The function f(a) = a
is the one-to-one correspondence from A
to itself.
If A ~ B,
there is a function f(a) = b
where each b B
appears for exactly one a A
and the function is defined for every b B
and a A.
The inverse function g(b) = a
where b = f(a)
shows B ~ A.
If A ~ B
and B ~ C,
then there are one-to-one functions f(a) = b
and g(b) = c
for each mapping. The function h(a) = g(f(a))
then is a one-to-one mapping between A
and C.
For (a) and (b), the table is Pascal’s triangle, which we already have in the notes. For (c), we
want the entry {P}_{6,3}
in our earlier notation. There are 20 such ways.
Wow. Never, ever give someone a list of all your identifying numbers. Identity theft
is a serious problem. While quite often all these numbers are available for a little
work, at least make a criminal work for them.
I’m just going to show these as lines. Illustrating them on a “number line” is
equivalent. A better diagram for the last would wrap portions of the result in
parentheses to show how the line extended.
Any set that contains only 1 and some other number is closed because 1
is the multiplicative identity.
Any set that contains only 1 and some other number is closed because 1
is the multiplicative identity.
2 ⋅ 4 = 8\mathrel{∉}\{0,2,4\},
so this is not closed.
The product of even numbers always is even, so this is closed.
The product of odd numbers cannot be even, so they must be odd and
this set is closed.
{2}^{2} ⋅ {2}^{3} = {2}^{5}\mathrel{∉}\{1,2,{2}^{2},{2}^{3}\},
so this set is not closed.
The product of powers of two is a power of two, so this set is closed.
Similarly, the product of powers of seven is a power of seven so this set is
closed. Both of these are closed because \{0,1,2,\mathop{\mathop{…}}\}
is closed under addition; {a}^{i} ⋅ {a}^{j} = {a}^{i+j},
moving the property up to the superscript.
The operation is closed because no new shapes are introduced in the operator’s table
of results.
The operation is commutative because the operation table is symmetric across the diagonal
axis. Thus a ⋆ b = b ⋆ a.
Because ∘ ⋆x = x,
the ∘
symbol is ⋆’s
identity.
After the identity, there are only two symbols remaining. Thus the commutative
property combined with the identity means that this operator must be associative.
Draw two 1 × 2 × 3 boxes
made of 1 × 1 × 1 boxes. To
illustrate (1 ⋅ 2) ⋅ 3, show that
the 1 × 2 portion is stacked
3 times. To illustrate 1 ⋅ (2 ⋅ 3),
show that the 2 × 3
portion is stacked across 1 time.