We all have seen upto now, natural numbers ( \(\mathbb{N}\) ) being constructed from "Naive Set Theory" and integers ( \(\mathbb{Z}\) ) and rational numbers ( \(\mathbb{Q}\) ) further being constructed from natural numbers with an interesting pattern, i.e.

In a similar way, we will construct real numbers out of rational numbers which of course isn't obvious to see. But, all we need is a pattern among rational numbers which could be uniquely associated to every element in the set of real numbers ( \(\mathbb{R}\) ) (We do know what kinds of numbers to include; the numbers which we are using in the name of real numbers since childhood). In 1872, a German mathematician "Richard Dedekind" discovered one such pattern and popularly his approach is known as "Dedekind's Cuts"; and we will proceed the same way as contained in the books referred ( \(-\) list of reference is mentioned in the last). NOTE: The word "number" would mean rational numbers unless otherwise specified.

Also,

Now here comes a simple but important theorem;

Proof: Let's assume that \(X_1\) is not an upper number for \(\xi\) , which from Definition implies, \(X_1 \in \xi\) and also \(X_1 < X\) . But according to Theorem's conditions \(X_1 > X\) , which contradicts \(X_1 \in \xi\) . Hence, \(X_1\) is an upper number.

Proof: Let's assume that \(X_1\) is not a lower number for \(\xi\) , which from Definition implies, \(X_1 \notin \xi\) and also \(X_1 > X\) . But according to Theorem's conditions \(X_1 < X\) , which contradicts \(X_1 \notin \xi\) . Hence, \(X_1\) is a lower number. Now, if we wish to show that a given set of rational number is a cut, we need to show only the following:

1. The set is not empty and there is a rational number not belonging to it.

2. With every number it contains, the set also contains all numbers smaller than that number.

3. with every number it contains, the set also contains a greater one.

In order to substitute numbers (known till now) with "cuts", we should be able to do all operations defined on numbers, over cuts. Some of them are ordering, addition, multiplication etc.

With this, we can show that cuts too follow the rule of trichotomy under ordering, like numbers. And the same has been shown as a theorem.

Proof: Let's check whether can two of them be simultaneously true. The statements \(\xi = \eta\) & \(\xi < \eta\) , simultaneously are incompatible by Definition and . Similarly, the statements \(\xi = \eta\) & \(\xi < \eta\) simultaneously are incompatible. But if we had \(\xi>\eta\) & \(\xi<\eta\) then for sure we can say that there exists a lower number \(X\) and an upper number \(Y\) for \(\xi\) such that \(X\) is an upper number for \(\eta\) and \(Y\) is a lower number for \(\eta\) . Now, this implies \(X<Y\) and \(X>Y\) simultaneously; which is not possible. Hence we can say that two of them can't be true at the same time. Similarly, all of them can't be true simultaneously as two at a time is not possible. Therefore we can have at most one of the three cases true. If \(\xi \neq \eta\) , then it's obvious that their lower numbers do not coincides, i.e. either some lower number for \(\xi\) is an upper number for \(\eta\) , in which case, it follows that \(\xi > \eta\) ; or some lower number for \(\eta\) is an upper number for \(\xi\) , in which case, it follows that \(\xi < \eta\) . This implies, at least one of them is true. But among these three at most one can be true. Hence, one of them has to be true at a time.

Before defining addition of cuts, there is one theorem which is helpful in our approach towards defining it. From the axiomatic point of view of addition of numbers, it is necessary to obtain a unique number on addition of two numbers. Following theorem fulfils this condition.

Proof: Consider any lower numbers \(X_1\) &\ \(X_2\) for \(\xi\) and any lower numbers \(Y_1\) &\ \(Y_2\) for \(\eta\) . Also consider any upper number \(X_3\) for \(\xi\) and any upper number \(Y_3\) for \(\eta\) ; with following relation.

We can always find such numbers because of the properties of cuts. In order to prove \(\zeta\) a cut, it is sufficient to show that it satisfies the three points of Definition .

