Try each one on paper first, then check a single step. That is worth far more than reading a finished solution.
Example 1 — building a truth table
Construct a truth table for ¬p∨q and show that it is logically equivalent to p→q.
- 1Two propositions means 22=4 rows.List every combination of T and F systematically.
- 2Row p=T,q=T: ¬p=F, so ¬p∨q=F∨T=TA disjunction needs only one true part.
- 3Row p=T,q=F: ¬p=F, so F∨F=FThe only false row.
- 4Row p=F,q=T: ¬p=T, so T∨T=T
- 5Row p=F,q=F: ¬p=T, so T∨F=T
- 6The column is T,F,T,T — identical to p→q.Same value in every row, so p→q≡¬p∨q.
Example 2 — converse, inverse and contrapositive
For "if x>3 then x2>9", write the converse, inverse and contrapositive, and say which are true.
- 1Here p is "x>3" and q is "x2>9".Identify the two parts before rearranging anything.
- 2Converse q→p: if x2>9 then x>3.False — take x=−4: x2=16>9 but x<3.
- 3Inverse ¬p→¬q: if x≤3 then x2≤9.Also false, and by the same counterexample x=−4.
- 4Contrapositive ¬q→¬p: if x2≤9 then x≤3.True.
- 5Only the contrapositive matches the original in truth value.It is logically equivalent; the converse and inverse are not.
Example 3 — proof by induction
Prove that 1+2+⋯+n=2n(n+1) for all positive integers n.
- 1Base case n=1: left side =1, right side =21×2=1 ✓Always verify the smallest case explicitly.
- 2Assume the result for n=k: 1+2+⋯+k=2k(k+1)This is the inductive hypothesis.
- 3Add the next term: 2k(k+1)+(k+1)Here is where the assumption is used.
- 4=(k+1)(2k+1)=(k+1)⋅2k+2Take out the common factor (k+1).
- 5=2(k+1)(k+2)Which is the formula with n=k+1 ✓
- 6By induction, the result holds for all n≥1.The conclusion sentence is part of the proof, not decoration.