# 2.3: Implications

- Page ID
- 8388

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Most theorems in mathematics appear in the form of compound statements called conditional and biconditional statements. We shall study biconditional statement in the next section. Conditional statements are also called implications.

An ** implication** is the compound statement of the form “if \(p\), then \(q\).” It is denoted \(p \Rightarrow q\), which is read as “\(p\) implies \(q\).” It is false only when \(p\) is true and \(q\) is false, and is true in all other situations.

\(p\) | \(q\) | \(p \Rightarrow q\) |
---|---|---|

T | T | T |

T | F | F |

F | T | T |

F | F | T |

The statement \(p\) in an implication \(p \Rightarrow q\) is called its ** hypothesis**,

**, or**

*premise***, and \(q\) the**

*antecedent***or**

*conclusion***.**

*consequence*Implications come in many disguised forms. There are several alternatives for saying \(p \Rightarrow q\). The most common ones are

- \(p\) implies \(q\),
- \(p\) only if \(q\),
- \(q\) if \(p\),
- \(q\), provided that \(p\).

All of them mean \(p\Rightarrow q\).

Implications play a key role in logical argument. If an implication is known to be true, then whenever the hypothesis is met, the consequence must be true as well. This is why an implication is also called a ** conditional statement**.

Example \(\PageIndex{1}\label{eg:imply-01}\)

The quadratic formula asserts that \[b^2-4ac>0 \quad \Rightarrow \quad ax^2+bx+c=0 \mbox{ has two distinct real solutions}.\] Consequently, the equation \(x^2-3x+1=0\) has two distinct real solutions because its coefficients satisfy the inequality \(b^2-4ac>0\).

hands-on exercise \(\PageIndex{1}\label{he:imply-01}\)

More generally,

- If \(b^2-4ac>0\), then the equation \(ax^2+bx+c=0\) has two distinct real solutions. In fact, \(ax^2+bx+c = a(x-r_1)(x-r_2)\), where \(r_1\neq r_2\) are the two distinct roots.
- If \(b^2-4ac=0\), then the equation \(ax^2+bx+c=0\) has only one real solution \(r\). In such an event, \(ax^2+bx+c = a(x-r)^2\). Consequently, we call \(r\) a repeated root.
- If \(b^2-4ac=0\), then the equation \(ax^2+bx+c=0\) has no real solution.

Use these results to determine how many solutions these equations have:

- \(4x^2+12x+9=0\)
- \(2x^2-3x-4=0\)
- \(x^2+x=-1\)

Example \(\PageIndex{2}\label{eg:imply-02}\)

We have remarked earlier that many theorems in mathematics are in the form of implications. Here is an example:

- If \(|r|<1\), then \(1+r+r^2+r^3+\cdots = \text{F}rac{1}{1-r}\).
- It means, symbolically, \(|r|<1 \Rightarrow 1+r+r^2+r^3+\cdots = \text{F}rac{1}{1-r}\).

hands-on exercise \(\PageIndex{2}\label{he:imply-02}\)

Express the following statement in symbols:

If \(x>y>0\), then \(x^2>y^2\).

Example \(\PageIndex{3}\label{eg:imply-03}\)

If a father promises his kids, “If tomorrow is sunny, we will go to the beach,” the kids will take it as a true statement. Consequently, if they wake up the next morning and find it sunny outside, they expect they will go to the beach. The father breaks his promise (hence making the implication false) only when it is sunny but he does not take his kids to the beach.

If it is cloudy outside the next morning, they do not know whether they will go to the beach, because no conclusion can be drawn from the implication (their father’s promise) if the weather is bad. Nonetheless, they may still go to the beach, even if it rains! Since their father does not contradict his promise, the implication is still true.

Many students are bothered by the validity of an implication even when the hypothesis is false. It may help if we understand how we use an implication.

**Solution**

Assume we want to show that a certain statement \(q\) is true.

- First, we find a result of the form \(p\Rightarrow q\). If we cannot find one, we have to prove that \(p\Rightarrow q\) is true.
- Next, show that the hypothesis \(p\) is fulfilled.
- These two steps together allow us to draw the conclusion that \(q\) must be true.

