Theorem 0.1.4.1.
Let \(a\) and \(b\) be real numbers. Then
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\(\displaystyle (a+b)^2=a^2+2ab+b^2\)
-
\(\displaystyle (a-b)^2=a^2-2ab+b^2\)
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\(\displaystyle (a-b)(a+b)=a^2-b^2\)
Subsection 0.1.4 Arithmetic
We recall basic properties of the operations on real numbers. Since the natural numbers, integers, and rational numbers are in the set of real numbers, they also hold for these.
For all real numbers \(a\) and \(b\) we have:
- Commutative laws
- Associative laws
- Distributive law
- \(\displaystyle a\cdot (b+c)=(a\cdot b)+(a\cdot c)\)
- Identity laws
We call \(1\) the multiplicative identity. Note that the additive identity \(0\) is an element of the real and rational numbers and the integers, but not the natural numbers.
The following applies only to integers, rational numbers, and real numbers \(a\text{,}\) since neither \(0\) nor negative numbers are in the set of natural number. It only holds for \(a=0\) in \(\NN_0\text{.}\)
- Inverse law for addition
- \(\displaystyle a+(-a)=0\)
We call \(-a\) the additive inverse of \(a\) and write \(a+(-b)=a-b\text{.}\)
Because the sets of integers and natural numbers do not contain fractions, the following only holds for rational numbers and real numbers in general. The only integers for which it holds are \(a=1\) and \(a=-1\text{.}\) The only natural number for which it holds is \(a=1\text{.}\)
- Inverse law for multiplication
- \(\displaystyle a\cdot \frac{1}{a}=1\)
We call \(\frac{1}{a}\) the multiplicative inverse of \(a\text{.}\)
We indicate the order of operations with parenthesis (evaluate the innermost expression first), in addition we follow the precedence of operations:
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exponentiation and roots, since exponentiation is repeated multiplication (when the exponent is a natural number)
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multiplication and division, since multiplication by natural numbers is repeated addition and division by natural numbers is repeated subtraction
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addition and subtraction
Also see the first panel in Figure 0.1.4.2.
Recall that \(a^2=a\cdot a\text{.}\) With the laws above we prove:
Proof.
We prove the first formula and leave the proof of the second formula as an exercise. For the third formula see Theorem 1.3.2.13.
By the distributive law we have
\begin{equation*}
(a+b)(a+b) = a(a+b)+b(a+b).
\end{equation*}
Applying the distributive law twice we get
\begin{equation*}
a(a+b)+b(a+b)=(a^2+ab)+(ba+b^2).
\end{equation*}
By the commutative law of addition we have
\begin{equation*}
(a^2+ab)+(ba+b^2)=(a^2+ab)+(ab+b^2).
\end{equation*}
With the associative law of addition we get
\begin{equation*}
(a^2+ab)+(ab+b^2)=a^2+(ab+ab)+b^2=a^2+2ab+b^2.
\end{equation*}
Thus we have proven that
\begin{equation*}
(a+b)(a+b)=a^2+2ab+b^2.
\end{equation*}
https://xkcd.com/992/) ’Sailor Moon’s head exploded once’ and ’Some men have explosive orgasms’ both work for the Great Lakes from west to east (Paddle-to-the-Sea order).
