Cardinality and Countably Infinite Sets

Cardinality and Countably Infinite Sets

Cardinality is a term used to describe the size of sets. Set A has the same cardinality as set B if a bijection exists between the two sets. We write this as |A| = |B|. One important type of cardinality is called “countably infinite.” A set A is considered to be countably infinite if a bijection exists between A and the natural numbers ℕ. Countably infinite sets are said to have a cardinality of אo (pronounced “aleph naught”).

Remember that a function f is a bijection if the following condition are met:

1. It is injective (“1 to 1”): f(x)=f(y)⟹x=y

2. It is surjective (“onto”): for all b in B there is some a in A such that f(a)=b.

A set is a bijection if it is both a surjection and in injection.

A set is a bijection if it is both a surjection and an injection.

Example 1. Show that the set of integers ℤ is countably infinite.

To show that ℤ is countably infinite, we must find a bijection between ℕ and ℤ, i.e. we need to find a way to match up each element of ℕ to a unique element of ℤ, and this function must cover each element in ℤ.

We can start by writing out a pattern. One pattern we can use is to count down starting at 0, then going back and “picking up” each positive integer. This follows the pattern {0, -1, 1, -2, 2, -3, 3…}. We match to ℕ to ℤ as follows:

0 1 2 3 4
0 -1 1 -2 2

Notice that each even natural number is matched up to it’s half. It follows the function f(n) = n/2.

The odd numbers follow the function f(n) = -(n+1)/2

We can write this as a piecewise function as:

$f(n) =

\begin{cases}

n/2, & \text{if n is even} \\

-(n+1)/2, & \text{if n is odd}

\end{cases} $

Now we need to check if our function is a bijection.

Injectivity: Suppose the function is not injective. Then there exists some natural numbers x and y such that f(x)=f(y) but x≠y.

For even integers, x/2 = y/2 ⟹ x=y

For odd integers, Then (-x+1)/2 = -(y+1)/2 ⟹ x=y

These are contradictions, so the function is injective.

Surjectivity: Suppose the function is not surjective. Then there is some integer k such that there is no n in ℕ for which f(n) = k.

For k≥0, k=n/2 ⟹2k=n ⟹ n is an even number in ℕ

For k<0, k=-(n+1)/2 ⇒ -2k-1 = n ⇒ n is an odd number in ℕ

These are contradictions, so the function is surjective.

Since f is both injective and surjective, it is a bijection. Therefore, |ℤ| = |N| =אo. ∎

This result is often surprising to students because the set ℕ is contained in the set ℤ.

Example 2. For A = {2n | n is a number in ℕ}, show that |A| = |ℤ| (the set of integers has the same cardinality as the set of even natural numbers).

We can either find a bijection between the two sets or find a bijection from each set to the natural numbers. Since we already found a bijection from ℤ to ℕ in the previous example, we will now find a bijection from A to ℕ.

One function that will work is f(n) = n/2. Checking that it is a bijection is very similar to Example 1.

Since |A| = |ℕ| and |ℤ| = |ℕ|, then |A| = |ℤ| = אo.∎

There are many sets that are countably infinite, ℕ, ℤ, 2ℤ, 3ℤ, nℤ, and ℚ. All of the sets have the same cardinality as the natural numbers ℕ. Some sets that are not countable include ℝ, the set of real numbers between 0 and 1, and ℂ.

Georg Cantor was a pioneer in the field of set theory and was the first to explore countably infinite sets

Georg Cantor was a pioneer in the field of set theory and was the first to explore countably infinite sets

 

Completing the Square

Completing the Square

– HT Goodwill

What Is Completing the Square?

Completing the Square is a technique in algebra that allows us to rewrite a quadratic expression that is in standard form in vertex form. That is

$ax^2 + bx + c$

so that $ax^2 + bx + c = a(x-h)^2 + k$.

Why Do We Complete the Square?

It should be noted that many times, students think (are taught?) that completing the square is just a way to solve a quadratic equation that has the form:

$latex ax^2 + bx + c = 0$.

However, in other areas of mathematics, we sometimes need to express a quadratic in the vertex form. The following are two examples of this.

Please note, you don’t have to understand these examples! That’s not the point of them!
\begin{itemize}
\item \textit{Integral Calculus:}
\begin{center}
\begin{tikzpicture}
\node [align=center] at (-1,0.75) {\scriptsize Cannot\\[-2mm]\scriptsize Integrate};
\node [align=center] at (1,0.75) {\scriptsize Can Now\\[-2mm] \scriptsize Integrate};
\node at (0,0) {\(\int \frac{dx}{x^2 + 4x + 5} = \int \frac{dx}{(x+2)^2 + 1}\)};
\draw [-latex] (-1,-0.3) arc [start angle=-180, end angle=0, x radius=1cm, y radius=0.5cm];
\node at (0,-1) [anchor=north,inner sep=0pt,align=center] {\scriptsize Complete the Square\\[-1mm] \scriptsize in Denominator};
\node at (3,-2) [anchor=east] {\tiny www.mathtutoringacademy.com};
\end{tikzpicture}
\end{center}
\item \textit{Differential Equations (Laplace Transforms):}
\begin{center}
\begin{tikzpicture}
% \draw[help lines,step=5mm] (-2,-1) grid (2.5,1);
\node [align = center] at (-1.25,1) {\scriptsize Cannot Find\\[-2mm] \scriptsize Transform};
\node [align=center] at (1.25,1) {\scriptsize Can Now Find\\[-2mm] \scriptsize Transform};
\node at (0,0) {\(\mathcal{L}^{-1}\left\{\frac{5}{s^2 + 4s + 5}\right\} = \mathcal{L}^{-1}\left\{\frac{5}{(s+2)^2 + 1}\right\}\)};
\draw [-latex] (-1.25,-0.3) arc [start angle=-180, end angle=0, x radius=1.5cm, y radius=0.5cm];
\node at (0,-1) [anchor=north,inner sep=0pt,align=center] {\scriptsize Complete the Square\\[-1mm] \scriptsize in Denominator};
\node at (3,-2) [anchor=east] {\tiny www.mathtutoringacademy.com};
\end{tikzpicture}
\end{center}