Consequently, if \(p\) is false, we are not expected to use the implication \(p\Rightarrow q\) at all. Since we are not are going to use it, we can define its truth value to anything we like. Nonetheless, we have to maintain [pg:consistence] consistency with other logical connectives. We will give a justification of our choice at the end of the next section.

Example \(\PageIndex{4}\label{eg:imply-04}\)

To show that “if \(x=2\), then \(x^2=4\)” is true, we need not worry about those \(x\)-values that are not equal to 2, because the implication is immediately true if \(x\neq 2\). It suffices to assume that \(x=2\), and try to prove that we *will* get \(x^2=4\). Since we do have \(x^2=4\) when \(x=2\), the validity of the implication is established.

In contrast, to determine whether the implication “if \(x^2=4\), then \(x=2\)” is true, we assume \(x^2=4\), and try to determine whether \(x\) *must* be 2. Since \(x = -2\) makes \(x^2=4\) true but \(x=2\) false, the implication is false.

In general, to disprove an implication, it suffices to find a counterexample that makes the hypothesis true and the conclusion false.

hands-on exercise \(\PageIndex{3}\label{he:imply-0}\)

Determine whether these two statements are true or false:

- If \((x-2)(x-3)=0\), then \(x=2\).
- If \(x=2\), then \((x-2)(x-3)=0\).

Explain.

Example \(\PageIndex{5}\label{eg:imply-05}\)

Although we said examples can be used to disprove a claim, examples alone can *never* be used as proofs. If you are asked to show that

\[\mbox{if $x>2$, then $x^2>4$},\]

you cannot prove it by *checking* just a few values of \(x\), because you may find a counterexample after trying a few more calculations. Therefore, examples are only for illustrative purposes, they are *not* acceptable as proofs.

Example \(\PageIndex{6}\label{eg:imply-06}\)

The statement

“If a triangle \(PQR\) is isosceles, then two of its angles have equal measure.”

takes the form of an implication \(p\Rightarrow q\), where

\[\begin{array}{l@{\quad}l} p: & \mbox{The triangle $PQR$ is isosceles} \\ q: & \mbox{Two of the angles of the triangle $PQR$ have equal measure} \end{array}\] I

n this example, we have to rephrase the statements \(p\) and \(q\), because each of them should be a stand-alone statement. If we leave \(q\) as “two of its angles have equal measure,” it is not clear what “its” is referring to. In addition, it is a good habit to spell out the details. It helps us focus our attention on what we are investigating.

Example \(\PageIndex{7}\label{eg:isostrig}\)

The statement

“A square must also be a parallelogram.”

can be expressed as an implication: “if the quadrilateral \(PQRS\) is a square, then the quadrilateral \(PQRS\) is a parallelogram.”

Likewise, the statement

“All isosceles triangles have two equal angles.”

can be rephrased as “if the triangle \(PQR\) is isosceles, then the triangle \(PQR\) has two equal angles.” Since we have expressed the statement in the form of an implication, we no longer need to include the word “all.”

hands-on exercise \(\PageIndex{1}\label{he:imply-04}\)

Rewrite each of these logical statements:

- Any square is also a parallelogram.
- A prime number is an integer.
- All polynomials are differentiable.

as an implication \(p\Rightarrow q\). Specify what \(p\) and \(q\) are.

Example \(\PageIndex{8}\label{eg:imply-08}\)

What does “\(p\) unless \(q\)” translate into, logically speaking? We know that \(p\) is true, provided that \(q\) does not happen. It means, in symbol, \(\overline{q}\Rightarrow p\). Therefore,

The quadrilateral \(PQRS\) is not a square unless the quadrilateral \(PQRS\) is a parallelogram

is the same as saying

If a quadrilateral \(PQRS\) is not a parallelogram, then the quadrilateral \(PQRS\) is not a square.

Equivalently, “\(p\) unless \(q\)” means \(\overline{p}\Rightarrow q\), because \(q\) is a necessary condition that prevents \(p\) from happening.