1. Since \(\xi\) and \(\eta\) are cuts, means both are non-empty. And by given condition, sum of their elements belongs to the set \(\zeta\) . Also \(\exists\) a number \(Z=X_3+Y_3\) which doesn't belong to \(\zeta\) (according to given condition).

2. Now consider all numbers less than \(X_1+Y_1\) , i.e. \(Z<X_1+Y_1\) or \(\frac{Z}{X_1+Y_1} < 1\) . This statement can be used to draw the following :

   \[X_1\dfrac{Z}{X_1+Y_1} < X_1\cdot1 \qquad \&\ \qquad Y_1\dfrac{Z}{X_1+Y_1} < Y_1\cdot1 \nonumber \]

   By Theorem and by given condition, the number

   \[( X_1\dfrac{Z}{X_1+Y_1} + Y_1\dfrac{Z}{X_1+Y_1} ) \in \zeta \nonumber \]

   which implies that \(Z\in \zeta\) .

3. For every \(X_1\in \xi\) , \(\exists\) a greater lower number \(X_2\) and since it is a lower number for \(\xi\) , \((X_2+Y_1) \in \zeta\) even though \(X_2+Y_1 > X_1+Y_1\) .

From 1, 2 and 3, we can conclude that the set \(\zeta\) is a cut. Now consider lower numbers for \(\zeta\) , from the first part of this theorem it could be represented as the sum of the lower numbers of both \(\xi\) and \(\eta\) i.e. of the form \(X_1+Y_1\) . And any such number can't be represented by the form \(X_3+Y_3\) , i.e. \(X_1+Y_1 \neq X_3+Y_3\) as \(X_1+Y_1 < X_3+Y_3\) (from the given relation). So, this proves the theorem.

Proof: Every X+Y is a Y+X, and vice versa; where X and Y are lower numbers for \(\xi\) and \(\eta\) , respectively.

Proof: Every (X+Y)+Z is a X+(Y+Z), and vice versa; where X and Y are lower numbers for \(\xi\) and \(\eta\) , respectively. There is yet another interesting property of property of cuts, which may be useful result to use.

Proof: Let \(X_1\) be some lower number and \(Y_1\) be some upper number, and consider all rational numbers of the form \(X_1+nA\) (where n is an integer) such that \(X_1+nA > Y_1\) . By Theorem , this number is an upper number. Consider a set S={ \(n\in\mathbb{N}\) | \(X_1+nA\) is an uper number for \(\xi\) } and since it is a set of natural number then it must have a least natural number, say u. This implies \(X_1 + (u-1)A\) is a lower number. Therefore letting \(U=X_1+uA\) and \(X=X_1+(u-1)A\) proves this.

Proof: If \(\xi>\eta\) , then according to the definition there exists a Rational number such that \(Y\in\xi\) but \(Y\notin\eta\) . If \(\xi\) is cut then, \(\exists\) a rational number \(X\in\xi\) such that, \(X>Y\) . By Theorem we can let \(X-Y=U-Z\) ; where U &\ Z are suitable upper and lower number for \(\zeta\) . The equation \(X-Y=U-Z\) is equivalent to \(X+Z=U+Y\) and \(X+Z\) is a lower number for \((\xi+\zeta)\) and \(U+Y\) is an upper number for \((\eta+\zeta)\) (by theorem ). Hence, it is one such number which belongs to \((\xi+\zeta)\) but doesn't belong to \((\eta+\zeta)\) . Therefore \(\xi+\zeta > \eta+\zeta\) .

Before defining multiplication of cuts, there is one theorem which is helpful in our approach towards defining it. From the axiomatic point of view of multiplication of numbers, it is necessary to obtain a unique number on multiplication of two numbers. Following theorem fulfils this condition.

Proof: Consider any lower numbers \(X_1\) &\ \(X_2\) for \(\xi\) and any lower numbers \(Y_1\) &\ \(Y_2\) for \(\eta\) . Also consider any upper number \(X_3\) for \(\xi\) and any upper number \(Y_3\) for \(\eta\) ; with following relation.

We can always find such numbers because of the properties of cuts. In order to prove \(\zeta\) a cut, it is sufficient to show that it satisfies the three points of Definition .