\end{itemize}

The two examples above are \textit{only} to show you that completing the square is used for \textit{other reasons} than to solve equations.\vskip 1.5cm

\textbf{\large Background (Leading up to \textit{How To\ldots})}
\vskip 2mm

Suppose we have a perfect square binomial that we expand using FOIL:

\begin{align*}
(x + B)^2 & = (x + B)(x + B)\\
& = x^2 + Bx + Bx + B^2\\
& = x^2 + 2Bx + B^2
\end{align*}

If we reverse this string of equations, we see that any quadratic that has this pattern: \(x^2 + 2Bx + B^2\) is a perfect square and we can factor it as \((x + B)^2\).

\[%
\underbrace{x^2 + 2Bx + B^2}_{\mbox{\parbox{3cm}{\centering \scriptsize The Perfect-Square\\Pattern\ldots}}} = \underbrace{(x+B)^2}_{\mbox{\parbox{3cm}{\centering\scriptsize \ldots always factors\\this way.}}}
\]\vskip 2mm

To Complete the Square, we adjust a quadratic expression so that it exhibits the perfect-square pattern. Here is an overview of the rest of the article so you can focus on what you need most.\vskip 1.5cm

\textbf{\large Introductory Examples}\vskip 2mm

\textbf{Example 1 How To Complete the Square}\vskip 2mm

Complete the square on \(x^2 + 14x + 40\).
\vskip 5mm

\textbf{Solution}\vskip 2mm

Match up the quadratic expression with the perfect square pattern, starting with the \(x^2\) term.

\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad x^2 + 2Bx + B^2\\
\mbox{Our Quadratic:} & \quad x^2 + 14x + 40
\end{align*}
\vskip 5mm

Step 1: Matching The Quadratic Terms\vskip 2mm

The quadratic terms match since they have the same coefficient.

\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad x^2 + 2Bx + B^2\\
\mbox{Our Quadratic:} & \quad x^2 + 14x + 40\\
\mbox{Match?} & \quad \checkmark
\end{align*}
\vskip 5mm

Step 2: Matching the Linear Terms\vskip 2mm

The \textit{value} of \(2B\) is assigned be us and is whatever we need it to be so we can match the perfect-square pattern. In this example, they match, if we assign \({\color{blue} 2B = 14}\).

\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad x^2 + {\color{blue} 2B}x + B^2\\
\mbox{Our Quadratic:} & \quad x^2 + {\color{blue}14}x + 40\\
\mbox{Match?} & \quad \checkmark\quad\quad \checkmark
\end{align*}
\vskip 5mm

Step 3: Matching the Constant Term\vskip 2mm

In order to match the perfect square pattern, the constant term has to be equal to \(B^2\). Since we defined \(2B = 14\) in Step 2, we know that \(B = 7\) which implies \({\color{red} B^2 = 7^2 = 49}\).
\vskip 2mm

To adjust our constant term, we add zero (which won’t change the value of anything) and choose to rewrite the zero as \(B^2 – B^2\).

\begin{align*}
x^2 + 14x + 40 & = x^2 + 14x + {\color{red} 0} + 40 &&\mbox{Adding }{\color{red} 0}.\\
& = x^2 + 14x + {\color{red} (49 – 49)} + 40 &&\mbox{Since } B^2 – B^2 = {\color{red} 49-49 = 0}\\
& = x^2 + 14x + 49 – 49 + 40\\
& = (x^2 + 14x + 49) – 49 + 40&&\mbox{Regrouping}
\end{align*}

Matching up to the pattern:

\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad x^2 + 2Bx + B^2\\
\mbox{Our Quadratic:} & \quad (x^2 + 14x + 49) – 49 + 40\\
\mbox{Match?} & \quad \checkmark\quad\quad \checkmark\quad\quad \checkmark
\end{align*}

Since the first three terms (the ones in the parentheses) match the perfect-square pattern, we know those three terms will factor into \((x+B)^2\).

\begin{align*}
{\color{blue}\overbrace{(x^2 + 14x + 49)}^{\mbox{Perfect Square}}} – 49 + 40
& = {\color{blue}\overbrace{(x+7)^2}^{\mbox{Factored}}} {\color{red}- 49 + 40}\\
& = (x+7)^2 {\color{red}- 9}
\end{align*}

\textbf{Answer:} \(x^2 + 14x + 40 = (x+7)^2 – 9\)
\vskip 1.5cm

\textbf{Example 2 (A \textit{Little} More Complicated)}\vskip 2mm
Complete the square on \(x^2 – 9x + 3\).
\vskip 5mm

\textbf{Solution} As before, we match up our quadratic with the perfect-square pattern. We’ll do this one a bit more quickly.\vskip 2mm

Step 1: Quadratic and Linear Terms

Both quadratic terms have a coefficient of \(1\) so they already match. The linear terms will match if we assign \({\color{blue}2B = -9}\). So without any real work or effort we match the first two terms already.