Given an implication \(p \Rightarrow q\), we define three related implications:

- Its
is defined as \(q \Rightarrow p\).*converse* - Its
is defined as \(\overline{p} \Rightarrow \overline{q}\).*inverse* - Its
is defined as \(\overline{q} \Rightarrow \overline{p}\).*contrapositive*

Among them, the contrapositive \(\overline{q}\Rightarrow\overline{p}\) is the most important one. We shall study it again in the next section.

Example \(\PageIndex{9}\label{eg:imply-09}\)

The converse, inverse, and contrapositive of “\(x>2\Rightarrow x^2>4\)” are listed below. \[% \arraygap{1.25} \begin{array}{l@{\quad}rcl} \mbox{converse:} & x^2>4 &\Rightarrow& x>2, \\ \mbox{inverse:} & x\leq2 &\Rightarrow& x^2\leq4, \\ \mbox{contrapositive:}& x^2\leq4 &\Rightarrow& x\leq2. \end{array}\] We can change the notation when we negate a statement. If it is appropriate, we may even rephrase a sentence to make the negation more readable.

Example \(\PageIndex{5}\label{he:imply-05}\)

List the converse, inverse, and contrapositive of the statement “if \(p\) is prime, then \(\sqrt{p}\) is irrational.”

The inverse of an implication is seldom used in mathematics, so we will only study the truth values of the converse and contrapositive.

\[\begin{array}{|*{7}{c|}} \hline p & q & p\Rightarrow q & q\Rightarrow p & \overline{q} & \overline{p} & \overline{q}\Rightarrow\overline{p} \\ \hline \text{T} & \text{T} & \text{T} & \text{T} & \text{F} & \text{F} & \text{T} \\ \text{T} & \text{F} & \text{F} & \text{T} & \text{T} & \text{F} & \text{F} \\ \text{F} & \text{T} & \text{T} & \text{F} & \text{F} & \text{T} & \text{T} \\ \text{F} & \text{F} & \text{T} & \text{T} & \text{T} & \text{T} & \text{T} \\ \hline \end{array}\]

An implication and its contrapositive always have the same truth value, but this is not true for the converse. What this means is, even though we know \(p\Rightarrow q\) is true, there is no guarantee that \(q\Rightarrow p\) is also true. This is an important observation, especially when we have a theorem stated in the form of an implication. So let us say it again:

\[\fbox{The converse of a theorem in the form of an implication may not be true.}\]

Accordingly, if you only know that \(p\Rightarrow q\) is true, do not assume that its converse \(q\Rightarrow p\) is also true. Likewise, if you are asked to prove that \(p\Rightarrow q\) is true, do not attempt to prove \(q\Rightarrow p\), because these two implications are not the same.

Example \(\PageIndex{10}\label{eg:imply-provingID}\)

We know that \(p\Rightarrow q\) does not necessarily mean we also have \(q\Rightarrow p\). This important observation explains the invalidity of the “proof” of \(21=6\) in Example [eg:wrongpf2].

\[\begin{eqnarray*}

21 &=& 6 \\

6 &=& 21 \\

27 &=& 27

\end{eqnarray*}\]

The argument we use here consists of three equations, but they are not individual unrelated equations. They are connected by implication.

\[\begin{eqnarray*}

\phantom{\Rightarrow\qquad} 21 &=& 6 \\

\Rightarrow\qquad\phantom{2} 6 &=& 21 \\

\Rightarrow\qquad 27 &=& 27

\end{eqnarray*}\]

Since implications are not reversible, even though we do have \(27=27\), we cannot use this fact to prove that \(21=6\). After all, an implication is true if its hypothesis is false. Therefore, having a true implication does not mean that its hypothesis must be true. In this example, the logic is sound, but it does not prove that \(21=6\).

There are two other ways to describe an implication \(p\Rightarrow q\) in words. They are completely different from the ones we have seen thus far. They focus on whether we can tell one of the two components \(p\) and \(q\) is true or false if we know the truth value of the other.