1. Since \(\xi\) and \(\eta\) are cuts, means both are non-empty. And by given condition, product of their elements belongs to the set \(\zeta\) . Also \(\exists\) a number \(Z=X_3\cdot Y_3\) which doesn't belong to \(\zeta\) (according to given condition).

2. Now consider all numbers less than \(X_1\cdot Y_1\) , i.e. \(Z<X_1\cdot Y_1\) or \(\frac{Z}{X_1} < Y_1\) . This implies \(\frac{Z}{X_1}\) is a lower number for \(\eta\) . Therefore its product with \(X_1\) will belong to \(\zeta\) , which implies \(Z\in \zeta\) (as \(X_1\dfrac{Z}{X_1} = Z\) ).

3. For every \(X_1\in \xi\) , \(\exists\) a greater lower number \(X_2\) and since it is a lower number for \(\xi\) , \((X_2\cdot Y_1) \in \zeta\) even though \(X_2\cdot Y_1 > X_1\cdot Y_1\) .

From 1, 2 and 3, we can conclude that the set \(\zeta\) is a cut. Now consider lower numbers for \(\zeta\) , from the first part of this theorem it could be represented as the product of the lower numbers of both \(\xi\) and \(\eta\) i.e. of the form \(X_1\cdot Y_1\) . And any such number can't be represented by the form \(X_3\cdot Y_3\) , i.e. \(X_1\cdot Y_1 \neq X_3\cdot Y_3\) as \(X_1\cdot Y_1 < X_3\cdot Y_3\) (from the given relation). So, this proves the theorem.

Proof: Every XY is a YX, and vice versa; where X and Y are lower numbers for \(\xi\) and \(\eta\) , respectively.

Proof: Every (XY)Z is a X(YZ), and vice versa; where X,Y&\ Z are lower numbers for \(\xi\) , \(\eta\) and \(\zeta\) .

Proof: Let X,Y&\ Z be the lower numbers for \(\xi\) , \(\eta\) and \(\zeta\) . From Definition and , lower number for \(\xi\cdot (\eta+\zeta)\) will be of the form \(X(Y+Z)\) which is equal to \(XY + XZ\) (as Distributive Law holds over Rational number) which is the lower number for \(\xi\cdot \eta + \xi\cdot \zeta\) (by Definition and ). this implies that every lower number for \((\xi\cdot(\eta+\zeta))\) is a lower number for \((\xi\cdot \eta + \xi\cdot \zeta)\) . Hence, \((\xi\cdot (\eta+\zeta)) \subset (\xi\cdot \eta + \xi\cdot \zeta)\) . From Definition and , lower number for \((\xi\cdot \eta + \xi\cdot \zeta)\) will be of the form \(XY+X_1Z\) (where \(X_1\in \xi\) and may be different from \(X\) ). Now if we let \(X=min(X,X_1)\) then, \(XY+XZ\) is also a lower number for \(\xi\) which is equal to \(X(Y+Z)\) , where \(X(Y+Z)\) is a lower number for \(\xi\cdot(\eta+\zeta)\) (by Definition and ). Hence, \((\xi\cdot \eta+\xi\cdot \zeta) \subset (\xi\cdot(\eta+\zeta))\) From above two paragraph we can conclude that \((\xi\cdot (\eta+\zeta)) = (\xi\cdot \eta + \xi\cdot \zeta)\) .

We are almost one step far from defining real Numbers. In the previous section we have seen how to construct a cut, which just ask for a set of Rational number to satisfy three axioms of Definition ; but the question is "how to associate uniquely a cut to a rational and what sort of cut will be suitable?". May be following theorem would help.

Proof: Let the set defined above be \(S\) .In order to prove \(\zeta\) a cut, it is sufficient to show that it satisfies the three points of Definition .

1. For every R, \(\exists\) a \(X<R\) &\ \(X_1>R\) (where \(X,X_1\) \(\in\) \(\mathbb{Q}\) ) which makes \(X\) to belong in \(S\) and \(X_1\) not to belong in \(S\) .

2. consider number less than any number \(X\in S\) ; i.e. \(Z<X\) , but \(X<R\) , which implies \(Z\in S\) for every \(Z<X\) .

