Kim says that under f(x) equal to x squared, the preimage of {9} always has exactly one element. Is Kim right?

Let f map real numbers to real numbers by f(x) equal to x squared. Which fact shows f is not injective?

• f(2) equals f(-2), both equal to 4
• f(0) equals 0
• f(3) equals 9
• f(1) equals 1

To prove that two sets A and B are equal by double inclusion, which two containments do you show?

• Only B inside A, shown twice in a row
• A inside B and B inside A
• That both sets are empty

With f(x) equal to x squared, what is the image of the set containing minus 2 and 3?

• The set containing minus 2 and 3
• The set containing 2 and 3
• The set containing 4 and 9

Kim says that under f(x) equal to x squared, the preimage of the set containing 9 holds exactly one element.

With f(x) equal to x squared, what is the image of the set {-2, 3} and the preimage of the set {4}?

• Image {4, 9}; preimage {-2, 2}
• Image {-2, 3}; preimage {4}
• Image {4, 9}; preimage {2}
• Image {2, 3}; preimage {-4, 4}

To prove that A union (B intersect C) equals (A union B) intersect (A union C) by double inclusion, what is the correct first half?

• Take an arbitrary x in the left hand side and show it lies in the right hand side
• List every element of A, B and C
• Assume both sides are empty
• Prove the converse statement instead

You have shown that every element of the left side lies on the right side. What still remains in a double inclusion proof?

• The reverse chase back to the left
• A longer Venn picture
• A check that both sides are nonempty

Suppose f and g are injective and f(g(a)) equals f(g(b)). What follows?

• Nothing follows here without several more checks first
• The composition of f and g must be constant
• a equals b by injectivity at both layers

Lee proved that the left side sits inside the right side and stopped, then claimed the sets are equal. What is wrong?

• The reverse containment is still unproved
• Nothing is wrong, half the proof is enough
• The two circle Venn picture is required