Surreal Numbers and Games/Simple Arithmetic

Addition

We have created 15 surreal numbers and given them familiar names. Their relative order is right, but how can we be sure that their values are appropriate? We could have named them ridiculous things like

-50, -49, -10, 2, 3, 3.14159, 4, 10, 11, 12, 50, 999.999, 1000, 1000.000001, 2^{10,000,000,000}

and the ordering would still be perfectly fine. One way we might justify the names we have given them is to show that they behave the right way under arithmetic operations- for example, show that $1+1=2$ and the like. We therefore require the method for doing addition with surreal numbers.

Template:Robox If $x,y$ are surreal numbers then

$x+y:=\{X_L+y,x+Y_L|X_R+y,x+Y_R\}$ Template:Robox/Close

The notation $X_L+y$ means the set of surreal numbers formed by taking all members of $X_L$ and adding (via Axiom 3) $y$ to them. This is another recursive definition but, since each step involves older sets, it is guaranteed to terminate. Let us calculate $0+0$:

$\begin{array}{ccc}0+0& =& \{\varnothing +0,0+\varnothing |\varnothing +0,0+\varnothing \}\\ & =& \{\varnothing ,\varnothing |\varnothing ,\varnothing \}\\ & =& \{\varnothing |\varnothing \}\\ & =& 0\end{array}$

Template:Robox Show that $x+0=0+x=x$ for all $x$. Template:Robox/Close This shows that $0$ is the additive identity for surreal numbers, just as it should be. Our designation for it seems justified. But before we start adding numbers together willy-nilly, we had better verify that our definition of addition gives it all the properties we want from it. We need it to produce well-formed numbers when two well-formed numbers are added, and we would like it to be both commutative $(x+y=y+x)$ and associative $(x+[y+z]=[x+y]+z)$.

Template:Robox If $x,y$ are any two surreal numbers, then $x+y=y+x$. Template:Robox/Close

$x+y=\{X_L+y,x+Y_L|X_R+y,x+Y_R\}$

$y+x=\{Y_L+x,y+X_L|Y_R+x,y+X_R\}$

It is plain to see that these are the same if:

$X_L+y=y+X_L$

$X_R+y=y+X_R$

$Y_L+x=x+Y_L$

$Y_R+x=x+Y_R$.

All these just amount to the commutative law on older numbers. You proved in Exercise 1 that commutativity holds when one of the numbers is $0$ so, by induction, commutativity holds in general.

Template:Robox The commutative law is actually stronger than mere equality; show that $x+y\equiv x+y$, provided that the $x$ on the left hand side is identical to the one on the right and similar for $y$. You may need to adapt Exercise 1 to finish the proof. Template:Robox/Close

• (2A) If $a,a^{\prime },b,b^{\prime }$ are any four surreal numbers such that $a\le a^{\prime }$ and $b\le b^{\prime }$, then $a+b\le a^{\prime }+b^{\prime }$.
• (2B) If $v,v^{\prime },w,w^{\prime }$ are any four surreal numbers such that $v+w\le v^{\prime }+w^{\prime }$ and $w^{\prime }\le w$ then $v\le v^{\prime }$.

It turns out that both parts of this theorem need to be proved in tandem. Let us introduce the shorthand notation $P_{2A}(a,a^{\prime },b,b^{\prime })$ to mean that (2A) is valid for these four numbers and $P_{2B}(v,v^{\prime },w,w^{\prime })$ to mean that (2B) is valid for $v,v^{\prime },w,w^{\prime }$. Now, $P_{2A}(a,a^{\prime },b,b^{\prime })$ is true only if:

• (2Aa) $A{\text{'}}_R+b^{\prime }\le ̸a+b$ and
• (2Ab) $a^{\prime }+B{\text{'}}_R\le ̸a+b$ and
• (2Ac) $a^{\prime }+b^{\prime }\le ̸A_L+b$ and
• (2Ad) $a^{\prime }+b^{\prime }\le ̸a+B_L$.

By contradiction, suppose (2Aa) is false. In that case we would have $A{\text{'}}_R+b^{\prime }\le a+b$, and we also know that $b\le b^{\prime }$. Now observe that, if we are allowed to assume $P_{2B}(a{\text{'}}_R,a,b^{\prime },b)$ is valid, then $A{\text{'}}_R<a$. This we know to be absurd, and so (2Aa) would have to be true.

Now let us find when $P_{2B}(v,v^{\prime },w,w^{\prime })$ is valid. Again, the proof is by contradiction. Let us assume that $w^{\prime }\le w$ and $v+w\le v^{\prime }+w^{\prime }$ but $v\le ̸v^{\prime }$. Then some $v{\text{'}}_R\le v$ or $v^{\prime }\le$ some $v_L$. But from the addition we get

• $V{\text{'}}_R+w^{\prime }\le ̸v+w$ and
• $v^{\prime }+w^{\prime }\le ̸V_L+w$.