3. since there exists infinite rational number between any two Rational numbers, i.e. for every \(X<R\) , \(\exists\) a \(X_2\in \mathbb{Q}\) such that \(X<X_2<R\) and by given condition \(X_2\in S\) .

From 1, 2 and 3 we can conclude that \(S\) is a cut. Also, it's clear that this cut \(S\) uniquely identifies R and hence, it is denoted by R* .

Here are some interesting properties of rational cuts which are among the motivations in choosing cuts for defining all sorts of numbers (means rational) and hence are stated as theorems.

Proof: If \(X<Y\) then for sure \(X\in Y\) * and by Definition \(X\notin X\) *. Hence, by Definition \(X\) * \(<\) \(Y\) *. Similarly, result for other two cases can also be proved.

Proof: Let \(X_1\) and \(Y_1\) be arbitrary lowers number for \(X\) * and \(Y\) *, respectively.

1. From definition , lower number for ( \(X\) * + \(Y\) *) will be of the form \(X_1+Y_1\) , where \(X_1<X\) &\ \(Y_1<Y\) . Because of the property of rational number, \(X_1+Y_1 < X+Y\) . Also by Definition \((X_1+Y_1)\) is a lower number for ( \(X+Y\) )*. This implies \((X\text{*} + Y\text{*}) \subset (X+Y)\text{*}\) .

2. Let \(Z\) be a arbitrary lower number for ( \(X+Y\) )* for which \(Z<(X+Y)\) or in other sense \(\frac{Z}{X+Y}<1\) . This statement can be used to draw the following :

   \[X_1\dfrac{Z}{X_1+Y_1} < X_1\cdot1 \qquad \&\ \qquad Y_1\dfrac{Z}{X_1+Y_1} < Y_1\cdot1 \nonumber \]

   Then obviously \(X_1\dfrac{Z}{X_1+Y_1}\) &\ \(Y_1\dfrac{Z}{X_1+Y_1}\) are lower number for \(X\) * and \(Y\) *, respectively and their sum must belong to \((X\text{*}+Y\text{*})\) . Since \(Z = X\dfrac{Z}{X+Y}+Y\dfrac{Z}{X+Y}\) , \(Z \in (X\text{*}+Y\text{*})\) . Hence, \((X+Y)\text{*} \subset (X\text{*} + Y\text{*})\) .

3. From Definition , lower number for ( \(X\) * \(\cdot\) \(Y\) *) will be of the form \(X_1\cdot Y_1\) for which \(X_1\cdot Y_1 < X\cdot Y\) (as \(X_1<X\) and \(Y_1<Y\) ). By Definition , \((X_1\cdot Y_1)\) is a lower number for ( \(X\cdot Y\) )*. Hence, \((X\text{*}\cdot Y\text{*}) \subset (X\cdot Y)\text{*}\) .

4. Let \(Z\) be a arbitrary lower number for ( \(X\cdot Y\) )* i.e. \(Z<(X\cdot Y)\) . Since ( \(X\cdot Y\) )* is a cut, therefore \(\exists\) a greater lower \(Z_1\in (X\cdot Y)\) * for every \(Z\) . This implies \(Z<Z_1<(X\cdot Y)\) and this relation can be written as

   \[\frac{Z}{Z_1} < 1 \quad \&\ \quad \frac{Z_1}{Y} < X \nonumber \]

   The number \(\frac{Z_1}{Y}\) is a lower number for \(X\) * whereas \(Y\frac{Z}{Z_1}\) is for \(Y\) * (as \(\frac{Z}{Z_1} < 1\) ). This implies \(Z\in (X\) * \(\cdot Y\) *) and \((X\cdot Y)\text{*} \subset (X\text{*}\cdot Y\text{*})\) .

From 1 and 2, \((X+Y)\text{*} = (X\text{*} + Y\text{*})\) , and from 3 and 4, \((X\cdot Y)\text{*} = (X\text{*}\cdot Y\text{*})\) .

Now we get the intuition in our mind that cut is a good candidate to be set as a real number and following definition defines natural numbers to rational numbers, the same way.

