natural number

a natural number is any element of the set
$\text{[math]}$ which is the set of all the numbers created by starting with 0 and then counting forward indefinitely. we call $\text{[math]}$ the set of natural numbers.
[cite:@tao_analysis_1 definition 2.1.1]

in some texts the natural numbers start at 1 instead of 0, but this is a matter of notational convention more than anything else. in [cite:@tao_analysis_1] we refer to the set $\text{[math]}$ as the positive integers $\text{[math]}$ rather than the natural numbers.
natural numbers are sometimes also known as whole numbers.

in a sense, broken link: blk:def-natural-set-inf solves the problem of what the natural numbers are: a natural number is any element of the set $\text{[math]}$ . however, it is not really that satisfactory, because it begs the question of what $\text{[math]}$ is. This definition of "start at 0 and count indefinitely" seems like an intuitive enough definition of $\text{[math]}$ , but it is not entirely acceptable, because it leaves many questions unanswered. for instance: how do we know we can keep counting indefinitely, without cycling back to 0? Also, how do you perform operations such as addition, multiplication, or exponentiation? we can answer the latter question first: we can define complicated operations in terms of simpler operations. exponentiation is nothing more than repeated multiplication: $\text{[math]}$ is nothing more than three fives multiplied together. multiplication is nothing more than repeated addition; $\text{[math]}$ is nothing more than three fives added together. (subtraction and division are not be covered here, because they are not operations which are well-suited to the natural numbers; they are defined for the integers and rationals, respectively.) and addition? it is nothing more than the repeated operation of counting forward, or incrementing. if you add three to five, what you are doing is incrementing five three times. on the other hand, incrementing seems to be a fundamental operation, not reducible to any simpler operation; indeed, it is the first operation one learns on numbers, even before learning to add.

we define 1 to be the number $\text{[math]}$ , 2 to be the number $\text{[math]}$ , 3 to be the number $\text{[math]}$ , etc. (in other words, $\text{[math]}$ , $\text{[math]}$ , $\text{[math]}$ , etc.

there exists a number system \nnumset, whose elements we call natural numbers, for which the peano axioms are true.
[cite:@tao_analysis_1 assumption 2.6]

3 is a natural number.

by broken link: blk:axi-peano-zero, 0 is a natural number. by broken link: blk:axi-peano-inc, 0++ = 1 is a natural number. by broken link: blk:axi-peano-inc again, 1++ = 2 is a natural number. by broken link: blk:axi-peano-inc again, 2++ = 3 is a natural number.

the principle of induction gives us a way to prove that a property $\text{[math]}$ is true for every natural number $\text{[math]}$ . we see many proofs which have a form similar to this:

a certain property $\text{[math]}$ is true for every natural number $\text{[math]}$ .

we use induction. we first verify the base case $\text{[math]}$ , i.e., we prove $\text{[math]}$ . (insert proof of $\text{[math]}$ here.) now suppose inductively that $\text{[math]}$ is a natural number, and $\text{[math]}$ has already been proven. we now prove $\text{[math]}$ . (insert proof of $\text{[math]}$ , assuming that $\text{[math]}$ is true, here.) this closes the induction, and thus $\text{[math]}$ is true for all numbers n.

suppose for each natural number $\text{[math]}$ , we have some function $\text{[math]}$ from the natural numbers to the natural numbers. let $\text{[math]}$ be a natural number. then we can assign a unique natural number $\text{[math]}$ to each natural number $\text{[math]}$ , such that $\text{[math]}$ and $\text{[math]}$ for each natural number $\text{[math]}$ .

we use induction. we first observe that this procedure gives a single value to $\text{[math]}$ , namely $\text{[math]}$ . (none of the other definitions $\text{[math]}$ will redefine the value of $\text{[math]}$ , because of broken link: blk:axi-peano-diff) now suppose inductively that the procedure gives a single value to $\text{[math]}$ . then it gives a single value to $\text{[math]}$ , namely $\text{[math]}$ . (none of the other definitions $\text{[math]}$ will redefine the value of $\text{[math]}$ , because of broken link: blk:axi-peano-induction) this completes the induction, and so an is defined for each natural number $\text{[math]}$ , with a single value assigned to each $\text{[math]}$ .

the definition we proposed broken link: blk:pro-natural-rec isnt circular because the concept of a function doesnt require the peano axioms.

let $\text{[math]}$ be a natural number. to add zero to $\text{[math]}$ , we define $\text{[math]}$ . now suppose inductively that we have defined how to add $\text{[math]}$ to $\text{[math]}$ . then we can add $\text{[math]}$ to $\text{[math]}$ by defining $\text{[math]}$ .
[cite:@tao_analysis_1 definition 2.2.1]

it is implicitly defined in broken link: blk:def-natural-add that the addition of two natural number yields a third natural number, as $\text{[math]}$ yields $\text{[math]}$ which is assumed to be a natural number, and by the induction hypothesis $\text{[math]}$ is defined and also yields a natural number, then $\text{[math]}$ is defined to be $\text{[math]}$ which is a natural number by broken link: blk:axi-peano-inc and the inductive hypthesis.
this means that any operations that apply to natural numbers also apply to a parenthesized addition of natural numbers such as $\text{[math]}$ , we can increment them, e.g. $\text{[math]}$ , or add additions of natural numbers to other additions of natural numbers, e.g. $\text{[math]}$ , and so on.

