My lesson Topic.md
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Math 454 Teaching Item 1: Texting Lesson.md
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My lesson Topic: CORE M2
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<style>
body {
background-color: #eeeeee;
}

h1 {
color: maroon;
margin-left: 40px;
}

.gray {
margin-left: 50px ;
margin-right: 29%;
font-weight: 500;
color: #000000;
background-color: #cccccc;
border-color: #aaaaaa;
}

.blue {
display: inline-block;
margin-left: 29% ;
margin-right: 0%;
width: -webkit-calc(70% - 50px);
width: -moz-calc(70% - 50px);
width: calc(70% - 50px);
font-weight: 500;
color: #fff;
border-color: #336699;
background-color: #337799;
}

.left {
content:url("https://i.imgur.com/rUsxo7j.png");
width:50px;
border-radius: 50%;
float:left;
}
.right{
content:url("https://i.imgur.com/5ALcyl3.png"); width:50px;
border-radius: 50%;
display: inline-block;
vertical-align:top;
}
</style>

<div id="container" style=" padding: 6px;
color: #fff;
border-color: #336699;
background-color: #337799;
display: flex;
justify-content: space-between;
margin-bottom:3px;">
<div>
<i class="fa fa-envelope fa-2x"></i>
</div>
<div>
<i class="fa fa-camera fa-2x"></i>
</div>

<div>
<i class="fa fa-comments fa-2x"></i>
</div>
<div>
<i class="fa fa-address-card fa-2x" aria-hidden="true"></i>
</div>
<div>
<i class="fa fa-phone fa-2x" aria-hidden="true"></i>
</div>
<div>
<i class="fa fa-list-ul fa-2x" aria-hidden="true"></i>
</div>
<div>
<i class="fa fa-user-plus fa-2x" aria-hidden="true"></i>
</div>
</div>

<div><img class="left"/><div class="alert gray">

Hi, Jane. Do you have a minute to clarify tonight's homework for me?

</div></div>

<div><div class="alert blue">

Sure, man. What's up?

</div><img class="right"/></div>

<div><img class="left"/><div class="alert gray">

We are given a couple of sets along with operations and we have to determine which sets form a group. Any idea what this means?

</div></div>

<div><div class="alert blue">

Ah, yes! Recall that certain sets form a group when we apply operations to them. We refer to the Group Axioms to determine whether or not a set forms a group under an operation. Do you recall the Group Axioms?

</div><img class="right"/></div>

<div><img class="left"/><div class="alert gray">

The Group Axioms sound familiar. Let me look for them in my notes...

</div></div>

<div><img class="left"/><div class="alert gray">

The Group Axioms:
* The set is closed under its operation
* The operation is associative
* The set contains an identity element
* The set contains an inverse for each of its elements

</div></div>

<div><div class="alert blue">

Yes, those are the Group Axioms! If we perform an operation on a set and the set meets all of those requirements, then the set is a group under that operation.

</div><img class="right"/></div>

<div><img class="left"/><div class="alert gray">

Ah, that kind of makes sense now! Do you mind working through an example with me?

</div></div>

<div><div class="alert blue">

Not at all! I'm totally fine with helping you as long as you make an honest effort. Show me what you've got so far.

</div><img class="right"/></div>

<div><img class="left"/><div class="alert gray">

Alright. I am currently working with the set of real numbers under the operation of multiplication. I guess I can start by showing that this set is closed under the operation:

Suppose we apply the operation to real numbers $a$ and $b$. Then we have $a \times b = ab$ where $ab$ is a real number.

</div></div>

<div><div class="alert blue">

That looks good to me. Keep going...

</div><img class="right"/></div>

<div><img class="left"/><div class="alert gray">

The Group Axioms say that the operation must also be associative.

So lets suppose we have real numbers $a, b,$ and $c$. Then we can see that $(a \times b) \times c = ab \times c = abc$. We can also see that $a \times (b \times c) = a \times bc = abc$. Since both methods give us the same result, we can see that the operation is associative.

</div></div>

<div><div class="alert blue">

Perfect. Now go on a show that the last two axioms are true.

</div><img class="right"/></div>

<div><img class="left"/><div class="alert gray">

Alright. Third, we will show that there is an identity element in the real numbers. This one is pretty easy becasue we can see that $a \times 1 = a$ where $a$ and $1$ are both real numbers.

The last axiom is the one where I get confused sometimes. What exactly is the inverse again?

</div></div>

<div><div class="alert blue">

The last axiom says that every element has an inverse. Essentially, this means that we can apply the operation to any element in the real numbers and our result will be the identity element.

</div><img class="right"/></div>

<div><img class="left"/><div class="alert gray">

Oh, gotcha. Since our identity element is 1, then we have to show that there is an inverse, $a^{-1}$, for all $a$ such that $a \times a^{-1} =1$.

Suppose we have real number $a$, notice that $a \times (1/a) = 1$ where $(1/a)$ is a real number. Therefore, we can see that every element has an inverse.

</div></div>

<div><div class="alert blue">

There you go! You have shown that all four Group Axioms are true for the set of real numbers under the operation of multiplication. Congratulations!

</div><img class="right"/></div>

<div><img class="left"/><div class="alert gray">

Yay! That was not as bad as I thought it would be. I guess I just needed a little guidance. Thank you so much! IOU

</div></div>

<div><div class="alert blue">

You are so welcome!

</div><img class="right"/></div>

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