\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad x^2 + {\color{blue} 2B}x + B^2\\
\mbox{Our Quadratic:} & \quad x^2\,{\color{blue}-\,9}x + 3\\
\mbox{Match?} & \quad \checkmark \quad\,\, \checkmark
\end{align*}

Step 2: Constant Term

Since we assigned \({\color{blue} 2B = -9}\), we divide by 2 to get \(B = -\frac 9 2\). And we know our constant term needs to be
\[%
{\color{red}B^2 = \left(-\frac 9 2\right)^2 = \frac{81} 4},
\]

and we introduce it the same way we did before, by adding zero.

\begin{align*}
x^2 – 9x + 3
& = x^2 – 9x + {\color{red} 0} + 3 && \mbox{Adding }{\color{red} 0}\\
& = x^2 – 9x + {\color{red}\left(\frac{81} 4 – \frac{81} 4\right)} + 3 && \mbox{Since } B^2 – B^2 = {\color{red} \frac{81} 4-\frac{81} 4 = 0}\\
& = x^2 – 9x + \frac{81} 4 – \frac{81} 4 + 3\\
& = \left(x^2 – 9x + \frac{81} 4\right) – \frac{81} 4 + 3 && \mbox{Regrouping}
\end{align*}

Making sure we match the pattern, we have

\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad\quad x^2 + 2Bx + B^2\\
\mbox{Our Quadratic:} & \quad \left(x^2 + 14x + \frac{81} 4\right) – \frac{81} 4 + 40\\
\mbox{Match?} & \quad\,\,\, \checkmark\quad\quad \checkmark\quad\quad \checkmark
\end{align*}

Since the three terms in the parentheses match the perfect-square pattern, we know those three terms factor as a perfect square.

\begin{align*}
{\color{blue}\overbrace{\left(x^2 – 9x + \frac{81} 4\right)}^{\mbox{Perfect Square}}} {\color{red}- \frac{81} 4 + 3}
& = {\color{blue} \overbrace{\left(x – \frac 9 2\right)^2}^{\mbox{Factored}}} {\color{red} – \frac{81} 4 + \frac{12} 4}\\
& = \left(x – \frac 9 2\right)^2 {\color{red} – \frac{69} 4}
\end{align*}
\vskip 5mm

\textbf{Answer:} \(x^2 – 9x + 3 = \left(x – \frac 9 2\right)^2 – \frac{69} 4\)
\vskip 1.5cm

\textbf{Example 3: Dealing With Leading Coefficients}\vskip 2mm

Complete the square on \(2x^2 + 12x – 5\).
\vskip 5mm

\textbf{Solution} In order to complete the square, we first need to reduce the leading coefficient to 1.
\vskip 2mm

Step 1: Factor out the leading coefficient out of the first two terms.

\[%
{\color{blue} 2}x^2 + 12x – 5 = {\color{blue} 2}[x^2 + 6x] – 5
\]
\vskip 5mm

Step 2: Complete the Square \textit{Inside the Braces}\vskip 2mm

\qquad Quadratic Term: \textit{Inside} the braces, our quadratic term has a coefficient of 1.\vskip 2mm

\qquad Linear Term: \textit{Inside} the braces, we set \({\color{blue} 2B = 6}\).\vskip 2mm

A quick check and we see that we are matching the perfect-square pattern \textit{inside the braces}.

\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad\,\,\,\, x^2 + {\color{blue}2B}x + B^2\\
\mbox{Our Quadratic (Inside Braces):} & \quad 2[x^2 + {\color{blue}6}x] – 5\\
\mbox{Match?}&\quad\quad \checkmark\quad\,\,\,\checkmark
\end{align*}

Let’s move on to the constant term.\vskip 2mm

\qquad Constant Term: \textit{Inside} the braces we have \({\color{blue} 2B = 6}\) which means \(B = 3\) and so \({\color{red} B^2 = 3^2 = 9}\). As before, adjust our constant term by adding {\color{red} 0}, but now it is \textit{inside the braces}.

\begin{align*}
2[x^2 + {\color{blue} 6}x] – 5
& = 2[x^2 + {\color{blue} 6}x + {\color{red} 0}] – 5 && \mbox{Adding } {\color{red} 0} \mbox{ \textit{inside} the braces}\\
& = 2[x^2 + {\color{blue} 6}x + {\color{red} (9 – 9)}] – 5 && \mbox{Since } B^2 – B^2 = {\color{red} 9 – 9 = 0}\\
& = 2[x^2 + 6x + 9 – 9] – 5\\
& = 2[(x^2 + 6x + 9) – 9] – 5 && \mbox{Regrouping \textit{inside} the braces}
\end{align*}
\vskip 2mm

Still keeping our attention focused \textit{inside} the braces, we check to see if we match the perfect-square pattern.