for any natural number $\text{[math]}$ , $\text{[math]}$ .
[cite:@tao_analysis_1 lemma 2.2.2]

we induct on $\text{[math]}$ , for the base case, $\text{[math]}$ by broken link: blk:def-natural-add, we assume $\text{[math]}$ (the induction hypothesis) and need to show $\text{[math]}$ , by broken link: blk:def-natural-add we know $\text{[math]}$ and by the induction hypothesis we know $\text{[math]}$ so $\text{[math]}$ which means $\text{[math]}$ .

the base case $\text{[math]}$ follows since we know that $\text{[math]}$ for every natural number $\text{[math]}$ , and 0 is a natural number. now suppose inductively that $\text{[math]}$ . we wish to show that $\text{[math]}$ . but by definition of addition, $\text{[math]}$ is equal to $\text{[math]}$ , which is equal to $\text{[math]}$ since $\text{[math]}$ . this closes the induction.
[cite:@tao_analysis_1 lemma 2.2.2]

for any natural numbers $\text{[math]}$ and $\text{[math]}$ , $\text{[math]}$

we induct on $\text{[math]}$ , keeping $\text{[math]}$ fixed, for the base case $\text{[math]}$ we need to show $\text{[math]}$ , by broken link: blk:def-natural-add we know $\text{[math]}$ and $\text{[math]}$ so $\text{[math]}$ which means the base case $\text{[math]}$ is true as both equal $\text{[math]}$ , we assume the induction hypothesis, $\text{[math]}$ , and have to show $\text{[math]}$ , by broken link: blk:def-natural-add we know $\text{[math]}$ and $\text{[math]}$ by the induction hypothesis and $\text{[math]}$ by broken link: blk:def-natural-add.

$\text{[math]}$

for any natural numbers $\text{[math]}$ and $\text{[math]}$ , $\text{[math]}$ .

we induct on $\text{[math]}$ , keeping $\text{[math]}$ fixed, for the base case, we must show $\text{[math]}$ , but we already know both equal $\text{[math]}$ by broken link: blk:def-natural-add and broken link: blk:lem-natural-add-zero. we assume the inductive hypotheis $\text{[math]}$ and we have to show $\text{[math]}$ : by broken link: blk:def-natural-add $\text{[math]}$ and by broken link: blk:def-natural-add-suc $\text{[math]}$ and by the inductive hypothesis $\text{[math]}$ .

for any natural numbers $\text{[math]}$ , we have $\text{[math]}$ .

we induct on $\text{[math]}$ , keeping $\text{[math]}$ and $\text{[math]}$ fixed, for the base case $\text{[math]}$ , we need to show $\text{[math]}$ , by broken link: blk:def-natural-add we know $\text{[math]}$ so $\text{[math]}$ and we know $\text{[math]}$ , so both sides are equal. we assume the inductive hypothesis $\text{[math]}$ and have to show $\text{[math]}$ , by broken link: blk:def-natural-add, $\text{[math]}$ , again by broken link: blk:def-natural-add $\text{[math]}$ and by the inductive hypothesis $\text{[math]}$ , by broken link: blk:def-natural-add, $\text{[math]}$ .

let $\text{[math]}$ be natural numbers such that $\text{[math]}$ , then $\text{[math]}$ .

we fix $\text{[math]}$ and induct on $\text{[math]}$ , for the base case, $\text{[math]}$ implies $\text{[math]}$ by broken link: blk:def-natural-add, we assume the inductive hypothesis, $\text{[math]}$ implies $\text{[math]}$ , and need to show $\text{[math]}$ implies $\text{[math]}$ , by broken link: blk:def-natural-add, $\text{[math]}$ and again by broken link: blk:def-natural-add $\text{[math]}$ , so we have $\text{[math]}$ , by broken link: blk:axi-peano-diff $\text{[math]}$ , and so by the inductive hypothesis $\text{[math]}$ , which closes the induction.

a natural number $\text{[math]}$ is said to be positive iff it is not equal to 0.

if $\text{[math]}$ is a positive natural number, and $\text{[math]}$ is a natural number, then $\text{[math]}$ is positive (and hence $\text{[math]}$ is also, by broken link: blk:pro-natural-add-com).

we fix $\text{[math]}$ and induct on $\text{[math]}$ , for the base case $\text{[math]}$ , $\text{[math]}$ is positive, we assume the inductive hypothesis; i.e. that $\text{[math]}$ is positive, and have to show that $\text{[math]}$ is positive, by broken link: blk:lem-natural-add-suc $\text{[math]}$ , and by broken link: blk:axi-peano-nosuc $\text{[math]}$ is positive as zero isnt the successor of any natural number, this closes the induction.

if $\text{[math]}$ and $\text{[math]}$ are natural numbers such that $\text{[math]}$ , then $\text{[math]}$ and $\text{[math]}$ .

assume in contradiction that $\text{[math]}$ or $\text{[math]}$ (that one of them is non-zero, i.e. one of them is positive), then we would get a contradiction by broken link: blk:pro-natural-pos-add which says $\text{[math]}$ would be positive but we have $\text{[math]}$ .

let $\text{[math]}$ be a positive number. then there exists exactly one natural number $\text{[math]}$ such that $\text{[math]}$ .
[cite:@tao_analysis_1 lemma 2.2.10]