- \(p\) is a
for \(q\)*sufficient condition* - \(q\) is a
for \(p\).*necessary condition*

They are difficult to remember, and can be easily confused. You may want to visualize it pictorially:

\[\fbox{$\mbox{sufficient condition} \Rightarrow

\mbox{necessary condition}$.}\]

The idea is, assuming that \(p\Rightarrow q\) is true, then

- For \(q\) to be true, it is enough to know or show that \(p\) is true. Hence, knowing \(p\) is true alone is sufficient for us to draw the conclusion the \(q\) must also be true.
- For \(p\) to be true, it is necessary to have \(q\) be true as well. Thus, knowing \(q\) is true does not necessarily mean that \(p\) must be true.

Example \(\PageIndex{11}\label{eg:imply-11}\)

Consider the implication \[x=1 \Rightarrow x^2=1.\] If \(x=1\), we must have \(x^2=1\). So, knowing \(x=1\) is enough for us to conclude that \(x^2=1\). We say that \(x=1\) is a sufficient condition for \(x^2=1\).

If \(x=1\), it is necessarily true that \(x^2=1\), because, for example, it is impossible to have \(x^2=2\). Nonetheless, knowing \(x^2=1\) alone is not enough for us to decide whether \(x=1\), because \(x\) can be \(-1\). Therefore, \(x^2=1\) is *not* a sufficient condition for \(x=1\). Instead, \(x^2=1\) is only a necessary condition for \(x=1\).

hands-on exercise \(\PageIndex{6}\label{he:imply-06}\)

Write these statements:

- For \(x^2>1\), it is sufficient that \(x>1\).
- For \(x^2>1\), it is necessary that \(x>1\).

in the form of \(p\Rightarrow q\). Be sure to specify what \(p\) and \(q\) are.

## Summary and Review

- An implication \(p\Rightarrow q\) is false only when \(p\) is true and \(q\) is false.
- This is how we typically use an implication. Assume we want to show that \(q\) is true. We have to find or prove a theorem that says \(p\Rightarrow q\). Next, we need to show that hypothesis \(p\) is met, hence it follows that \(q\) must be true.
- An implication can be described in several other ways. Can you name a few of them?
- Converse, inverse, and contrapositive are obtained from an implication by switching the hypothesis and the consequence, sometimes together with negation.
- In an implication \(p\Rightarrow q\), the component \(p\) is called the sufficient condition, and the component \(q\) is called the necessary condition.

Exercise \(\PageIndex{1}\label{ex:imply-01}\)

Let \(p\), \(q\), and \(r\) represent the following statements:

\(p\): | Sam had pizza last night. |

\(q\): | Chris finished her homework. |

\(r\): | Pat watched the news this morning. |

Give a formula (using appropriate symbols) for each of these statements:

- If Sam had pizza last night then Chris finished her homework.
- Pat watched the news this morning only if Sam had pizza last night.
- Chris finished her homework if Sam did not have pizza last night.
- It is not the case that if Sam had pizza last night, then Pat watched the news this morning.
- Sam did not have pizza last night and Chris finished her homework implies that Pat watched the news this morning.

Exercise \(\PageIndex{2}\label{ex:imply-02}\)

Define the propositional variables as in Problem 1. Express in words the statements represented by the following formulas.

- \(q\Rightarrow r\)
- \(p\Rightarrow(q\wedge r)\)
- \(\overline{p}\Rightarrow (q\vee r)\)
- \(r\Rightarrow(p\vee q)\)

Exercise \(\PageIndex{3}\label{ex:imply-03}\)

Consider the following statements:

\(p\): | Niagara Falls is in New York. |

\(q\): | New York City is the state capital of New York. |

\(r\): | New York City will have more than 40 inches of snow in 2525. |

The statement \(p\) is true, and the statement \(q\) is false. Represent each of the following statements by a formula. What is their truth value if \(r\) is true? What if \(r\) is false?

- If Niagara Falls is in New York, then New York City is the state capital of New York.
- Niagara Falls is in New York only if New York City will have more than 40 inches of snow in 2525.
- Niagara Falls is in New York or New York City is the state capital of New York implies that New York City will have more than 40 inches of snow in 2525.
- For New York City to be the state capital of New York, it is necessary that New York City will have more than 40 inches of snow in 2525.e
- For Niagara Falls to be in New York, it is sufficient that New York City will have more than 40 inches of snow in 2525.

