# Multiplying Even and Odd Numbers

Today I’m going to be showing you what would happen if you were to multiply:

a) An even number by an even number;

b) An odd number by an odd number;

c) An even number by an odd number.

Firstly, let us define what an even number is:

An even number can be described using the expression $2n$, whereby (n) would be a whole number ranging from 0 upwards.

Next, let us define what an odd number is:

An odd number can be described using the expression $2n+1$, and similarly (as is the case with even numbers), (n) would be a whole number ranging from 0 upwards.

Now, since we’ve defined how both even numbers and odd numbers can be described in terms of mathematical expressions, let’s focus our attention on multiplying even numbers by even numbers, odd numbers by odd numbers and even numbers by odd numbers…

Multiplying even numbers by even numbers:

Let’s produce two whole numbers which could be equal to one another or not equal to one another… Let’s call these numbers ${ n }_{ 1 }$ and ${ n }_{ 2 }$.

Using these two whole numbers we can multiply two unknown even numbers by each other in such a manner: $2{ n }_{ 1 }\cdot 2{ n }_{ 2 }$

This would invariably give us the result $4{ n }_{ 1 }{ n }_{ 2 }=2\cdot 2{ n }_{ 1 }{ n }_{ 2 }$. Now, the product of ${ n }_{ 1 }{ \cdot n }_{ 2 }$ would be a whole number and since this is the case, you would have to say that an even number multiplied by an even number would produce an even number. Let’s not forget that even numbers are multiples of 2.

Multiplying odd numbers by odd numbers:

Once again, let’s come up with two whole numbers which could be equal to one another or not equal to one another… These whole numbers will be ${ n }_{ 3 }$ and ${ n }_{ 4 }$.

This would mean that two odd numbers being multiplied by one another would produce an expression as such: $\left( 2{ n }_{ 3 }+1 \right) \left( 2{ n }_{ 4 }+1 \right)$

And if we expand the expression above, we’ll get: $4{ n }_{ 3 }{ n }_{ 4 }+2{ n }_{ 3 }+2{ n }_{ 4 }+1$

Now if we re-arrange the expression above, we can get: $2\left( 2{ n }_{ 3 }{ n }_{ 4 }+{ n }_{ 3 }+{ n }_{ 4 } \right) +1$

Since the expression $2{ n }_{ 3 }{ n }_{ 4 }+{ n }_{ 3 }+{ n }_{ 4 }$ must be a whole number, you would be forced to conclude that an odd number multiplied by an odd number would produce an odd number.

Multiplying even numbers by odd numbers:

We will for the last time come up with two whole numbers ${ n }_{ 5 }$ and ${ n }_{ 6 }$.

An even number and an odd number being multiplied by one another could be shown using the mathematical expression below: $2{ n }_{ 5 }\cdot \left( 2{ n }_{ 6 }+1 \right)$

Lazily, we could conclude that an even number multiplied by an odd number would produce an even number. This is because even numbers are all multiples of 2.

Ok… So, let’s summarise what we’ve discovered:

i) An even number multiplied by an even number would produce an even number;

ii) An odd number multiplied by an odd number would produce an odd number;

iii) An even number multiplied by an odd number would produce an even number.

Knowing this we can further strengthen our mathematical reasoning. 🙂

# How To Expand (a+b+c)(d+e+f)(g+h+i)

Firstly you have to know what (a+b+c)(d+e+f) is. You can expand this expression using a rectangle: So you know that: $\left( a+b+c \right) \left( d+e+f \right) =ad+ae+af+bd+be+bf+cd+ce+cf$

