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I am working my way through "Mathematical Analysis" by Apostol.


What I am attempting to prove is that if there exist $q_{1}$ and $q_{2}$ such that $x + q_1 = x$ and $y+q_2=y$, then $q_1=q_2$


Sometimes I will be using $q$ to denote $0$

I am using the first $4$ field axioms from the book:

Axiom 1: Commutative Laws

$x+y=y+x$, $xy=yx$

Axiom 2: Associative Laws

$x+(y+z)=(x+y)+z$, $x(yz)=(xy)z$

Axiom 3: Distributive Law

$x(y+z)=xy+yz$

Axiom 4:

Given any two real numbers $x$ and $y$, there exists a real number $z$ such that $x+z=y$. This $z$ is denoted by $y-x$; the number $x-x$ is denoted by $0$ (it can be proved that $0$ is independent of $x$.) We write $-x$ for $0-x$ and call $-x$ the negative of $x$.


Lemma 1: $x + 0 = x$

From axiom 4 we are guaranteed a $z$ such that $x+z=x$. This $z$ is denoted by $x-x$, which is denoted by $0$

Therefore $x+z=x \Longrightarrow$ $x+0=x$

Lemma 2: $x+(-x)=0$

We can rewrite the above as $x + (0-x)$, which, from definition in axiom 4, evaluates to $0$

Lemma 3: If $x+q_a=x$, and $x+q_b=x$, then $q_a=q_b$

From the first sentence of axiom 4 we are guaranteed at least one $q_a$ such that

$x + q_a = x$

If $q_a$ is not unique, then we will also have

$x+q_b=x$

$x+q_a=x+q_b$

Add $(-x)$ to both sides

$(x+q_a)+(-x)=(x+q_b)+(-x)$

Commutative Propriety

$(q_a+x)+(-x)=(q_b+x)+(-x)$

Associative Proprety

$q_a+(x+(-x))=q_b+(x+(-x))$

Lemma 2

$q_a + 0=q_b+0$

Lemma 1

$q_a=q_b$


Now for the actual proof:

We are trying to prove that if there exist $q_{1}$ and $q_{2}$ such that $x + q_1 = x$ and $y+q_2=y$, then $q_1=q_2$

By the first sentence in axiom 4, we are guaranteed $q_1$ and $q_2$ such

$x + q_1 = x$

$x+q_2=x$

By lemma 3, we are guaranteed that if

By axiom 4 guaranteed the existence of a $z$ such that

$x+z=y$

Now we substitute

$(x+z)+q_2=(x+z)$

Add $(-z)$ to both sides

$((x+z)+q_2)+(-z)=(x+z)+(-z)$

Associative and commutative proprieties lead to

$x+q_2 + (z+(-z))=x+(z+(-z))$

$x+q^2=x$

By lemma 3 we are guaranteed that if $x+q_1=x$, and $x+q_2=x$, then $q_1=q_2$

Therefore $q_1=q_2$ so we have completed the proof.

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    $\begingroup$ Let $e_1,e_2$ denote zeros. Then $e_1=e_1+ e_2 = e_2$. $\endgroup$ – Rubertos Dec 21 '15 at 21:08
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This is correct, but as noted in the comments, this is massive overkill when it comes to complexity. $q_1=q_1+q_2=q_2$ where the first equality holds because $q_2$ is the additive identity and the second holds because $q_1$ is the additive identity. Thus by the transitive property of equality, we are done.

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  • $\begingroup$ That is definitely much cleaner, though I believe we still still need lemma 1 for this $\endgroup$ – Ovi Dec 21 '15 at 22:03
  • $\begingroup$ Are there any sources of problems like this to prove/do you have any suggestions? $\endgroup$ – Ovi Dec 21 '15 at 23:34
  • $\begingroup$ You don't need Lemma 1 because this holds by definition. You are supposing that $q_1$ and $q_2$ are both additive identities, and therefore it immediately follows that these equations hold, because what it means for $q_i$ to be an additive identity is that $\forall a,a+q_i=a$ $\endgroup$ – Stella Biderman Dec 23 '15 at 5:29
  • $\begingroup$ But we haven't defined "additive identity" anywhere in the axioms, don't we need to define it and prove it's existence with lemma 1? $\endgroup$ – Ovi Dec 23 '15 at 6:02
  • $\begingroup$ You're right that you need to prove existence, but Lemma 1 doesn't properly tell you that $0$ is an additive identity because the quantifiers are wrong. Since you proved Lemma 1 with Axiom 4, Lemma 1 says $\forall x\exists y$ such that $x+y=x$ whereas you need $\exists y \forall x$ such that $x+y=x$ $\endgroup$ – Stella Biderman Dec 23 '15 at 7:22
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You can one-line this (see comment by Rubertos). That said, the only thing you write that I might call "incorrect" is when you use q1, q2, and 0. If you are assuming that q1 and q2 are the identities under addition and your goal is to show that they are equal, you don't want to throw a third into the mix. Since both q1 and q2 have the same properties as 0, use q1 or q2 in place of 0 all the time and the same arguments will work.

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  • $\begingroup$ Are there any sources of problems like this to prove/do you have any suggestions? $\endgroup$ – Ovi Dec 21 '15 at 23:34
  • $\begingroup$ This type of reasoning is typical of abstract algebra and this result in particular is explicitly addressed in the group theory sections of a number of books. I started with Abstract Algebra by I.N. Herstein. I thought the book was very well written. It has plenty of exercises and explicitly tells the reader which ones are easier or harder than others. Unfortunately, it is also criminally priced by the publisher. Introductory group theory should be in any elementary treatment of abstract algebra and I'm sure there are many people in these fora who know much more than I about where to look. $\endgroup$ – Brendan Hodis Dec 22 '15 at 3:13

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