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The 2-adic valuation is the P-adic valuation at the prime number $p=2$.

An additive nontrivial valuation on the rational field is equivalent to exactly one P-adic valuation. Integers have nonnegative values; some prime must have positive value, and Bézout's identity prevents two distinct primes from doing so. Unique factorization then gives the formula, with $c>0$ in the real-valued case. The trivial valuation is a separate possibility if admitted by the definition.

Two p-adic integers agree in their first $n$ low-order p-adic digits precisely when their difference is divisible by $p^n$. Equivalently,If the first differing digit has index $j$, the remaining leading coefficient is nonzero modulo $p$, giving exact P-adic valuation $j$ and p-adic metric $p^{-j}$. If no digit differs, the two elements are equal and their distance is zero. Truncating at $r_n$ therefore gives $|x-r_n{|}_p\le p^{-n}$, with a stricter bound possible when further digits are zero.

After clearing denominators in a matrix, for every disjoint pair of index sets $I,B$ with $B$ nonempty and ${\sum }_{i\in B}{a}_i$ outside the linear span of the $I$-columns, choose an integer linear functional vanishing on those columns but not on the indicated sum. Choose a prime number dividing none of the finitely many nonzero evaluations. The last nonzero digit coloring for this prime, applied to a monochromatic positive solution, groups coordinates into blocks by P-adic valuation. Applying the appropriate functional and reducing modulo the prime shows that each block sum lies in the linear span of all earlier columns. This proves necessity in Rado's theorem without an asymptotic limiting argument.

For a prime number $p$ and integers $0\le k\le n$, the P-adic valuation of the binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ equals the number of arithmetic carries in the base-$p$ addition of $k$ and $n-k$.^[ref] Indeed, subtract the two denominator factorial valuations in the Legendre formula:Each arithmetic carry accounts for one unit of this quotient by subadditivity of the base-p digit sum. In particular, the binomial coefficient is divisible by $p$ exactly when at least one carry occurs.

For $n=50,k=17,p=3$, the factorial counts giveThe matching base-p expansions are $17=(0122{)}_3$ and $33=(1020{)}_3$. Their units add to two; the next column gives $2+2=4=1+3$, sending one carry into the nines column. That column then gives $1+0+1=2$, and no further carry occurs. The result is $(1212{)}_3=50$. The base-p digit sums check the same result: $(5+3-6)/(3-1)=1$. Consequently $\left(\genfrac{}{}{0ex}{}{50}{17}\right)$ is divisible by three but not by nine.

For a nonnegative integer $n$ and a prime number $p$, the Legendre formula computes the prime exponent in a factorial without forming that huge product:Here $\lfloor x\rfloor$ is the floor function. The sum is finite: every term with $p^j>n$ is zero. When $n=0$, all terms vanish, agreeing with $0!=1$ and $v_p(1)=0$. The prime number hypothesis matters: a composite number does not give the same P-adic valuation or additive prime exponent. For example, the floor sum with $n=6,p=4$ is one, whereas $4^2$ divides $6!=720$.

Why count multiples repeatedly? Each multiple of $p$ in $1,\dots ,n$ contributes one factor of $p$; each multiple of $p^2$ contributes another; and so on. A factor $m$ with P-adic valuation $a$ occurs in exactly the first $a$ counts. Adding those contributions proves the formula.^[ref]

For the original example $n=50,p=3$,The five multiples $9,18,27,36,45$ each contribute a second factor of three. The factor $27$ contributes a third, so it appears in all three rows. Equivalently, among $1,\dots ,50$ there are eleven factors of P-adic valuation one, four of valuation two, and one of valuation three: $11+2\cdot 4+3\cdot 1=22$. Thus $3^{22}$ divides $50!$, while $3^{23}$ does not.

Repeated Euclidean division provides a practical evaluation: replace $q$ by $\lfloor q/p\rfloor$, starting from $q=n$, and add each new quotient until it is zero. This uses only the successive quotients $16,5,1,0$ for the example. The same calculation handles divisibility of binomial coefficients, where the P-adic valuations of the denominator factorials are subtracted.

