Science:Math Exam Resources/Courses/MATH312/December 2005/Question 11

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MATH312 December 2005

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Question 11
Prove In this section, prove the statement given to you. Suppose a and N are integers with $N\geq 0$. Show that $(1+a)^{N}\equiv 1+Na\mod a^{2}$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
This can be solved either by induction or using the binomial theorem.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution 1
We proceed by induction. The case $\displaystyle N=1$ is easy. For $\displaystyle N=2$, we have $\displaystyle (1+a)^{2}=1+2a+a^{2}\equiv 1+2a\mod {a^{2}}$ Suppose the claim holds for N. Then for $\displaystyle N+1$, we have $\displaystyle (1+a)^{N+1}\equiv (1+a)^{N}(1+a)\equiv (1+Na)(1+a)\equiv (1+(N+1)a+Na^{2})\equiv 1+(N+1)a\mod {a^{2}}$ and so by mathematical induction we are done.

Solution 2
We can use the binomial theorem to see that $\displaystyle (1+a)^{N}=\sum _{i=0}^{N}{\binom {N}{a}}a^{i}(1)^{N-i}\equiv {\binom {N}{0}}+{\binom {N}{1}}a+\sum _{i=2}^{N}{\binom {N}{a}}a^{i}(1)^{N-i}\equiv 1+Na\mod {a^{2}}$ where all the terms in the right most summation are divisible by $\displaystyle a^{2}$.

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