Next you’d have to multiply (a+b+c)(d+e+f) by (g+h+i) using another rectangle: And from here you’d figure out that: $\left( a+b+c \right) \left( d+e+f \right) \left( g+h+i \right) \\ \\ =\left[ \left( a+b+c \right) \left( d+e+f \right) \right] \left( g+h+i \right) \\ \\ =\left[ ad+ae+af+bd+be+bf+cd+ce+cf \right] \left( g+h+i \right) \\ \\ =adg+aeg+afg+bdg+beg+bfg+cdg+ceg+cfg\\ \\ +adh+aeh+afh+bdh+beh+bfh+cdh+ceh+cfh\\ \\ +adi+aei+afi+bdi+bei+bfi+cdi+cei+cfi\\ \\ =ag\left( d+e+f \right) +bg\left( d+e+f \right) +cg\left( d+e+f \right) \\ \\ +ah\left( d+e+f \right) +bh\left( d+e+f \right) +ch\left( d+e+f \right) \\ \\ +ai\left( d+e+f \right) +bi\left( d+e+f \right) +ci\left( d+e+f \right) \\ \\ =\left( d+e+f \right) \left[ ag+bg+cg+ah+bh+ch+ai+bi+ci \right] \\ \\ =\left( d+e+f \right) \left[ g\left( a+b+c \right) +h\left( a+b+c \right) +i\left( a+b+c \right) \right] \\ \\ =\left( d+e+f \right) \left[ \left( a+b+c \right) \left( g+h+i \right) \right] \\ \\ =\left( a+b+c \right) \left( d+e+f \right) \left( g+h+i \right)$

# How To Multiply Surds Contained within Fractions

First Scenario: $\frac { 1 }{ a+\sqrt { b } } =\frac { 1 }{ \left( a+\sqrt { b } \right) } \cdot \frac { \left( a-\sqrt { b } \right) }{ \left( a-\sqrt { b } \right) } \\ \\ =\frac { a-\sqrt { b } }{ { a }^{ 2 }-a\sqrt { b } +a\sqrt { b } -b } =\frac { a-\sqrt { b } }{ { a }^{ 2 }-b }$

Second Scenario: $\frac { 1 }{ a-\sqrt { b } } =\frac { 1 }{ \left( a-\sqrt { b } \right) } \cdot \frac { \left( a+\sqrt { b } \right) }{ \left( a+\sqrt { b } \right) } \\ \\ =\frac { a+\sqrt { b } }{ { a }^{ 2 }+a\sqrt { b } -a\sqrt { b } -b } =\frac { a+\sqrt { b } }{ { a }^{ 2 }-b }$

Notice that what we’re ultimately doing in both cases is multiplying the surd within a fraction by 1. When the value of the numerator is exactly the same as the value of the denominator in a fraction, what you have is 1.

You should know that 1/1, 2/2, 3/3, (a+b)/(a+b) are all equal to 1.

# How To Multiply Surds

In order to multiply surds, you should first know these rules: ${ a }^{ m }\cdot { a }^{ n }={ a }^{ m+n }\\ \\ { a }^{ m }\div { a }^{ n }={ a }^{ m-n }$

You should also know that: ${ a }^{ \frac { 1 }{ 2 } }=\sqrt [ 2 ]{ { a }^{ 1 } } =\sqrt { a }$

So, knowing these rules, what would you get if you multiplied: $\sqrt { 3 } \cdot \sqrt { 3 }$?

Well, $\sqrt { 3 } \cdot \sqrt { 3 } ={ 3 }^{ \frac { 1 }{ 2 } }\cdot { 3 }^{ \frac { 1 }{ 2 } }={ 3 }^{ \frac { 1 }{ 2 } +\frac { 1 }{ 2 } }={ 3 }^{ 1 }=3$.

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Now how about $\sqrt { 3 } \cdot \left( -\sqrt { 3 } \right)$? $\sqrt { 3 } \cdot \left( -\sqrt { 3 } \right) =\sqrt { 3 } \cdot \left( -1 \right) \cdot \sqrt { 3 } \\ \\ =\sqrt { 3 } \cdot \sqrt { 3 } \cdot \left( -1 \right) =3\cdot \left( -1 \right) =-3$

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What about $\left( -\sqrt { 3 } \right) \left( -\sqrt { 3 } \right)$? $\left( -\sqrt { 3 } \right) \left( -\sqrt { 3 } \right) =\left( -1 \right) \cdot \sqrt { 3 } \cdot \left( -1 \right) \cdot \sqrt { 3 } \\ \\ =\left( -1 \right) \left( -1 \right) \sqrt { 3 } \sqrt { 3 } =1\cdot 3=3$

# (a+b)(a-b)=a^2-b^2 (The Real Proof) (Difference Of Two Squares Demystified)

Ever wondered why (a+b)(a-b)=a^2-b^2? The videos below will reveal to you why we accept this fact. Enjoy!

1) Why (a+b)(a+b)=a^2+2ab+b^2

2) Why (a+b)(a-b)=a^2-b^2