Fix a prime number $p$. Write a nonzero rational number as $x=p^{k}a/b$, where $k\in \mathbb{Z}$ and the nonzero integers $a,b$ are not divisible by $p$. The exponent $k$ is uniquely determined. Its P-adic valuation is this signed exponent $v_p(x)=k$: equivalently, for any nonzero integer numerator $A$ and denominator $B$,Factors of $p$ in the numerator decrease the p-adic absolute value; factors in the denominator increase it. For $p=2$ the computations are:

This is an absolute value on a field measuring divisibility, rather than the magnitude on a real number number line. It satisfiesThe proof comes directly from P-adic valuations: exponents add in a product of numbers, and extracting a common power of $p$ from a sum gives $v_p(x+y)\ge \min \{v_p(x),v_p(y)\}$. If the two P-adic valuations differ, the remaining term with smaller exponent cannot cancel modulo $p$, so equality holds. Thus the p-adic absolute value is a Non-Archimedean absolute value. ^[ref]

The metric $d_p(x,y)=|x-y{|}_p$ makes two rational numbers close when their difference contains a high power of $p$. This packages successive modular congruences into a numerical error estimate, allowing convergence in a metric space and convergent series to encode exact arithmetic information.

The necessity direction in Rado's theorem colors an integer using initial data from a P-adic valuation. Applying partition regularity and grouping the coordinates of a monochromatic solution by valuation yields the blocks in the columns property; reduction modulo successively higher powers of $p$ supplies the required linear dependences.

For a prime number $p$, a p-adic number is an element of the field ${\mathbb{Q}}_p$ obtained by completing the rational numbers for the metric $d_p(x,y)=|x-y{|}_p$. Here $|\cdot {|}_p$ is the p-adic absolute value. The Cauchy-sequence construction of a metric completion supplies limits for every Cauchy sequence, and addition, multiplication, and the p-adic absolute value extend to this complete metric space. The rational numbers form a dense subset. ^[ref]

For a concrete example with $p=2$, consider the partial sums of a geometric series:The first values $1,3,7,15,31$ have distances $1/2,1/4,1/8,1/16,1/32$ from $-1$ in the 2-adic absolute value. ConsequentlyThis is a statement about convergence in a metric space with the 2-adic absolute value. In the ordinary real number absolute value, the same partial sums grow without bound. Choosing a metric changes which approximation errors shrink.

More generally every nonzero P-adic number has a unique digit expansionOnly finitely many negative powers occur, while positive powers can continue indefinitely. Written with the highest powers on the left, the digits extend infinitely to the left; for example the P-adic number $-1$ at $p=2$ is written $\dots 111_2$. The p-adic integers are exactly the elements with $|x{|}_p\le 1$, whose digit expansions use only nonnegative powers. ^[ref]

The construction can be understood through modular arithmetic. Successively choose the low-order digits to agree modulo $p$, $p^2$, and higher powers. After fixing digits below $p^n$, the remaining error has p-adic absolute value at most $p^{-n}$; the partial sums form a Cauchy sequence, so completeness supplies a limit. If two expansions first differ at position $j$, their difference has P-adic valuation $j$, proving uniqueness. This turns compatible finite congruence classes into one exact number. In particular, Hensel's lemma upgrades suitable approximate roots of a polynomial modulo powers of $p$ into roots in the p-adic integers.

For a prime number $p$ and a nonzero integer $N$, the P-adic valuation counts the exponent of $p$ in its prime factorization:For $72=2^3\cdot 3^2$, it gives $v_2(72)=3$, $v_3(72)=2$, and $v_5(72)=0$. The sign of a nonzero integer does not change these prime exponents. Relative to an ordered factor base, the valuations supply the coordinates of its exponent vector.

The P-adic valuation extends to nonzero rational numbers by $v_p(a/b)=v_p(a)-v_p(b)$. For example, $v_5(72/25)=-2$ records a factor in the denominator instead of the numerator. Multiplication adds valuations, $v_p(xy)=v_p(x)+v_p(y)$; this is the arithmetic reason why adding exponent vectors can make a product a square. In particular, a positive integer is a square of a number precisely when all its prime exponents are even integers, the criterion used by a factor base.

No prime works. The P-adic valuation induces a well-defined homomorphismThe class of $p$ maps to one, whereas the class of $p^p$ maps to zero. They therefore cannot be equal. Equivalently, $p$ cannot be a $p$th power because its valuation is not divisible by $p$.

