Cards In This Deck

• $\text{What is the very first step when asked to prove a statement by contradiction?}$
  $\text{Write down the negation of the original statement (assume the statement is not true).}$
• $$\begin{gathered} \text{True or False?} \\ \text{The negation of the statement "All multiples of 5 are odd" is "No multiples of 5 are odd".} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Negation: "At least one multiple of 5 is even."} \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \text{To prove by contradiction that "If }{n}^2\text{ is even, then }n\text{ must be even", the assumption should be "If }{n}^2\text{ is odd, then }n\text{ is odd".} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Assumption: "There exists a number }n\text{ such that }{n}^2\text{ is even but }n\text{ is odd."} \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \text{If your proof by contradiction leads to a result like }4n^2=2b^2\text{ (meaning }2n^2=b^2\text{), this automatically means }b\text{ is even.} \end{gathered}$$
  $$\begin{gathered} \text{✅ True} \\ \text{If }{b}^2=2\times (aninteger)\text{, then }{b}^2\text{ is even. If a square number is even, its root (}b\text{) must also be even.} \end{gathered}$$
• $\text{In the proof that }\sqrt{2}\text{ is irrational, after assuming }\sqrt{2}=\frac{a}{b}\text{ and showing that }{a}^2=2b^2\text{ implies }a\text{ is even (}a=2k\text{), what is the next logical step?}$
  $\text{Substitute }a=2k\text{ back into the equation }{a}^2=2b^2\text{ to show that }b\text{ must also be even.}$
• $\text{When defining a rational number }\frac{a}{b}\text{ at the start of a proof by contradiction (e.g., for }\sqrt{2}\text{), what two conditions must }a\text{ and }b\text{ satisfy?}$
  $$\begin{gathered} \text{1. }a\text{ and }b\text{ are integers.} \\ \text{2. }a\text{ and }b\text{ have no common factors (the fraction is in simplest form).} \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \text{In the proof of the infinity of primes, the constructed number }N=p_1{p}_2...p_n+1\text{ is always a prime number itself.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Conclusion: }N\text{ is either prime itself OR it has a prime factor that is not in the original list.} \end{gathered}$$
• $\text{When proving the infinity of primes by contradiction, if you assume there is a finite list of primes }{p}_1,p_2,...,p_n\text{, what specific number }N\text{ must you construct?}$
  $N=(p_1\times p_2\times ...\times p_n)+1$
• $$\begin{gathered} \text{You are proving by contradiction that "if }{n}^2\text{ is a multiple of 3, then }n\text{ is a multiple of 3". } \\ \text{What algebraic forms should you use for }n\text{ to represent the assumption that }n\text{ is not a multiple of 3?} \end{gathered}$$
  $n=3k+1\text{ or }n=3k+2$