\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad x^2 + 2Bx + B^2\\
\mbox{Our Quadratic (Inside Braces):} & \quad 2[(x^2 + 6x + 9) – 9] – 5\\
\mbox{Match?}&\quad\quad \checkmark\quad\,\,\,\checkmark\quad\,\,\checkmark
\end{align*}

Since the three terms inside the parentheses match the pattern, we know those three terms form a perfect square, so they will easily factor into \((x + B)^2\).

\begin{align*}
2\big[{\color{blue}\overbrace{\left(x^2 + 6x + 9\right)}^{\mbox{Perfect Square}}} – 9\big] – 5
& = 2\big[{\color{blue}\overbrace{\left(x + 3\right)^2}^{\mbox{Factored}}} – 9\big] – 5
\end{align*}
\vskip 5mm

Step 3: Adjusting the Final Form\vskip 2mm

Start by multiplying the leading coefficient \textit{through the square braces} (NOT the parentheses!), and simplify.

\begin{align*}%
{\color{blue} 2}\big[\left(x + 3\right)^2 – 9\big] – 5
& = \big[{\color{blue} 2}(x+3)^2 – {\color{blue} (2)} 9\big] – 5 && \mbox{Multiply through braces}\\
& = \big[2(x + 3)^2 – 18\big] – 5\\
& = 2(x + 3)^2 – 18 – 5 && \mbox{Braces no longer needed}\\
& = 2(x+3)^2 – 23 && \mbox{Combine the constants}
\end{align*}
\vskip 5mm

\textbf{Answer:} \(2x^2 + 12x – 5 = 2(x+3)^2 – 23\)
\vskip 1.5cm

\textbf{\large Quick Example}\vskip 2mm

\textbf{Example 4: A Quick Example}\vskip 2mm

Complete the Square on \(3x^2 – 4x + 8\).\vskip 5mm

\textbf{Solution:}\vskip 2mm

\begin{align*}%
3x^2 – 4x + 8 & = 3\left[x^2 – \frac 4 3 x\right] + 8 && \mbox{Factor out leading coefficient}\\[6pt]
& = 3\left[x^2 {\color{blue}- \frac 4 3} x\right] + 8 && \mbox{Identify } {\color{blue} 2B = -\frac 4 3} \longrightarrow {\color{red} B^2 = \left(-\frac 2 3\right)^2 = \frac 4 9}\\[6pt]
& = 3\left[x^2 {\color{blue}- \frac 4 3} x + {\color{red} \left(\frac 4 9 – \frac 4 9\right)}\right] + 8 && \mbox{Add } {\color{red} 0}\\[6pt]
& = 3\left[\left(x^2 {\color{blue}- \frac 4 3} x + {\color{red}\frac 4 9}\right)\,\,{\color{red}- \frac 4 9}\right] + 8 && \mbox{Regrouping}\\[6pt]
& = 3\left[\left(x – \frac 2 3\right)^2 – \frac 4 9\right] + 8 && \mbox{Factor}\\[6pt]
& = 3\left(x – \frac 2 3\right)^2 – \frac 4 3 + 8 && \mbox{Distribute}\\[6pt]
& = 3\left(x – \frac 2 3\right)^2 – \frac 4 3 + \frac{24} 3\\[6pt]
& = 3\left(x – \frac 2 3\right)^2 + \frac{20} 3 && \mbox{Combine like terms}
\end{align*}
\vskip 5mm

\textbf{Answer:} \(3x^2 – 4x + 8 = 3\left(x – \frac 2 3\right)^2 + \frac{20} 3\)
\vskip 1.5cm

\textbf{\large The Quick Method}\vskip 2mm

The method shown in the examples can be made more efficient if we recognize that the pattern is always the same. For a simple quadratic with a leading coefficient of \(1\), the completed square form looks like this:

\begin{center}
\begin{tikzpicture}
% Help Lines
% \draw [step=0.25cm,help lines] (-2,-1) grid (2,1);
% \draw (-2,-1) grid (2,1);
% Equation
\node {%
\(%
x^2 + {\color{blue} 2B}x + c = (x + {\color{blue} B})^2 – {\color{blue} B^2} + c
\)};
% Arrows
\node (linear) [inner sep=0pt] at (-1.625,-0.15){};
\node (BLow) [inner sep=0pt] at (0.75, -0.15){};
\node (BUp) [inner sep=0pt] at (0.75, 0.2){};
\node (B2) [inner sep=0pt] at (1.75,0.2){};
\draw [-latex] (linear.south) to [out=270,in=270] (BLow.south);
\draw [-latex] (BUp.north) to [out=90,in=90] (B2.north);
% Labels
\node at (-0.375, -0.625) {\(\div 2\)};
\node at (1.5, 0.75) {\scriptsize Minus the Square};
\node at (3,-1) [anchor=east] {\tiny www.mathtutoringacademy.com};
\end{tikzpicture}
\end{center}

Inside the final parentheses we always end up with \(x + B\), where \(B\) is half of the coefficient of the original \(x\) term.\vskip 2mm

Next, we subtract \(B^2\) \textit{outside} the parentheses. Let’s try it with one of our previous examples to see it in action.\vskip 5mm

\textbf{Example 5: Simple Quick Method}\vskip 2mm

Use the quick version of Completing the Square on \(x^2 + 14x + 40\).

