = k + 1, a calculation mistake is made in showing that 12*2^k is divisible by 5. A simpler rule for the difference of two terms with the same power is suggested, but it is noted that this method is not commonly taught in school and the mathematical induction method is still used.
  • #1
Monoxdifly
MHB
284
0
Prove by mathematical induction that \(\displaystyle 7^n-2^n\) is divisible by 5.What I've done so far:For n = 1

\(\displaystyle 7^1-2^1=7-2=5\) (true that it is divisible by 5)

For n = k

\(\displaystyle 7^k-2^k=5a\) (assumed to be true that it is divisible by 5)

For n = k + 1

\(\displaystyle 7^{k+1}-2^{k+1}=7^k\cdot7-2^k\cdot2=7(7^k-2^k)+12\cdot2^k=7(5a)+12\cdot2^k\)

This is where the problem lies. How can I show that \(\displaystyle 12\cdot2^k\) is divisible by 5?
 
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  • #2
Monoxdifly said:
Prove by mathematical induction that \(\displaystyle 7^n-2^n\) is divisible by 5.What I've done so far:For n = 1

\(\displaystyle 7^1-2^1=7-2=5\) (true that it is divisible by 5)

For n = k

\(\displaystyle 7^k-2^k=5a\) (assumed to be true that it is divisible by 5)

For n = k + 1

\(\displaystyle 7^{k+1}-2^{k+1}=7^k\cdot7-2^k\cdot2=7(7^k-2^k)+12\cdot2^k=7(5a)+12\cdot2^k\)

This is where the problem lies. How can I show that \(\displaystyle 12\cdot2^k\) is divisible by 5?
Hi Monoxdifly,

$12\cdot2^k$ is certainly not divisible by $5$ (look at the prime factors).

You made a mistake in your calculation: I get:
$$
7^k\cdot7 - 2^k\cdot2 = 7(7^k-2^k) + 5\cdot2^k
$$
and this should clear things up.
 
  • #3
Ah, I see. Thank you very much!
 
  • #4
It seems a bit overkill to use Induction, when there's a simple rule for the Difference of Two Terms With the Same Power...

$\displaystyle \begin{align*} a^n - b^n \equiv \left( a - b \right) \sum_{r = 0}^{n - 1}{ a^{n - 1 - r}\,b^r } \end{align*}$

so in your case the factor would be (7 - 2) which equals 5.
 
  • #5
Prove It said:
It seems a bit overkill to use Induction, when there's a simple rule for the Difference of Two Terms With the Same Power...

$\displaystyle \begin{align*} a^n - b^n \equiv \left( a - b \right) \sum_{r = 0}^{n - 1}{ a^{n - 1 - r}\,b^r } \end{align*}$

so in your case the factor would be (7 - 2) which equals 5.

Well, the school curriculum doesn't teach this rule, so yes, we were supposed to solve it with the "overkill" method.
 
  • #6
Monoxdifly said:
Well, the school curriculum doesn't teach this rule, so yes, we were supposed to solve it with the "overkill" method.

Then you can consider it as something new that you have learnt. The most concise method of proof is always the best.
 

FAQ: [ASK] Mathematical Induction: Prove 7^n-2^n is divisible by 5.

1. What is mathematical induction?

Mathematical induction is a proof technique used to prove statements about natural numbers. It involves showing that a statement is true for a base case, typically n=1, and then showing that if the statement is true for some arbitrary value of n, it must also be true for n+1.

2. How do you use mathematical induction to prove a statement?

To prove a statement using mathematical induction, you first show that it is true for the base case, typically n=1. Then, you assume that the statement is true for some arbitrary value of n, and use this assumption to prove that it must also be true for n+1. This completes the inductive step, and by repeating this process, you can show that the statement is true for all natural numbers.

3. How does mathematical induction relate to divisibility?

Mathematical induction can be used to prove statements about divisibility, such as the statement "7^n-2^n is divisible by 5". By showing that the statement is true for the base case n=1, and then using the inductive step to show that if it is true for some arbitrary value of n, it must also be true for n+1, we can prove that the statement holds for all natural numbers.

4. How can you prove that 7^n-2^n is divisible by 5 using mathematical induction?

To prove that 7^n-2^n is divisible by 5 using mathematical induction, we first show that the statement is true for the base case n=1, since 7^1-2^1 = 5, which is divisible by 5. Then, we assume that the statement is true for some arbitrary value of n, and use this assumption to show that it must also be true for n+1. This can be done by expanding (7^(n+1)-2^(n+1)) as (7^n-2^n)*7 - (7^n-2^n)*2, and using the fact that 7^n-2^n is divisible by 5 (by our assumption) to show that (7^(n+1)-2^(n+1)) is also divisible by 5. Therefore, by mathematical induction, we have proven that 7^n-2^n is divisible by 5 for all natural numbers.

5. Can mathematical induction be used to prove statements about other types of numbers?

Yes, mathematical induction can be used to prove statements about other types of numbers, such as integers, rational numbers, or even complex numbers. However, the base case and inductive step may differ depending on the type of numbers being considered. For example, to prove a statement about integers, the base case would typically be n=0 instead of n=1, and the inductive step would involve showing that if the statement is true for some arbitrary integer n, it must also be true for n+1.

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