The strict inequality means $v_3(x^2+2)>2$. Taking $x=5$ gives $x^2+2=27$, soFor the other expression, if $3\nmid x$ then $x^2+3\not\equiv 0(\mathrm{mod}3)$ and its P-adic valuation is zero. If $3\mid x$, then $x^2+3=3(3(x/3{)}^2+1)$ has P-adic valuation exactly one. It can never have valuation greater than two. Thus no such integer exists for the second strict inequality.

For a rational $\ell \times n$ matrix $A$, write its columns as $c_1,\dots ,c_n$. It is partition regular if every finite colouring of $\mathbb{N}$ admits positive integers $x_1,\dots ,x_n$, all of one colour, with $A\mathbf{x}=0$. Repeated coordinates are allowed. The columns condition is an ordered partitioninto nonempty blocks such thatRado's theorem states that a rational matrix is partition regular if and only if it satisfies the columns condition. We prove both implications.

For necessity, clear denominators so the columns are integral. For every disjoint pair of index sets $I,B$ with $B\ne \varnothing$ andchoose a rational linear functional annihilating all columns in $I$ but not $v_B$. Such a functional exists by extending a basis of the indicated linear span to include $v_B$, prescribing zero on the former and one on the latter. Clearing its denominators produces an integer-coefficient functional $f_{I,B}$ with $f_{I,B}(v_B)$ a nonzero integer. There are only finitely many pairs $I,B$. Choose a prime number $p$ dividing none of these nonzero evaluations; if there are no bad pairs, choose any prime.

Use the last nonzero digit coloring for this prime. Namely, write $x=p^{{\nu }_p(x)}u$ with $p\nmid u$ and colour $x$ by $u\mathrm{mod}p\in \{1,\dots ,p-1\}$. By partition regularity, obtain a positive monochromatic solution. Let the distinct P-adic valuations of its coordinates beand put $B_j=\{i:{\nu }_p(x_i)=v_j\}$. All unit parts $u_i$ have a common nonzero residue $t$ modulo $p$.

For a block $B_j$, let $I=B_1\cup \cdots \cup B_{j-1}$. If its column sum were outside the earlier linear span, apply $f_{I,B_j}$ to the equation ${\sum }_i{c}_i{x}_i=0$. Earlier-block contributions are exactly zero. Divide the remaining equation by $p^{v_j}$ and reduce modulo $p$. Higher-valuation terms vanish, leavingThis contradicts the choice of $p$ and $t\not\equiv 0$. Consequently every block sum lies in the earlier linear span. For $j=1$ that span is $\{0\}$, so its sum is zero. This proves the columns condition by the finite separating-functional proof of the columns condition.

For sufficiency, suppose the blocks satisfy the columns condition. Choose rational coefficients ${\alpha }_{ij}$, for $i\in B_1\cup \cdots \cup B_{j-1}$, such thatTake a positive integer $c$ clearing all their denominators and an integer $p\ge 1$ at least as large as every $|c{\alpha }_{ij}|$. An M-p-c set with generators $z_1,\dots ,z_s$ consists of all valueswith every such value positive. The generators are therefore chosen with $c{z}_h>p{\sum }_{j>h}{z}_j$. Positivity is part of the configuration, not achieved by deleting inconvenient values.

Use the permitted monochromatic m-p-c set theorem to obtain one such monochromatic set. For each $i\in B_h$, putEvery $x_i$ is a positive element of that set, because its non-leading coefficients are integers of magnitude at most $p$. In the sum ${\sum }_i{c}_i{x}_i$, the coefficient of $z_1$ is $c{\sum }_{i\in B_1}{c}_i=0$, and the coefficient of each $z_j$ with $j>1$ isThus $A\mathbf{x}=0$. This is an explicit Rado solution inside an m-p-c set and proves sufficiency.

For the Finite sums theorem, defineTo deduce its monochromaticity from Rado's theorem, introduce variables $z_I$ for all nonempty $I\subseteq [k]$ and equationsLet $A_k$ be their coefficient matrix, and partition its columns by $B_j=\{I:\min I=j\}$. For a fixed $j$, the vector with $I$-coordinate $1$ when $j\in I$, and $0$ otherwise, solves this homogeneous system: in every row both sides count whether $j$ belongs to its indexing set. HenceThe first block sum is zero, and each later block sum lies in the linear span of earlier columns. This proves the columns partition for finite-sums systems. The positive monochromatic solution supplied by Rado's theorem has $x_i=z_{\{i\}}>0$, and its other coordinates are exactly all the nonempty finite sums. Therefore $FS(x_1,\dots ,x_k)$ is monochromatic. For $k=1$ a singleton is immediate. Equivalently, the full positive M-p-c set with $m=k,p=c=1$ directly contains every such sum, by taking its leading index to be the least index in $I$.

