Section 2.6 Chapter Review

Negation: The power goes off and the food does not spoil.

Converse: If the food spoils, then the power went off.

Contrapositive: If the food does not spoil, then the power did not go off.

Negation: The door is closed and the light is on.

Converse: If the light is off then the door is closed.

Contrapositive: If the light is on then the door is open.

Negation: \(\exists x (x \lt 1 \wedge x^2 \ge 1)\)

Converse: \(\forall x( x^2 \lt 1 \to x \lt 1)\)

Contrapositive: \(\forall x (x^2 \ge 1 \to x \ge 1)\text{.}\)

Negation: There is a natural number \(n\) which is prime but not solitary.

Converse: For all natural numbers \(n\text{,}\) if \(n\) is solitary, then \(n\) is prime.

Contrapositive: For all natural numbers \(n\text{,}\) if \(n\) is not solitary then \(n\) is not prime.

Negation: There is a function which is differentiable and not continuous.

Converse: For all functions \(f\text{,}\) if \(f\) is continuous then \(f\) is differentiable.

Contrapositive: For all functions \(f\text{,}\) if \(f\) is not continuous then \(f\) is not differentiable.

Negation: There are integers \(a\) and \(b\) for which \(a\cdot b\) is even but \(a\) or \(b\) is odd.

Converse: For all integers \(a\) and \(b\text{,}\) if \(a\) and \(b\) are even then \(ab\) is even.

Contrapositive: For all integers \(a\) and \(b\text{,}\) if \(a\) or \(b\) is odd, then \(ab\) is odd.

Negation: There are integers \(x\) and \(y\) such that for every integer \(n\text{,}\) \(x \gt 0\) and \(nx \le y\text{.}\)

Converse: For every integer \(x\) and every integer \(y\) there is an integer \(n\) such that if \(nx > y\) then \(x > 0\text{.}\)

Contrapositive: For every integer \(x\) and every integer \(y\) there is an integer \(n\) such that if \(nx \le y\) then \(x \le 0\text{.}\)

Negation: There are real numbers \(x\) and \(y\) such that \(xy = 0\) but \(x \ne 0\) and \(y \ne 0\text{.}\)

Converse: For all real numbers \(x\) and \(y\text{,}\) if \(x = 0\) or \(y = 0\) then \(xy = 0\)

Contrapositive: For all real numbers \(x\) and \(y\text{,}\) if \(x \ne 0\) and \(y \ne 0\) then \(xy \ne 0\text{.}\)

Negation: There is at least one student in Math 321 who does not understand implications but will still pass the exam.

Converse: For every student in Math 321, if they fail the exam, then they did not understand implications.

Contrapositive: For every student in Math 321, if they pass the exam, then they understood implications.

The statement is true. If \(n\) is an even integer less than or equal to 7, then the only way it could not be negative is if \(n\) was equal to 0, 2, 4, or 6.

There is an integer \(n\) such that \(n\) is even and \(n \le 7\) but \(n\) is not negative and \(n \not\in \{0,2,4,6\}\text{.}\) This is false, since the original statement is true.

For all integers \(n\text{,}\) if \(n\) is not negative and \(n \not\in\{0,2,4,6\}\) then \(n\) is odd or \(n > 7\text{.}\) This is true, since the contrapositive is equivalent to the original statement (which is true).

For all integers \(n\text{,}\) if \(n\) is negative or \(n \in \{0,2,4,6\}\) then \(n\) is even and \(n \le 7\text{.}\) This is false. \(n = -3\) is a counterexample.

For any number \(x\text{,}\) if it is the case that adding any number to \(x\) gives that number back, then multiplying any number by \(x\) will give 0. This is true (of the integers or the reals). The “if” part only holds if \(x = 0\text{,}\) and in that case, anything times \(x\) will be 0.

The converse in words is this: for any number \(x\text{,}\) if everything times \(x\) is zero, then everything added to \(x\) gives itself. Or in symbols: \(\forall x (\forall z (x \cdot z = 0) \to \forall y (x + y = y))\text{.}\) The converse is true: the only number which when multiplied by any other number gives 0 is \(x = 0\text{.}\) And if \(x = 0\text{,}\) then \(x + y = y\text{.}\)

The contrapositive in words is: for any number \(x\text{,}\) if there is some number which when multiplied by \(x\) does not give zero, then there is some number which when added to \(x\) does not give that number. In symbols: \(\forall x (\exists z (x\cdot z \ne 0) \to \exists y (x + y \ne y))\text{.}\) We know the contrapositive must be true because the original implication is true.

The negation: there is a number \(x\) such that any number added to \(x\) gives the number back again, but there is a number you can multiply \(x\) by and not get 0. In symbols: \(\exists x (\forall y (x + y = y) \wedge \exists z (x \cdot z \ne 0))\text{.}\) Of course since the original implication is true, the negation is false.

If you have lost weight, then you exercised.

If you exercise, then you will lose weight.

If you are American, then you are patriotic.

If you are patriotic, then you are American.

If a number is rational, then it is real.

If a number is not even, then it is prime. (Or the contrapositive: if a number is not prime, then it is even.)

If the Broncos don’t win the Super Bowl, then they didn’t play in the Super Bowl. Alternatively, if the Broncos play in the Super Bowl, then they will win the Super Bowl.

Direct proof.

Proof.

Let \(n\) be an integer. Assume \(n\) is odd. So \(n = 2k+1\) for some integer \(k\text{.}\) Then

\begin{equation*} 7n = 7(2k+1) = 14k + 7 = 2(7k +3) + 1. \end{equation*}

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Since \(7k + 3\) is an integer, we see that \(7n\) is odd.

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The converse is: for all integers \(n\text{,}\) if \(7n\) is odd, then \(n\) is odd. We will prove this by contrapositive.

Proof.

Let \(n\) be an integer. Assume \(n\) is not odd. Then \(n = 2k\) for some integer \(k\text{.}\) So \(7n = 14k = 2(7k)\) which is to say \(7n\) is even. Therefore \(7n\) is not odd.

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