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| The following example (from ring theory) illustrates the one aspect of proofs in mathematics: proving the existence of certain mathematical \PMlinkescapetext{objects or properties}. |
The following example (from ring theory) illustrates the one aspect of proofs in mathematics: proving the existence of certain mathematical \PMlinkescapetext{objects or properties}. |
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| \textbf{Statement:} Let $R$ be a ring such that $1-ab$ is right invertible, with $a,b\in R$. Then $1-ba$ is right invertible. |
\textbf{Statement:} Let $R$ be a ring such that $1-ab$ is right invertible, with $a,b\in R$. Then $1-ba$ is right invertible. |
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| This statement will be proven here using two methods. The first method is called an \emph{existential proof} (also known as an \emph{existence proof}), in which one only seeks to prove that the mathematical \PMlinkescapetext{object or property} in question exists, \emph{not} to show how to obtain it. The second method is called a \emph{constructive proof}, in which one actually shows how to obtain the mathmetical \PMlinkescapetext{object or property} in question. |
This statement will be proven here using two methods. The first method is called an \emph{existential proof} (also known as an \emph{existence proof}), in which one only seeks to prove that the mathematical \PMlinkescapetext{object or property} in question exists, \emph{not} to show how to obtain it. The second method is called a \emph{constructive proof}, in which one actually shows how to obtain the mathmetical \PMlinkescapetext{object or property} in question. |
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| \textbf{Existential proof:} |
\textbf{Existential proof:} |
| Since $1-ab\in R$ is right invertible, $(1-ab)R=R$. Now, $$(1-ba)R\supseteq(1-ba)bR=b(1-ab)R=bR.$$ So $$(ba)R=b(aR)\subseteq bR\subseteq\ (1-ba)R,$$ and consequently, $$R=(1-ba)R+(ba)R\subseteq (1-ba)R,$$ showing that $1\in(1-ba)R$. \qed |
Since $1-ab\in R$ is right invertible, $(1-ab)R=R$. Now, $$(1-ba)R\supseteq(1-ba)bR=b(1-ab)R=bR.$$ So $$(ba)R=b(aR)\subseteq bR\subseteq\ (1-ba)R,$$ and consequently, $$R=(1-ba)R+(ba)R\subseteq (1-ba)R,$$ showing that $1\in(1-ba)R$. \qed |
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| Notice, we merely demonstrated the existence of a right inverse of $1-ba$ without actually finding such an inverse. The next proof in fact finds a right inverse of $1-ba$. |
Notice, we merely demonstrated the existence of a right inverse of $1-ba$ without actually finding such an inverse. The next proof in fact finds a right inverse of $1-ba$. |
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| \textbf{Constructive proof:} |
\textbf{Constructive proof:} |
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Since $1-ab\in R$ is right invertible, let $c\in R$ be a right inverse so that $1=(1-ab)c$. We seek to construct a right inverse of $1-ba$ in terms of $a,b,$ and $c$. Rewriting the equation, we have $abc=c-1$. Then, $$(1-ba)bc=bc-babc=bc-b(c-1)=b.$$ We have just expressed $b$ in terms of $1-ba$. Next, multiply $a$ on the right to each term on both sides of the equation, to get $$ba=(1-ba)bca.$$ Then, negate both terms and add 1, to get
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Since $1-ab\in R$ is right invertible, let $c\in R$ be a right inverse so that $1=(1-ab)c$. We seek to construct a right inverse of $1-ba$ in \PMlinkescapetext{terms} of $a,b,$ and $c$. Rewriting the equation, we have $abc=c-1$. Then, $$(1-ba)bc=bc-babc=bc-b(c-1)=b.$$ We have just expressed $b$ in terms of $1-ba$. Next, multiply $a$ on the right to each term on both sides of the equation, to get $$ba=(1-ba)bca.$$ Then, negate both terms and add 1, to get
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| $$1-ba=1-(1-ba)bca.$$ Finally, rearranging the terms and we have $$1=(1-ba)+(1-ba)bca=(1-ba)(1+bca),$$ showing that |
$$1-ba=1-(1-ba)bca.$$ Finally, rearranging the terms and we have $$1=(1-ba)+(1-ba)bca=(1-ba)(1+bca),$$ showing that |
| a right inverse of $1-ba$ exists by explicitly constructing one. \qed |
a right inverse of $1-ba$ exists by explicitly constructing one. \qed |
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| Many other techniques are used in proving mathematical statements. Proof by mathematical induction, proof by contradiction, proof by exhaustion, and proof by similarity are just some of the major techniques. |
Many other techniques are used in proving mathematical statements. Proof by mathematical induction, proof by contradiction, proof by exhaustion, and proof by similarity are just some of the major techniques. |
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| As this entry is still in its very rough form, PM users are welcome and encouraged to refine and provide additional techniques with interesting and illustrative examples! |
As this entry is still in its very rough form, PM users are welcome and encouraged to refine and provide additional techniques with interesting and illustrative examples! |