For a rational matrix $A$ with columns $a_1,\dots ,a_n$, partition regularity means that every finite colouring of the positive integers admits a positive monochromatic vector $x$ with $Ax=0$. Repeated coordinates are allowed. The columns condition is an ordered partition $\{1,\dots ,n\}=B_1\bigsqcup \cdots \bigsqcup B_s$ into nonempty blocks such thatRado's theorem says that $A$ is partition regular if and only if it satisfies the columns condition. We prove both directions.

For necessity, clear denominators so the columns are integer vectors. For every pair of disjoint index sets $I,B$ with $B\ne \varnothing$ for which ${\sum }_{i\in B}{a}_i$ is not in the rational linear span of the I-columns, choose an integer linear functional ${\ell }_{I,B}$ that vanishes on those I-columns but not on that sum. Such a functional exists by finite-dimensional linear algebra; clear its rational denominators. There are only finitely many pairs, so choose a prime number $p$ dividing none of the nonzero integersIf there are no offending pairs, any prime number suffices.

Colour a positive integer $x=p^{v_p(x)}u$, $p\nmid u$, by $u\mathrm{mod}p$. This is the last nonzero digit coloring, using $p-1$ colours. Partition regularity gives a positive monochromatic solution. Group its coordinate indices by their distinct P-adic valuations $e_1<\cdots <e_s$, writing $B_j=\{i:v_p(x_i)=e_j\}$. All units $x_i/p^{e_j}$ have the same nonzero residue $\alpha$ modulo $p$.

For each $j$, let $I=B_1\cup \cdots \cup B_{j-1}$. If its block sum failed the required span condition, apply the chosen ${\ell }_{I,B_j}$ to ${\sum }_i{a}_i{x}_i=0$. The earlier terms vanish exactly, not just modulo a power of $p$. Divide by $p^{e_j}$; terms with later valuations are still divisible by $p$. Reduction modulo $p$ then yieldscontrary to the choice of $p$. Thus every required span relation holds, and with $I=\varnothing$ the first block sum is zero. This is the finite separating-functional proof of the columns condition; it avoids an unjustified passage from congruences to rational equality.

For sufficiency, assume the columns condition. Choose rational coefficients ${\lambda }_{ij}$, for $j\ge 2$ and $i\in B_1\cup \cdots \cup B_{j-1}$, such thatChoose a positive integer $c$ clearing every denominator, and a positive integer $p\ge \max |c{\lambda }_{ij}|$, taking $p=1$ if the list is empty. By the allowed monochromatic m-p-c set theorem, there are positive generators $z_1,\dots ,z_s$ such that all numbersare positive and have one colour. This is a full m-p-c set; positivity is part of its definition, not a rule discarding negative expressions. For $i\in B_h$, putThese coordinates all belong to that monochromatic m-p-c set. Collecting coefficients of $z_j$ givesThus the Rado solution inside an m-p-c set proves partition regularity and completes the theorem.

To obtain the Finite sums theorem from Rado's theorem, introduce a positive variable $y_F$ for every nonempty $F\subseteq \{1,\dots ,k\}$ and imposePartition the columns of this system into $B_j=\{F:\min F=j\}$. The $B_1$-column sum is zero: in a row indexed by $F$ containing $1$, the contributions of $y_F$ and $y_{\{1\}}$ cancel, and other rows receive neither. For $j>1$, the $B_j$-column sum iswhere $a_F$ is the column of $y_F$. Those columns lie in earlier blocks. Hence the columns condition holds. A monochromatic positive solution gives $x_i=y_{\{i\}}$ and $y_F={\sum }_{i\in F}{x}_i$, provingFor $k=1$ the system has no rows and the conclusion is immediate. The construction is also visible directly in the sufficiency proof: a monochromatic $(k,1,1)$-set contains every nonempty sum of its generators, classified by the smallest chosen index.

