MAT102 Test 1 Prep

• 1
  • 1
    • Here is an example:
    • Let set $A=\{1,2,3\}$
    • It is bounded above as there is a $y$ which for all $x\in A$, $x\le y$.
    • Let $y=4$.
    • $4\in \mathbb{R}$
    • $A\subset \mathbb{R}$
    • $\mathrm{\forall }x\in A,x\le 4$
    • This can also hold true for $y=3$, as the condition is $x\le y$ not $x<y$
  • 2
    • We want to show that there is a set $A$ which is bounded above with a unique upper bound.
    • Unique meaning only a single upper bound for the set $A$.
    • Let's assume this is true.
    • Definition of an upper bound is $\mathrm{\forall }x\in A,x\le y$
    • Let $z=y+1$
    • If $y$ is an upper bound for $A$, and $z>y$, then $z$ must be an upper bound too.
    • $\mathrm{\forall }x\in A,x\le y\le z$
    • We assumed that there was a unique upper bound for the set $A$. We have just found another bound $z$ which is also an upper bound for the set $A$.
    • This is a contradiction, we can't have only one upper bound when we have clearly found two upper bounds.
  • 3
    • To be bounded below means that:
      • $\mathrm{\forall }x\in A,\mathrm{\exists }y\in \mathbb{R}(x\ge y)$
• 2
  • 1
    • We want to compute:
      • $A\cap B$
      • $A\cap C$
      • $B\cup C$

      ---

    • We want to show that $A\cap (B\cup C)=(A\cap B)\cup (A\cap C)$

    ---

    • $A=\{1,2,3,4,5\}$
      • This is true regardless whether we include $0\in \mathbb{N}$ or not.
    • $B=\{\text{all even integers}\}$ or $B=\{x\in \mathbb{Z}:x\text{ is even}\}$
    • $C=\{-2,-1,0,1,2\}$

    ---

    • $A\cap B$
      • This is the even integers from $A$.
      • $A\cap B=\{2,4\}$
    • $A\cap C$
      • $A\cap C=\{1,2\}$
    • $B\cup C$
      • $B\cup C=\{x\in \mathbb{Z}:x\text{ is even}\}\cup \{-1,1\}$
    • Now let's show that $A\cap (B\cup C)=(A\cap B)\cup (A\cap C)$
    • Let $x\in A\cap (B\cup C)$
    • This means $x\in A$ and $x\in B\cup C$
    • If $x\in A$ and it's also in $B\cup C$, that means that $x\in A\cap C$ or that $x\in A\cap B$
    • So $A\cap (B\cup C)=(A\cap B)\cup (A\cap C)$.
    • Let's show this another way.
    • $A\cap (B\cup C)$
      • $A=\{1,2,3,4,5\}$
      • $B=\{\text{all even integers}\}$ or $B=\{x\in \mathbb{Z}:x\text{ is even}\}$
      • $C=\{-2,-1,0,1,2\}$
      • $B\cup C=\{x\in \mathbb{Z}:x\text{ is even}\}\cup \{-1,1\}$
      • $A\cap (B\cup C)=\{1,2,4\}$
      • $(A\cap B)=\{2,4\}$
      • $(A\cap C)=\{1,2\}$
      • $(A\cap B)\cup (A\cap C)=\{1,2,4\}$
  • 2
    • Prove that $A\cap (B\cup C)=(A\cap B)\cup (A\cup C)$
    • Let's proceed with double subset inclusion.
    • Let $x\in A\cap (B\cup C)$
    • This means $x\in A$ and $x\in B\cup C$
    • Thus follows $x\in B$ or $x\in C$
    • However $x$ must be in $A$ as well as either $B\cup C$
    • So this means $x\in A\cap C$ or that $x\in A\cap B$
    • So $x\in (A\cap B)\cup (A\cap C)$

    ---

    • $x\in A\cap B$ or $x\in A\cap C$
    • $x\in A$ and $x\in B\cup C$
    • Which means $x\in A\cap (B\cup C)$
• 3
• 3.1
• $P⟹Q\equiv \mathrm{\neg }P\vee Q$

• 3.2

• 4
  • 1
    • $\mathrm{\forall }m\in \mathbb{Z},\mathrm{\exists }n\in \mathbb{Z},m+n\text{ is even}$
    • For all integer numbers, there exists another integer where the sum of both numbers is even.
    • Is this true? I would say so, no matter what integer you have, you can always add something to make it even.
    • Example:
      • Let $m\in Z$.
      • For any value of $m$. If $m$ is even, we are done, because we can set $n=0$ and have $m+0$ resulting in an even number.
      • If $m$ is odd, this means that $\frac{m}{2}\notin \mathbb{Z}$.
      • Let's set $n=1$ in the case that $m$ is odd.
      • let's set $m=2k+n$
      • $m=2k+1$ which is the definition of an odd integer.
      • Sub back into the original.
      • We get $2k+1+n$
      • $2k+1+1$
      • $2k+2$
      • $(2)(k+1)$
      • If we are able to factor out 2, this means that $m$ is even. So we found a number which results in an even $m$.
  • 2
    • There exists an integer number where for every integer number, the sum of both are even.
    • This is false
    • Example:
      • Let's negate what we are trying to prove, and then prove that.
      • $\mathrm{\neg }(\mathrm{\exists }m\in \mathbb{Z},\mathrm{\forall }n\in \mathbb{Z},m+n\text{ is even})$
      • $\mathrm{\forall }m\in \mathbb{Z},\mathrm{\exists }n\in \mathbb{Z},m+n\text{ is odd}$
      • Now we are trying to prove that for all $m$ there is an $n$ where the sum $m+n$ is odd.
      • Case 1: $m$ is odd
        • We are done, set $n=0$ and now $\text{odd}+0=\text{odd}$
      • Case 2: $m$ is even
        • This means that $m=2k$ for some $k\in \mathbb{Z}$.
        • If this is the case, we can set $n=1$.
        • Now $m+n=2k+1$
        • $2k+1$ is the definition of an odd number for some $k\in \mathbb{Z}$.
      • Thus we have found a counter example for our original statement.
      • We have found an $n$ value where $m+n$ is NOT even.

Multiple Choice

• 1
  • $A\setminus B\subseteq C$
• 2
  • b
• 3
• 4
  • $\mathrm{\forall }x\in A,\mathrm{\forall }y\in B,\mathrm{\exists }z\in A\cap B,|x-y|<1⟹|x-z|<\frac{1}{2}$
  • $\mathrm{\exists }x\in A,\mathrm{\exists }y\in B,\mathrm{\forall }z\in A\cap B,\mathrm{\neg }(|x-y|<1⟹|x-z|\ge \frac{1}{2})$
  • $\mathrm{\exists }x\in A,\mathrm{\exists }y\in B,\mathrm{\forall }z\in A\cap B,\mathrm{\neg }(\mathrm{\neg }|x-y|<1\vee |x-z|\ge \frac{1}{2})$
  • $\mathrm{\exists }x\in A,\mathrm{\exists }y\in B,\mathrm{\forall }z\in A\cap B,|x-y|<1\wedge \mathrm{\neg }|x-z|\ge \frac{1}{2})$
  • $\mathrm{\exists }x\in A,\mathrm{\exists }y\in B,\mathrm{\forall }z\in A\cap B,|x-y|<1\wedge |x-z|\ge \frac{1}{2})$
• 5

• 8