Because of such replacement, interesting results arises, some of which needed to be proved; hence, is stated as theorems.

Proof: By Definition , rational numbers are rational cuts, i.e. if the number is X , then X = \(X\) * , which implies for every \(Z<X\) , \(Z\in X\) * but \(X \notin X\) * (where \(Z\) is a rational number(old one)). This implies X is the least upper number.

Two irrational numbers could be added, multiplied and ordered as they are cuts. Now we are ready to define real numbers because we have already defined rational and irrational number.

There is yet another important definition which will enable us to identify cut as a number and element of a cut, i.e. number as a cut; any time.

Proof: Consider a cut \(\xi\) which consists of all the numbers of first class (except the greatest one, if any). This cut can be made as two requirements are already fulfilled according theorem's condition and the third one is fulfilled by excluding the greatest one. This implies that for every such division of all real numbers we can associate a cut. Also, by definition , the numbers \(< \xi\) , \(\in \xi\) ; and since, all lower numbers of \(\xi\) are from first class implies that every real \(H<\xi\) belongs to the first class. Also, the numbers \(> \xi\) , \(\notin \xi\) ; and since, all upper numbers of \(\xi\) are from second class implies that every real \(H>\xi\) belongs to the second class.

There is an interesting point still remaining towards defining real numbers and i.e. "Do natural cuts satisfies Peano's Axioms?" . And the answer is stated as a theorem, which if true could become the reason for identifying natural cuts as natural numbers.

Proof:

1. 0* is a natural number. Explanation : given condition

2. If \(x\) * is a natural number then \(\exists\) a unique successor denoted by ( \(x\) *) \('\) , which is also a natural number. Explanation : \(x\) * is a cut and ( \(x\) *) \('\) = \(x\) *+ \(1\) * = \((x+1)\) *; which is a natural cut, or in other words natural number.

3. If ( \(x\) *) \('\) = ( \(y\) *) \('\) , then ( \(x\) )* = ( \(y\) )* Explanation : if ( \(x\) *) \('\) = ( \(y\) *) \('\) then, \((\) x' \()\text{*} = (\) y' \()\text{*}\) (as ( \(x\) *) \('\) = \(x\) *+ \(1\) * = \((x+1)\) *); which by Theorem equivalent to the expression \(x'=y'\) or more precisely, \(x=y\) . This implies \(x\text{*}=y\text{*}\) .

4. ( \(x\) *) \('\) \(\neq\) 0* \qquad \(\forall\) \(x\) * Explanation : since \(x'\neq 0\) \(\forall\) \(x\) , therefore by Theorem , \((x')\text{*} \neq 0\text{*}\) . this implies \((x\text{*})' \neq 0\text{*} \qquad \forall x\text{*}\) .

5. Let \(\mathbb{N}\) * be the set of all natural cuts and \(M\) * be a subset of it, s.t. 0* \(\in\) \(M\) * &\ ( \(m\) *) \('\) \(\in\) \(M\) * , whenever \(m\) * \(\in\) \(M\) * ; then \(M\) * = \(\mathbb{N}\) * . Explanation : Consider a set M = {\ \(m\) \(\mid\) \(m\) * \(\in\) \(M\) *}\ . We can say, 0 \(\in\) \(M\) (since 0* \(\in\) \(M\) *) and also, \(m'\) \(\in\) \(M\) , whenever \(m\) \(\in\) \(M\) (as ( \(m\) *) \('\) \(\in\) \(M\) * , whenever \(m\) * \(\in\) \(M\) * ). From axioms of Natural number we can conclude that \(M=\mathbb{N}\) , which implies \(M\) * = \(\mathbb{N}\) *.

So with this, we can conclude that this so constructed real number set is an extension of rational number and the pattern mentioned as equation is restored, i.e.

The author wishes to thank the DST-Inspire Programme and the IISc Mathematics department for providing the resources that went into this study. Considerable use has been made of the J.R.D. Tata Library, for which the author is correspondingly grateful. The author especially thanks his project guide, Prof. M.K. Ghosh, for his guidance and help during the project.