For finite sets $A_1,\dots ,A_r$, the inclusion-exclusion principle isTo prove it, fix an element occurring in exactly $k$ of these sets. Its coefficient on the right is ${\sum }_{s=1}^k(-1{)}^{s+1}(\genfrac{}{}{0ex}{}{k}{s})=1$, by the binomial theorem applied to $(1-1{)}^k$. An element occurring in none has coefficient zero. Summing these elementwise coefficients proves the formula.

Write the prime factorization $n={\prod }_{i=1}^r{p}_i^{c_i}$ with distinct primes and positive exponents. In $\{1,\dots ,n\}$ let $A_i$ be the multiples of $p_i$. A number is coprime to $n$ exactly when it is outside their union. For a nonempty index set $J$, the intersection has $n/{\prod }_{j\in J}{p}_j$ elements, since that product divides $n$. Applying inclusion-exclusion to the complement therefore givesFor $n=1$, the empty product gives $\phi (1)=1$. If positive integers $a,b$ are coprime, their prime factors are disjoint, so the product separates and proves the multiplicativity of the Euler totient function:For a positive divisor $d\mid n$, write the exponent of a prime in $d$ as $b_i\le c_i$, allowing $b_i=0$. Its contribution to $\phi (n)/\phi (d)$ is $p_i^{c_i-b_i}$ if $b_i>0$, and $p_i^{c_i-1}(p_i-1)$ if $b_i=0$. Every contribution is an integer; multiplying proves totient divisibility along divisors:Finally suppose $\phi (n)\mid n$. If $n=1$ the required form is immediate. If $n>1$ were odd, some odd prime $p$ would divide it, and the even factor $p-1$ in $\phi (n)$ would make an even number divide an odd number, impossible. Thus write $n=2^c{\prod }_{j=1}^s{p}_j^{a_j}$ with odd primes and $c\ge 1$. Counting powers of two in the totient formula giveswhere $v_2$ is the $2$-adic P-adic valuation. Every term in the sum is at least one, so there is at most one odd prime $p$ and, if present, $v_2(p-1)=1$. Moreover $p-1\mid \phi (n)\mid n$. Since $p-1$ is coprime to $p$, and $n$ has no prime factors other than $2,p$, the integer $p-1$ must be a power of two. Its exponent is one, hence $p-1=2$ and $p=3$. This provesMore precisely, the integers whose totient divides them are $n=1$ and $2^c{3}^d$ with $c\ge 1$, $d\ge 0$; substitution in the product formula proves the converse. A positive power of $3$ alone does not satisfy the divisibility condition.

Van der Waerden theorem says that for every pair of positive integers $r,k$ there is $W(r,k)$ such that every finite colouring of $[W(r,k)]=\{1,\dots ,W(r,k)\}$ using at most $r$ colours contains a monochromatic arithmetic progression of length $k$ and positive common difference. Consequently every finite colouring of $\mathbb{N}$ contains arbitrarily long monochromatic arithmetic progressions.

Here is a colour-focusing proof of Van der Waerden theorem. Induct on $k$, simultaneously for all $r$. For $k=1$, take $W(r,1)=1$. Suppose $W(R,k)$ exists for every number of colours $R$, and fix $r$. We will construct finite bounds $F_t$, $1\le t\le r$, with the following property: every finite colouring of $[F_t]$ using at most $r$ colours contains either a monochromatic length-$(k+1)$ arithmetic progression, or $t$ length-$k$ arithmetic progressions of pairwise distinct colours, all focused at the same point $f\in [F_t]$. Specifically, their terms are“Focused” means that adjoining $f$ completes the corresponding length-$(k+1)$ arithmetic progression if $f$ has its colour.

For $t=1$, put $F_1=2W(r,k)$. Find a monochromatic length-$k$ arithmetic progression in the first $W(r,k)$ points. Its next point is at most $2W(r,k)$ and supplies the focus. For $k=1$, give the selected singleton common difference $1$.

Suppose $F_t=N$ has been constructed. SetDivide $[2NM]$ into $2M$ consecutive blocks of length $N$. Colour the first $M$ block indices by their full colour profiles, that is, the ordered list of their $N$ colours. There are at most $r^N$ profiles, so the induction hypothesis on $k$ gives equally profiled blocks with indicesWhen $k=1$, choose $D=1$. Otherwise $D>0$ and $j_0+kD\le 2M$, since the last selected block is at most $M$ and $D\le M$.