Exercise \(\PageIndex{4}\label{ex:imply-04}\)

Express each of the following compound statements symbolically:

- If the triangle \(ABC\) is equilateral, then it is isosceles.
- If \(\sqrt{47089}\) is greater than 200 and \(\sqrt{47089}\) is an integer, then \(\sqrt{47089}\) is prime.
- If \(\sqrt{47089}\) is greater than 200, then, if \(\sqrt{47089}\) is prime, it is greater than 210.
- The line \(L_1\) is perpendicular to the line \(L_2\) and the line \(L_2\) is parallel to the line \(L_3\) implies that \(L_1\) is perpendicular to \(L_3\).
- If \(x^3-3x^2+x-3=0\), then either \(x\) is positive or \(x\) is negative or \(x=0\).

Exercise \(\PageIndex{5}\label{ex:imply-05}\)

Express each of the following compound statements in symbols.

- \(x^3-3x^2+x-3=0\) only if \(x=3\).
- A necessary condition for \(x^3-3x^2+x-3=0\) is \(x=3\).
- A sufficient condition for \(x^3-3x^2+x-3=0\) is \(x=3\).
- If \(e^\pi\) is a real number, then \(e^\pi\) is either rational or irrational.
- All NFL players are huge.

Exercise \(\PageIndex{6}\label{ex:imply-06}\)

Find the converse, inverse, and contrapositive of the following implication:

If the quadrilateral \(ABCD\) is a rectangle, then \(ABCD\) is a parallelogram.

Exercise \(\PageIndex{7}\label{ex:imply-07}\)

Construct the truth tables for the following expressions:

- \((p\wedge q)\vee r\)
- \((p\vee q)\Rightarrow (p\wedge r)\)

**Hint**-
To help you get started, fill in the blanks.]

(a) \(\setlength{\arraycolsep}{3pt} \begin{array}[t]{|*{5}{c|}} \noalign{\vskip-9pt}\hline p & q & r & p\wedge q & (p\wedge q)\vee r \\ \hline \text{T} &\text{T} &\text{T} && \\ \text{T} &\text{T} &\text{F} && \\ \text{T} &\text{F} &\text{T} && \\ \text{T} &\text{F} &\text{F} && \\ \text{F} &\text{T} &\text{T} && \\ \text{F} &\text{T} &\text{F} && \\ \text{F} &\text{F} &\text{T} && \\ \text{F} &\text{F} &\text{F} && \\ \hline \end{array}\) (b) \(\begin{array}[t]{|c|c|c|c|c|c|} \noalign{\vskip-9pt}\hline p & q & r & p\vee q & p\wedge r & (p\vee q)\Rightarrow(p\wedge r) \\ \hline \text{T} &\text{T} &\text{T} &&& \\ \text{T} &\text{T} &\text{F} &&& \\ \text{T} &\text{F} &\text{T} &&& \\ \text{T} &\text{F} &\text{F} &&& \\ \text{F} &\text{T} &\text{T} &&& \\ \text{F} &\text{T} &\text{F} &&& \\ \text{F} &\text{F} &\text{T} &&& \\ \text{F} &\text{F} &\text{F} &&& \\ \hline \end{array}\)

Exercise \(\PageIndex{8}\label{ex:imply-08}\)

Construct the truth tables for the following expressions:

- \((p\Rightarrow q) \vee (\overline{p}\Rightarrow q)\)
- \((p\Rightarrow q) \wedge (\overline{p}\Rightarrow q)\)

Exercise \(\PageIndex{9}\label{ex:imply-09}\)

Determine (you may use a truth table) the truth value of \(p\) if

- \((p\wedge q)\Rightarrow (q\vee r)\) is false
- \((q\wedge r)\Rightarrow (p\wedge q)\) is false

Exercise \(\PageIndex{10}\label{ex:imply-10}\)

Assume \(p\Rightarrow q\) is true.

- If \(p\) is true, must \(q\) be true? Explain.
- If \(p\) is false, must \(q\) be true? Explain.
- If \(q\) is true, must \(p\) be false? Explain.
- If \(q\) if false, must \(p\) be false? Explain.