Now, if we could assume $P_{2A}(v{\text{'}}_R,v,w^{\prime },w)$ and $P_{2A}(v^{\prime },v_L,w^{\prime },w)$, these would give us the contradiction necessary to prove $P_{2B}(v,v^{\prime },w,w^{\prime })$. Write them out in full if this is not clear. In other words, we the validity of $P_{2B}$ depends on the validity of $P_{2A}$ applied to a simpler set of numbers. The structure of the inductive proof is becoming clearer. Returning to (2Aa), it is true if $P_{2B}(a{\text{'}}_R,a,b^{\prime },b)$ is valid; and this is valid if in turn

• $P_{2A}(a_R,a{\text{'}}_R,b,b^{\prime })$ and
• $P_{2A}(a,a{\text{'}}_{RL},b,b^{\prime })$

are valid. Inductively, (2Aa) is true. (2Ab,c,d) can be verified in the same manner. Observe that the proof of this theorem does not depend on the sum of two numbers actually being well-formed; that's just as well, because we have not yet proved that adding two numbers produces a well-formed number. But it is true, and we will prove it now:

Template:Robox If $x$ and $y$ are well-formed, then $x+y$ is well-formed. Template:Robox/Close

We require all of:

• $x_R+y\le ̸x+y$
• $x+y_R\le ̸x+y$
• $x+y\le ̸x_L+y$
• $x+y\le ̸x+y_L$.

By induction we may assume that $x_R+y$ etc. are all well-formed, which means we can replace the "$\le ̸$" inequalities with "$>$" inequalities to obtain

• $x+y<x_R+y$
• $x+y<x+y_R$
• $x_L+y<x+y$
• $x+y_L<x+y$.

Since $x,y$ are well-formed, $x<x_R$ and so on. Therefore Theorem 2A applies to everything, and the proof is complete.

Template:Robox If $x,y,z$ are any three numbers, then $x+(y+z)=(x+y)+z$. Template:Robox/Close

The proof is simple and requires little explanation. It is left as an exercise. Template:Robox Prove Theorem 4. Template:Robox/Close

Subtraction

• If $x=\{X_L|X_R\}$ then the negative of $x$ is denoted $-x=\{-X_R|-X_L\}$.
• $x-y=x+(-y)$.

That is, subtraction is the addition of the negative, as we would expect.

Template:Robox Look at all the numbers present on Day 2. Verify that, e.g., $-\frac{1}{2}$ is indeed the negative of $\frac{1}{2}$. What is the negative of zero? Template:Robox/Close

We need to make sure that the negative of a well-formed number is well-formed. To do this we first prove that, if $a<b$ then $-b<-a$ for any (not necessarily well-formed) $a,b$. Template:Robox Show that, if $a<b$ then $-b<-a$. Template:Robox/Close Template:Robox If $x$ is well-formed, then so is $-x$. Template:Robox/Close Since $x$ is well-formed, we know that $x_L<x<x_R$. From Exercise X, it follows that $-x<-(x_L)$ and $-(x_R)<x$. Therefore by the transitive law $-(x_R)<-x<-(x_L)$. But by the definition of the negative this is identical to $(-x{)}_L<-x<(-x{)}_R$, and so $-x$ is well-formed.

Interlude

Now that we have addition and subtraction, we can better answer the question of whether the names we have given surreal numbers are appropriate. Recall that, after day 2, we have created seven surreal numbers that we have named

$-2,-1,-(1/2),0,(1/2),1,2$.

We have also shown that the numbers created on any given day are created between adjacent numbers already known and at the upper and lower ends. Thus we might expect that after Day 3 we will have the following set of numbers:

$-3,-2,-(3/2),-1,-(3/4),-(1/2),-(1/4),0,1/4,1/2,3/4,1,3/2,2,3$. That turns out to be true. To show this we will need to demonstrate the following:

• The largest number created on a certain day is one more than the largest number from the previous day.
• If $x$ was created on a certain day, so was $-x$. This will show that numbers are distributed symmetrically about $0$.
• If $a,b$ are adjacent numbers on a given day, with no older or equally old numbers between them, then $\{a|b\}$ is half way between them. This will justify saying, for example, that $\{0|1\}$ is $1/2$ instead of 1/3 or 0.999.

Template:Robox If $x$ is the largest number known on some day n then $x+1=\{x|\varnothing \}$. Template:Robox/Close $\begin{array}{ccc}x+1& =& \{X_L+1,x+1_L|X_R+1,x+1_R\}\\ & =& \{X_L+1,x|X_R+1,\varnothing \}\end{array}$

This will be equal to $\{x|\varnothing \}$ if $X_R$ is empty and all $x_L+1=x$. But this is just Theorem X applied to simpler numbers, so it is inductively true.

Template:Robox If $x$ is created on a certain day, then $-x$ is created on the same day. Template:Robox/Close The proof is easy, and is left as an exercise. Template:Robox Prove Theorem 7. Template:Robox/Close

Template:Robox If $a<b$ are numbers such that there is no $w$ older than either of them and between them, then $\{a|b\}+\{a|b\}=a+b$. Template:Robox/Close

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