Assume the original finite colouring contains no monochromatic length-$(k+1)$ arithmetic progression. Applying the property of $F_t$ to the common profile gives $t$ focused length-$k$ arithmetic progressions at a position $f$, with common differences $d_i$ and distinct colours $q_i$. The colour $q$ of position $f$ differs from every $q_i$: otherwise the profile already contains a length-$(k+1)$ monochromatic arithmetic progression, and so does any selected block.

Put $j_{\ast }=j_0+kD$ and $F=N(j_{\ast }-1)+f$, a point in the original interval. For $1\le h\le k$, the pointlies in a selected block at profile position $f-h{d}_i$, so has colour $q_i$. Similarly $F-hND$ has colour $q$. Thus $t+1$ length-$k$ arithmetic progressions of distinct colours focus at $F$. This proves the step in $t$. At $t=r$, the focus necessarily has the colour of one of the $r$ progressions, giving a monochromatic length-$(k+1)$ arithmetic progression. Hence $W(r,k+1)$ exists, completing the mathematical induction and the proof of Van der Waerden theorem.

For the partition regularity deduction we first prove a strengthened progression lemma, the Brauer progression theorem: for positive $r,s,L$ with $L\ge 2$, there is $B(r,s,L)$ such that every finite colouring of $[B(r,s,L)]$ using at most $r$ colours contains a monochromatic setInduct on $r$. For $r=1$, take $B(1,s,L)=\max (s,L)$ and $a=d=1$. Given the result for $r-1$, putVan der Waerden theorem supplies a colour-$q$ arithmetic progression $a+j{d}_0$, $0\le j<K$, inside $[W]$. If some $sj{d}_0$, $1\le j\le N$, also has colour $q$, take $d=j{d}_0$ and use the $L$ terms $a,a+j{d}_0,\dots ,a+(L-1)j{d}_0$. They are in the progression because $(L-1)j\le K-1$.

Otherwise the integers $sj{d}_0$, $1\le j\le N$, use at most $r-1$ colours. Colour $j\in [N]$ by the colour of $sj{d}_0$ and apply the induction hypothesis. It supplies a monochromatic set $\{se,b,b+e,\dots ,b+(L-1)e\}$. Multiplying by $s{d}_0$ gives the required configuration with $a=s{d}_{0}b$ and $d=s{d}_{0}e$: its distinguished member is $s^2{d}_{0}e=sd$. All of these points are at most $sN{d}_0\le sNW$, so the finite bound is valid. This proves the Brauer progression theorem using only the just-proved Van der Waerden theorem.

A row is a partition regular matrix over the positive integers if every finite colouring of $\mathbb{N}$ admits a positive vector $x$ whose coordinates have one colour and whose product with the row is zero. Coordinates may repeat. Since the coefficients are rational numbers, multiplying them by a common positive denominator leaves the equation and every zero subset sum unchanged, so write the row as nonzero integer coefficients $b_1,\dots ,b_n$.

To prove necessity, suppose no nonempty subset of the $b_i$ sums to zero. Choose a prime $p>{\sum }_i|b_i|$ and use the last nonzero digit coloringwhere $v_p$ is the P-adic valuation. If positive $x_i$ all had the same colour $u$ and satisfied ${\sum }_i{b}_i{x}_i=0$, let $m={\min }_i{v}_p(x_i)$ and $I=\{i:v_p(x_i)=m\}$. After division by $p^m$ and reduction modulo $p$, the equation becomesAs $u\ne 0$, this says $p\mid {\sum }_{i\in I}{b}_i$. The latter sum is nonzero and has absolute value at most ${\sum }_i|b_i|<p$, a contradiction. This explicitly proves the necessary obstruction without invoking a general partition regularity criterion.

Conversely, suppose $\varnothing \ne I\subseteq \{1,\dots ,n\}$ and ${\sum }_{i\in I}{b}_i=0$. Select an index $j\in I$, and putFor any finite colouring, the Brauer progression theorem gives one-coloured $\{sd,a,a+d,\dots ,a+(L-1)d\}$. The indices $u$ and $u+q$ both lie between $0$ and $L-1$. Define positive, same-coloured coordinates byThenThis also covers $I$ equal to the whole index set and $q=0$. Reversing the clearing of denominators proves the exact criterion:

Here is the matrix form of Rado's theorem. A matrix $A$ with entries in the rational numbers and columns $a_1,\dots ,a_n$ is a partition regular matrix if every finite coloring of the positive integers admits a vector $x\in {\mathbb{N}}^n$ whose coordinates are monochromatic and for which $Ax=0$. Its columns condition asks for an ordered partition into nonempty blocks $B_1,\dots ,B_s$, withRado's theorem states that these two conditions are equivalent. Coordinates need not be distinct; positivity is required for every coordinate.