\begin{center}
\begin{tikzpicture}
% Help Lines
% \draw [step=0.25cm,help lines] (0,-1) grid (5,1);
% \draw (0,-1) grid (5,1);
% Equation
\node at (1.2,-0.25){%
\(%
x^2 + {\color{blue} 14}x + 40 = (x + {\color{blue} 7})^2 \underbrace{- {\color{blue} 7^2} + 40}_{\mbox{\scriptsize Add Together}} = (x + 7)^2 – 9
\)};
% Arrows
\node (linear) [inner sep=0pt] at (-1.625,-0.15){};
\node (BLow) [inner sep=0pt] at (0.75, -0.15){};
\node (BUp) [inner sep=0pt] at (0.75, 0.2){};
\node (B2) [inner sep=0pt] at (1.75,0.2){};
\draw [-latex] (linear.south) to [out=270,in=270] (BLow.south);
\draw [-latex] (BUp.north) to [out=90,in=90] (B2.north);
\draw [-latex] (2.85,-0.575) .. controls (3,-0.625) and (5,-0.75)..(5,-0.25);
% Labels
\node at (-0.375, -0.625) {\(\div 2\)};
\node at (1.5, 0.75) {\scriptsize Minus the Square};
\node at (5,-1) [anchor=east] {\tiny www.mathtutoringacademy.com};
\end{tikzpicture}
\end{center}
\vskip 5mm

\textbf{Answer:} \(x^2 + 14x + 40 = (x + 7)^2 – 9\)

Note: This is the same answer we got in Example 1.\vskip 1.5cm

\textbf{Example 6: Quick Method with Leading Coefficient}\vskip 2mm

The quick version can also be used when the leading coefficient isn’t \(1\). Like before, we just factor out the leading coefficient first, and then work inside the braces.\vskip 2mm

Let’s complete the square on \(5x^2 + 80x + 13\).

\noindent\textbf{Solution}\vskip 2mm

Step 1) Factor out the leading coefficient.

\[%
5x^2 + 80x + 13 = 5[x^2 + 16x] + 13
\]

Step 2) Now use the quick method to complete the square on the \textit{inside} of the braces.

\begin{center}
\begin{tikzpicture}
% Help Lines
% \draw [step=0.25cm,help lines] (-2,-2) grid (4,2);
% \draw (-2,-1) grid (4,1);
% \draw [red, step = 2] (0,-1) grid (3,1);
% Equation
\node at (1.2,0){%
\begin{minipage}{\textwidth}
\begin{align*}%
5[x^2 + {\color{blue} 16}x] + 13
& = 5[x^2 + {\color{blue} 16}x] + 13\\[5mm]
& = 5[(x + {\color{blue} 8})^2 – {\color{blue} 64}] + 13\\[6mm]
& = 5(x + {\color{blue} 8})^2 – 320 + 13\\
& = 5(x + {\color{blue} 8})^2 – 307%
\end{align*}%
\end{minipage}
};
% Arrows
\node (linear) [inner sep=0pt] at (2.375,0.95){};
\node (BLow) [inner sep=0pt] at (2.23, -0.05){};
\node (BUp) [inner sep=0pt] at (2.23, 0.25){};
\node (B2) [inner sep=0pt] at (3.23,-0.05){};
\draw [-latex] (linear.south) to [out=270,in=90] (BUp.south);
\draw [-latex] (BLow.south) to [out=270,in=270] (B2.south);
% Labels
\node at (2.625, 0.625) {\scriptsize \(\div 2\)};
\node at (3, -0.55) {\scriptsize Minus the Square};
\node at (4.5,-2) [anchor=east] {\tiny www.mathtutoringacademy.com};
\end{tikzpicture}
\end{center}
\vskip 5mm

\textbf{Answer:} \(5x^2 + 80x + 13 = 5(x + 8)^2 – 307\)

\vskip 1.5cm

\textbf{\large Solving Equations}\vskip 2mm

When solving a quadratic equation, I find the technique of completing the square is not very efficient, except in the simple cases where the leading coefficient is \(1\) and the linear coefficient is even.\vskip 2mm

But that doesn’t mean we shouldn’t know \textit{how} to solve more complicated quadratic equations with this technique.\vskip 5mm

\textbf{Two Approaches}\vskip 2mm

When solving an equation through completing the square, there are two basic approaches: (1) Complete the square first, \textit{then} solve the equation, or (2) Solve the equation \textit{as} you complete the square.\vskip 5mm

\textbf{Example 7: Solving a Quadratic Equation by Completing the Square}\vskip 2mm

Use Completing the Square to solve \(4x^2 + 3x – 10 = 0\).\vskip 5mm

\textbf{Solution (Method 1)}\vskip 2mm

For this approach, we’ll first complete the square, then solve the equation.\vskip 2mm

Step 1) Complete the Square on \(4x^2 + 3x – 10\).\vskip 2mm

Using the techniques we’ve discussed above, we get the following.

\[%
4x^2 + 3x – 10 = 4\left(x + \frac 3 8\right)^2 – \frac{169}{16}
\]
\vskip 5mm

Step 2) Solve the equation.

\begin{align*}
4x^2 + 3x – 10 & = 0\\[6pt]
4\left(x + \frac 3 8\right)^2 – \frac{169}{16} & = 0 && \mbox{Complete the Square}\\[6pt]
4\left(x + \frac 3 8\right)^2 & = \frac{169}{16} && \mbox{Add the constant}\\[6pt]
\left(x + \frac 3 8\right)^2 & = \frac{169}{64} && \mbox{Divide by \(4\)}\\[6pt]
x + \frac 3 8 & = \pm\sqrt{\frac{169}{64}} && \mbox{Take the square root}\\[6pt]
x + \frac 3 8 & = \pm\frac{13} 8 && \mbox{Simplify}\\[6pt]
x & = – \frac 3 8 \pm\frac{13} 8 && \mbox{Subtract}
\end{align*}

