Hello Class,

Congratulations on completing the midterm exam! Now that the grading is underway, it is the perfect time to review your work. One of the most effective ways to learn mathematics is to analyze your own thought process and compare it with the correct solutions. To assist you in this, I have uploaded the full Mid Term Answer Sheet below.

While simply checking if you got the right answer is helpful, understanding the process to arrive at that answer is where real learning happens. Below, I have highlighted a few key concepts from the exam that appeared in questions involving symbolic logic and mathematical proofs.

1. Symbolic Logic and Validity

Several questions on the exam (such as #15, #163, and #169) asked you to determine the validity of an argument or the equivalence of statements. Remember the following strategies:

  • Testing Validity: To show an argument is invalid (like in #163), you only need to find one counterexample. Specifically, you are looking for a scenario where the premises are True but the conclusion is False.
  • Quantifiers: In question #85, we dealt with translating sentences into logical notation using quantifiers. The solution $\forall y (M(y) \to \neg U(y))$ demonstrates the universal quantifier. Pay close attention to the scope of the negation symbol.
  • Contrapositives: Question #36 reminds us that a conditional statement $p \to q$ is logically equivalent to its contrapositive $\neg q \to \neg p$. If the original statement is "If you are not registered, then you cannot check out books," the equivalent contrapositive is "If you check out books, then you are registered."

2. Constructing Proofs: Even and Odd Integers

A significant portion of the exam focused on direct proofs involving number theory. Questions #178, #182, and #190 required you to use the formal definitions of even and odd integers. Let's review the standard structure for these proofs.

Definitions to memorize:

  • An integer $n$ is even if $n = 2k$ for some integer $k$.
  • An integer $n$ is odd if $n = 2k + 1$ for some integer $k$.

Example from Question #178:
The prompt asks to show the relationship between an odd and an even number. If we suppose $x$ is odd and $y$ is even, we write:
$$x = 2k + 1$$
$$y = 2l$$
When we add them together:
$$x + y = (2k + 1) + 2l = 2k + 2l + 1 = 2(k + l) + 1$$
Since $k+l$ is an integer, the result is in the form of an odd integer. This is a classic direct proof structure.

3. Proof by Cases

Question #190 required a slightly more complex approach: Proof by Cases. The statement regarding $n^2$ depending on whether $n$ is even or odd requires us to examine both possibilities separately.

  • Case 1 (Even): If $n = 2k$, then $n^2 = (2k)^2 = 4k^2 = 2(2k^2)$, which is clearly even.
  • Case 2 (Odd): If $n$ is odd, then $n = 2k+1$. Squaring this gives:
    $$n^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$$
    This final form represents an odd integer.

Please download the full PDF below to check all your answers, including those involving truth values and contradictions. If you have questions about a specific problem or if your reasoning differed from the key but you believe it is correct, please bring it up in the next office hours.

Keep up the hard work!

- Professor Baker