First prove necessity using the finite separating-functional proof of the columns condition. Clearing denominators does not change either the kernel or the columns condition, so assume that $A$ has integer entries. For each pair of disjoint index sets $I,B\subseteq \{1,\dots ,n\}$, with $B$ nonempty, for which $b_B={\sum }_{i\in B}{a}_i$ is outside the rational linear span of the $I$-columns, choose an integer linear functional ${\ell }_{I,B}$ such thatSuch a linear functional exists: extend a basis over the rational numbers of the indicated linear span by $b_B$, prescribe values zero on the former basis and one on $b_B$, extend to a basis of the ambient vector space over the rational numbers, and then clear the functional's denominators. There are only finitely many pairs $(I,B)$. Choose a prime number $p$ larger than the absolute values of all their nonzero evaluations; if there are none, choose any prime number.

Color $x>0$ by its last nonzero digit coloring value, namelywhere $v_p$ is the P-adic valuation. By partition regularity, choose a positive solution $Ax=0$ all of whose coordinates have the same nonzero digit $u$. Group its indices into nonempty blocks $B_1,\dots ,B_s$ according to increasing distinct P-adic valuations ${\nu }_1<\cdots <{\nu }_s$.

Suppose that a block sum $b_j$ is outside the rational linear span of the earlier columns, and put $I=B_1\cup \cdots \cup B_{j-1}$, taking $I=\varnothing$ for $j=1$. Apply the selected linear functional ${\ell }_{I,B_j}$ to $Ax=0$. The earlier terms vanish exactly. Divide the remaining integer equality by $p^{{\nu }_j}$ and reduce modulo $p$. Later blocks vanish modulo $p$, while each coordinate in $B_j$ contributes its common digit $u$. We obtainwhich is impossible by the choice of $p$ and because $u\not\equiv 0(\mathrm{mod}p)$. Thus each block sum lies in the earlier linear span. For the first block, this span is zero, so $b_1=0$. This proves the columns condition.

For sufficiency, assume the columns condition and choose rational coefficients ${\lambda }_{ij}$, for $i\in B_1\cup \cdots \cup B_{j-1}$, such thatChoose a positive integer $c$ clearing all these coefficients' denominators and a positive integer $p$ with $p\ge |c{\lambda }_{ij}|$ for every coefficient. Use the permitted monochromatic m-p-c set theorem, in the suffix convention for an m-p-c set, to obtain positive generators $z_1,\dots ,z_s$ such that the full setis positive and monochromatic. The positivity of the full m-p-c set is important: we do not discard expressions that happen to be negative. For $i\in B_h$, setAll these coordinates belong to $D$, so they are positive and monochromatic. On collecting coefficients of each $z_j$, their matrix product isThis explicit Rado solution inside an m-p-c set proves sufficiency and completes the proof of Rado's theorem.

To deduce the Finite sums theorem directly from the columns condition, introduce one positive variable $y_F$ for each nonempty $F\subseteq \{1,\dots ,k\}$ and imposeFor $k=1$ there are no equations and any positive integer works. For $k\ge 2$, write $a_F$ for the column belonging to $y_F$. Partition the columns into blocks $B_j=\{F:\min F=j\}$ in the order $j=1,\dots ,k$. Consider the row indexed by a set $H$ with $|H|\ge 2$. The sum of the columns in $B_j$ has row valueFor $j=1$ this is always zero. For $j>1$ it equals one exactly for those $H$ with $\min H<j$ and $j\in H$. Each such $a_H$ is the negative coordinate vector in row $H$, and it belongs to an earlier block. Thereforewhich verifies the columns condition, rather than assuming the desired Finite sums theorem. Apply Rado's theorem to this system. Its monochromatic positive solution satisfies $y_F={\sum }_{i\in F}{x}_i$ with $x_i=y_{\{i\}}$, and hence