The two solutions are

\[%
\begin{array}{rl}
x & = \displaystyle – \frac 3 8 + \frac {13} 8 = \frac{10} 8 = \frac 5 4\\[12pt]
x & = \displaystyle – \frac 3 8 – \frac{13} 8 = – \frac{16} 8 = – 2
\end{array}
\]
\vskip 5mm

\textbf{Solution (Method 2)}\vskip 2mm

This time, we’ll complete the square as we solve the equation. It should be noted that this approach is what many students think \textit{is} Completing the Square.\vskip 2mm

\begin{align*}
4x^2 + 3x – 10 & = 0\\[6pt]
4x^2 + 3x & = 10 && \mbox{Add the constant}\\[6pt]
x^2 + \frac 3 4 x & = \frac{10} 4 && \mbox{Divide by 4}\\[6pt]
x^2 + \frac 3 4 x + \frac 9 {64} & = \frac{10} 4 + \frac 9 {64} && \mbox{Add \(B^2\)}\\[6pt]
\left(x + \frac 3 8\right)^2 & = \frac{160}{64} + \frac 9 {64} && \mbox{Factor the left side}\\[6pt]
x + \frac 3 8 & = \pm \sqrt{\frac{169}{64}} && \mbox{Take square root}\\[6pt]
x & = -\frac 3 8 \pm \frac{13} 8 && \mbox{Subtract}
\end{align*}

Again, we see the two answers are \(x =-\frac 3 8 + \frac{13} 8 = \frac 5 4\) and \(x = -\frac 3 8 – \frac{13} 8 = -2\).
\vskip 5mm

\textbf{Answer:} \(x = \frac 5 4\) or \(x = -2\)
\vskip 1.5cm

\textbf{Conclusion}\vskip 2mm

Admittedly, most high school student will Complete the Square in the context of solving quadratic equations. However, it is important that they understand that the technique is more than just an equation-solving technique. Outside of high school math, much of its use is in changing the form of a quadratic function so that it can be used in higher mathematical techniques.

10 Ways Tutoring Can Help You

10 Ways Tutoring Can Help You

Finding a good math tutor can be essential to your success in school. Tutoring can do much more than just improving math grades. Here are 10 ways in which you can benefit from tutoring.

1. Improve your confidence

When working one-on-one outside of the classroom setting, you might find it easier to focus on the subject material and return to math class with increased confidence.

2. Improve your grades 

Supplementing your coursework with tutoring can help you understand the material more thoroughly so you can do better in school.

3. Work at a pace you are comfortable with 

Your teacher might be moving through the material at a speed that is too fast for you, but when you are studying with a tutor they will structure their sessions at a pace that makes you feel comfortable. Alternatively, if you feel that your class is moving too slow, your tutor can work ahead with you to prepare for future material.

4. Build mathematical foundation

It is essential that you understand the basics before you can succeed in more difficult math. While your teacher must follow a strict curriculum,a tutor will work with you to review the material that you didn’t understand before. Building a strong mathematical foundation now will allow you to tackle the more complicated topics later.

5. Provide supplemental work

If you are confused about a certain topic, your tutor can find you extra material and work through it with you.

6. Tailor lessons to your needs 

In a classroom setting, students work at different paces and understand things differently. In one-on-one tutoring, your lessons can be perfectly tailored to your own learning style, which can make learning math easier and more enjoyable.

7. Find a new challenge 

Is your math class too easy for you? Tutoring might be the perfect way to push your own boundaries. A tutor can expand and enrich upon the material you’re learning in school, or work on interesting material that’s not even offered at your school. Your tutor can also help you find local math competitions to challenge you in a fun way.

8. Improve SAT/ACT scores 

Tutoring is a great way to make sure you ace your college entrance exam. Tutors can help you learn test taking strategies, as well as helping you learn what kind of questions to expect.

9. Bring a fresh new perspective 

If you’re feeling stuck in a mathematical rut, working with a tutor can help. Your tutor can discuss math from a new point of view that might make more sense to you than the way your teacher approaches the material.

10. Make math more interesting 

Tutors are passionate about math and would love to share that passion with you. If there is a certain topic of interest to you, your tutor can build some lessons around that to keep you interested.

Modular Arithmetic & Fermat’s Little Theorem

Modular Arithmetic & Fermat’s Little Theorem

Modular Arithmetic and Fermat’s Little Theorem

Modular arithmetic is a way of counting in which the numbers wrap around after reaching a certain value. The clock is often used as an analogy. While time always progresses forward, the 12-hour clock “resets” to 1 after passing 12 (13 o’clock is equivalent to 1 o’clock). If we replace 12 with 0, we have the set known as ℤ12 which is made up of {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}. If we consider midnight as t0=0, then 5 hours later is t5=5. 17 hours later will also be t17=5. 84 hours later will be t84 = 0. 85 hours later will be t85 = 1. There is a pattern here. The value a in the set ℤ12 corresponding to any other integer b is the remainder when b is divided by 12. We can write this mathematically as b ≡ a (mod 12), and we say “b is congruent to a modulo 12”

