Discrete Math - The MKD Remix

In computing, integer data types are used to represent loop counters or indices into arrays, while floating-point data types are used to represent real numbers (like decimals) in scientific or financial calculations. In general, a number that can be stored using an integer type can also be stored using a floating-point type, but the ways in which that number can be used will depend on the data type. The following code examples show why it is important to keep this in mind when coding.

First, recall that the floor of x , written as \(\lfloor x \rfloor,\) is the greatest integer less than or equal to the real number x. The floor() function is available in both Python and Java as an implementation of \(\lfloor x \rfloor.\) In both programming languages floor() takes a double precision floating-point number as its input, but Python floor() returns a value of integer data type while Java floor() function returns a value of floating-point data type.

Notice that in the Python code below, the return value from the floor() function is an int which we can then use as an index into list L.

To step through the code, click on the "Next" button.

Now notice that in the Java code below, that the return value from the floor() function is a double which we cannot use as an index into array L without error. We must use a composition of functions (Java’s floor() function followed by the function that casts an input of type double to an output of type int ) to get the correct data type for an array index.

To step through the code, click on the "Next" button.

A function \(f\) from set \(A\) to set \(B\) is an ordered triple \((f,\, A,\, B)\) consisting of sets \(f,\) \(A,\) and \(B\) such that

That is, \(f \subseteq A \times B\) and for each element \(a \in A\) there is exactly one \(b \in B\) such that \((a,\, b) \in f\). The set \(f\) of ordered pairs is called the graph of the function. The set \(A\) is called the domain of the function and the set \(B\) is called the codomain of the function.
Note: This definition of a function as an ordered triple is based on the Bourbaki definition in the 1970 book Théorie des ensembles.

Consider functions \(f\), \(g\), and \(h\) defined as follows:

\(f : \mathbb{R} \rightarrow \mathbb{R}\) is defined by \(f = \{ (x,\,x^{2}) \mid x \in \mathbb{R} \}.\)

\(g : \mathbb{R} \rightarrow \mathbb R_{\ge 0}\) is defined by \(g = \{ (x,\,x^{2}) \mid x \in \mathbb{R} \}.\)

\(h : \mathbb{N} \rightarrow \mathbb{N}\) is defined by \(h = \{ (x,\,x^{2}) \mid x \in \mathbb{N} \}.\)

No two of these functions are equal even though they can all be described by the rule "the output is the square of the input" and have identical formulas: \(f(x) = x^{2},\) \(g(x) = x^{2},\) and \(h(x) = x^{2}.\) Each of the functions is defined for a different domain and/or codomain than the other two. In particular, \(f\) and \(g\) are not equal because they have different codomains, even though the two functions have the same graph and the same domain.

A function \(f: A \rightarrow B\) is invertible if and only if \(f\) is bijective.

Consider \(f\left(x\right)=x^2+1\) and \(g\left(x\right)=\sqrt x\) defined on \(f,\ g: \mathbb{R}_{\geq0} \rightarrow \mathbb{R}\). Find the rules for the functions \(\left(f+g\right)\), \(\left(f-g\right)\), \(\left(f\cdot\ g\right)\), and \(\left(\frac{f}{g}\right)?\)

The common domain is \(\mathbb{R}_{\geq0}\), since the square root is real valued only for \(\ x\ \geq0\).

\(\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)=x^2+1+\sqrt x\) , for \( x ≥ 0\)

\(\left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)=x^2+1- \sqrt x\) , for \( x ≥ 0\)

\(\left(f\cdot\ g\right)\left(x\right)=f\left(x\right)\cdot\ g\left(x\right)=\left(x^2+1\right)\cdot\ \sqrt x\), for \( x ≥ 0\)

\(\left(\frac{f}{g}\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}=\frac{x^2+1\cdot\ }{\ \sqrt x}\), for \( x > 0\).

Notice that the domain of \(\frac{f}{g}\) is \(x>0\), because \(g\left(0\right)=\sqrt0=0\), and division by \(0\) is not defined.

Consider \(f\left(x\right)=\frac{1}{x}\) and \(g\left(x\right)=2x-3\), defined on \(f,g:R\rightarrow R\). Notice that \(g\left(x\right)\) is defined for all real \(x\) and \(f\left(x\right)\) is defined for all real \(x\ \neq0\). Form the compositions, \(h\left(x\right)=\left(f \circ g\right)\left(x\right)\), and \(k\left(x\right)=\left(g \circ f\right)\left(x\right)\). Also determine their respective domains.

\(h\left(x\right)=\left(f \circ g\right)\left(x\right)=f\left(g\left(x\right)\right)=f\left(2x-3\right)=\frac{1}{2x-3}\). Here \(x\) needs to be in the domain of \(g\left(x\right)\), or all real \(x\), and \(g\left(x\right)\) needs to be in the domain of \(f\left(x\right)\). In particular \(g\left(x\right)\neq 0\), or \(2x-3\ \neq 0\), or \(x\ \neq\frac{3}{2}\).

By contrast, \(k\left(x\right)=\left(g\circ f\right)\left(x\right)=g\left(f\left(x\right)\right)=g\left(\frac{1}{x}\right)=2\left(\frac{1}{x}\right)-3=\frac{2}{x}-3\). Here \(x\) needs to be in the domain of \(f\left(x\right)\), or \(x\ \neq 0\), and \(f\left(x\right)\) needs to be in the domain of \(g\left(x\right)\), or \(f\left(x\right)\) can be any real number.

Consider \(f\left(x\right)=x^3+1\) and \$g(x) =root(3)(x-1)\$ defined on on \(f,g:R\rightarrow R\). Show that \(\left(g \circ f\right)\left(1\right)=1, \left(g \circ f\right)\left(2\right)=2, \left(g\circ f\right)\left(3\right)=3\), and \(\left(g\circ f\right)\left(x\right)=x\)

\(f\left(1\right)=1^3+1=2\)

\(f\left(2\right)=2^3+1=9\)

\(f\left(3\right)=3^3+1=28\)

\(f\left(x\right)=x^3+1\)

Therefore,

\( \left(g\circ f\right)\left(1\right)=g\left(f\left(1\right)\right)=g\left(2\right)=\) \$ root(3)(2-1)= root(3)(1)=1\$

\(\left(g\circ f\right)\left(2\right)=g\left(f\left(2\right)\right)=g\left(9\right)=\) \$ root(3)(9-1)= root(3)(8)=2\$

\(\left(g\circ f\right)\left(3\right)=g\left(f\left(3\right)\right)=g\left(28\right)=\) \$ root(3)(28-1)= root(3)(27)=3\$

\(\left(g\circ f\right)\left(x\right)=g\left(f\left(x\right)\right)=g\left(x^3+1\ \right)=\)\$ root(3)(x^3 +1 -1)= root(3)(x^3 )=x\$