A clock is a good analogy for modular arithmetic in Z12

A clock represents modular arithmetic in ℤ12

More generally, we can write the set of integers for any modulus n as ℤn = {0, 1, 2, 3, … n-1} . Each element a in ℤn represents not only a itself, but all of the b’s for which b ≡ a (mod n). These elements are called congruence classes, and the set ℤn is a set made up of these classes. Sometimes, it’s more convenient to write the relationship as b-a ≡ 0 (mod n), i.e. each b is assigned to a congruence class depending on how much needs to be subtracted to be divisible by n. Addition and multiplication can be done using modular arithmetic if you are using the same modulus. If b ≡ a (mod n) and d ≡ c (mod n), then

  1. (b+d) ≡ (a+c) (mod n)
  2. (b*d) ≡ (a*c) (mod n)

Example 1. What is 84,872 (mod 5)? A simple way to approach this problem is to find the closest multiple of 5 to 84,872. That is 84,870. Therefore, we can write 84,872 (mod 5) ≡ 84,870 (mod 5) + 2 (mod 5) ≡ 0 (mod 5) + 2 (mod 5) ≡ 2 (mod 5). Example 2. In modulo 10, what is 19,374 · 3,172? One way to attempt this problem is to multiply out these numbers and then find the remainder when dividing by 10. However, since we want our answer in modulo 10, we can instead multiply just the representatives from their congruence classes. The remainder when 19,374 is divided by 10 is 4, so 19,374 ≡ 4 (mod 10) and 3,172 ≡ 2 (mod 10), so (19,374 · 3,172) ≡ (2·4) (mod 10) ≡ 8 (mod 10). Example 3. In modulo 4, what is 180,700,247 + 64,987,422? Again we first find the congruence class for each number modulo 4. We can simplify each by breaking it into the sum of smaller multiples of 4. One possible way is to write 180,700,247 ≡ 180,700,200 (mod 4) + 44 (mod 4) + 3 (mod 4) ≡ 0 (mod 4) + 0 (mod 4) + 3 (mod 4) ≡ 3 (mod 4) Similarly, 64,987,422 ≡ 2 (mod 4). Therefore, (180,700,247+ 64,987,422) ≡ (3 + 2) (mod 4) ≡ 5 (mod 4). However, 5 is not in the set ℤ4 = {0, 1, 2, 3}, so we must reduce 5 (mod 4) to something in the set. We can simplify by noticing that 5 (mod 4) ≡ 4 (mod 4) + 1 (mod 4) ≡ 0 (mod 4) + 1 (mod 4) ≡ 1 (mod 4).

Fermat’s Little Theorem

One important application for modular arithmetic is Fermat’s Little Theorem which states that if p is a prime number and a is not divisible by p, then ap-1 ≡ 1 (mod p). This theorem is useful because allows you to find a remainder when dividing a really big number by a prime number. Example 4. What is the remainder when 2784 is divided by 11? 2784 is too big to put in most calculators and too time-consuming to multiply by hand. We can apply Fermat’s Little Theorem to make it easier to solve. From the theorem we know that 210 ≡ 1 (mod 11). And we also know that we can multiply congruence classes with the same modulus. Therefore, one strategy we can apply is to write 2784 as the product of as many 210’s as possible. We can simplify as follows: 2784= (210)78·(24) ≡ 178 · 24 (mod 11) ≡ 24 (mod 11) ≡16 (mod 11) ≡ 5 (mod 11). The remainder is 5.

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Algebraic Rings

Algebraic Rings

Introduction to Algebraic Rings

An algebraic ring is one of the most fundamental algebraic structures. It builds off of the idea of algebraic groups by adding a second operation  (For more information please review our article on groups). For rings we often use the notation of addition and multiplication because the integers are a good analogy for a ring, but the operations “+” and “⋅” do not necessarily represent addition and multiplication in the ring. A ring is defined as a set S combined with two operations (“+” and “⋅”) that has the following properties: I. The set S is a group under one operation (we will call this operation “+”): a. It is closed under this operation: For all elements a and b in the set S, a+b is also in S. b. It contains an additive identity element (which we will call “0”, although it does not necessarily represent the integer 0): There is some element 0 in the set S such that for every element a in S, a+0 = 0+a = a. c. The operation is associative: For all a, b, and c in the set S, (a+b)+c = a+(b+c). d. Inverses exist: For every element a in the set S, there is a (-a)  in S such that a+(-a)= (-a)+a = 0. II. The group operation is commutative: For all a and b in the set S, a + b = b + a. III. The second operation (written as “⋅”) follows these rules: a. It is closed under this operation. b. The set contains a multiplicative identity element (which we call “1”, although it does not necessarily represent the integer 1): such that for all elements a in the set S, a⋅1 = 1⋅a = a. c. It is associative: For all a, b, and c in the set S, (a⋅b)⋅c = a⋅(b⋅c). d. It is distributive over “+” on both the right and left side: For all elements a, b, and c in the set S, a⋅(b+c) = a⋅b + a⋅c and (b+c)⋅a = b⋅a + c⋅a. Note that inverses are not required for the “⋅” operation. If inverses do exist under “⋅”, we have a special kind of ring called a field.   We will now look at some examples of rings and sets that aren’t rings. Example 1. The integers under addition and multiplication. I. The integers are a group under addition (see our group article). I. Addition can be carried out in any order so it is commutative. III. Multiplication follows the additional rules laid out for rings. The integers are closed under multiplication. The identity element is 1. Multiplication is associative and it is also distributive. Therefore, the integers are a ring under addition and multiplication. ∎ This is the reason addition, multiplication, 0, and 1 show up in our notation. If a set “acts like” the integers by following the ring axioms, it is a ring. Example 2. The set of 2x2 matrices under addition and multiplication. I. The set of 2x2 matrices form a group under addition. a. For any 2x2 matrices A and B, A+B is also a 2x2 matrix. b. The additive identity “0” is the zero matrix $\begin{bmatrix} 0 & 0 \\ 0 & 0 \\ \end{bmatrix}$. c. Matrix addition is associative. d. The additive inverse of a matrix A is just -A. Observe $\begin{bmatrix} a & b \\ c & d \\ \end{bmatrix} + \begin{bmatrix} -a & -b \\ -c & -d \\ \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \\ \end{bmatrix}$. II. Addition of 2x2 matrices is commutative, A+B = B+A. III. Multiplication of 2x2 matrices satisfies the following conditions: a. It is closed under the operation. The multiplication of two 2x2 matrices results in another 2x2 matrix. b. The identity matrix is I = $\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}$ and for any 2x2 matrix A, A⋅I = I⋅A = A. c. Matrix multiplication is associative. d. Matrix multiplication distributes over addition on the left and right side (This can be easily proven, but to save space it will not be proven here). Therefore, the set of 2x2 matrices is a ring. ∎   Example 3. The set of 2x2 matrices with non-negative entries under addition and multiplication. This set satisfies all of the requirements for the multiplication operation, but it is not a group under addition. Without negative entries, most matrices in this set do not have additive inverses. Therefore, this set is not a ring. ∎ Like groups, rings do not have to be composed of real numbers or matrices. Some examples include sets of polynomials, the complex numbers, and integers modulo n. The word “ring” comes from the German word “Zahlring,” which means a collection of numbers that circle back on themselves. However, now that the field has been studied more extensively, we know that cyclic rings are just one type of rings. The symbol ℤ for the integers is also named after a German word for numbers “Zahlen.” The word “ring” comes from the German word “Zahlring,” which means a collection of numbers that circle back on themselves. However, now that the field has been studied more extensively, we know that cyclic rings are just one type of rings. The symbol ℤ for the integers is also named after a German word for numbers “Zahlen.”

Algebraic Groups

Introduction to Algebraic Groups

One of the most fundamental algebraic structures in mathematics is the group. A group is a set of elements paired with an operation that satisfies the following four conditions: I. It is closed under an operation (represented here by “+”, although it does not necessarily mean addition): For all elements a and b in the set S, a+b is also in S. II. It contains an identity element (often written as “e”): There is some element e in the set S such that for every element a in S, a+e = e+a = a. III. The operation is associative: For all a, b, and c in the set S, (a+b)+c = a+(b+c). IV. Inverses exist: For every element a in the set S, there is an a-1 in S such that a+a-1 = a-1+a = e We will look at some examples of groups and sets that aren’t groups. Example 1. Integers under addition, (ℤ, +). I. The group operation is addition. For any two integers n and m, n+m is also an integer. II. The identity element is 0 because for any integer n, n+0 = 0+n = n. III. Addition is associative. IV. Inverses exist. The inverse of any integer n is -n, and n + (-n) = (-n) + n = 0. Therefore, (ℤ, +) is a group. ∎ Example 2. Integers under multiplication, (ℤ, ⋅) I. The group is closed under multiplication, because for any two integers n and m, n⋅m is also an integer. II. The identity element is 1. For any integer n, n⋅1 = 1⋅n = n. III. Multiplication is associative. IV. Inverses do NOT exist. For any integer n, 1/n is not an integer except when n=1. Since the the set does not meet all four criteria, (ℤ, ⋅) is not a group. ∎ Example 3. The set of rational numbers not including zero, under multiplication (ℚ – 0, ) I. The group is closed under multiplication. II. 1 is the identity element. III. Multiplication is associative. IV. Inverses exist in this group. If q is a rational number, 1/q is also a rational number. And q⋅ (1/q) = (1/q) ⋅ q = 1. Notice that if 0 were in this set, it would not be a group because 0 has no inverse. Therefore, ℚ – 0 is a group. ∎ Example 4. The set {0, 1, 2} under addition I. This set is not closed under addition because 1 + 2 = 3, and 3 is not part of the set. Therefore, the set cannot be a group. ∎ Example 5. The trivial group {e} This group consists only of the identity element. We don’t need to specify the operation here because it works for both multiplication and addition. For addition, e=0. 0+0=0 so the group is closed under addition, has an identity element, and is closed under inverses (and addition is associative). For multiplication, e=1 and similarly, 1⋅1 = 1. ∎ The examples here involved only numbers, but there are many different types of groups. For example, the ways to transform a triangle is called a Dihedral Group, with the group operation being the act of rotating or reflecting the shape around an axis. The possible permutations of a Rubik’s Cube are also a group, with the operation being a sequence of moves. It’s powerful to know that a set is a group because it gives you an understanding of how the elements will always behave. Groups have many other properties that are useful to mathematicians, and there is a whole field of study built off of this knowledge called Group Theory.

The set of permutations of a Rubik’s Cube is considered an algebraic group.

The set of permutations of a Rubik’s Cube is considered an algebraic group.