Assume a Number Is a Fraction and Watch the Assumption Collapse
Sort numbers into rational and irrational, plot a square root on the number line by construction, prove root 2 irrational by contradiction, and turn any recurring decimal into a fraction.
How can you prove a number is not a fraction when you cannot test every fraction?
There are infinitely many fractions. So if somebody claims that cannot be written as a fraction, how could that ever be checked? You cannot try them all.
The answer is to stop trying and argue instead. Assume that is a fraction, follow the consequences carefully, and show that they contradict each other. If the assumption leads to something impossible, the assumption itself must be false.
That is proof by contradiction, and it is the first genuinely powerful argument in the whole of school algebra. It settles in six lines a question that no amount of calculation could settle.
The rest of this chapter is about the two families of numbers the proof separates.
- A rational number can be written as with and integers and
- An irrational number cannot, and its decimal expansion goes on for ever without ever repeating
Together they make up the real numbers, and every point on the number line is one or the other.
This page covers the first part of the ICSE Class 9 Mathematics chapter on rational and irrational numbers: classifying numbers, representing them on the number line by construction, the contradiction proofs, and converting decimals to form.
The answer is to stop trying and argue instead. Assume that is a fraction, follow the consequences carefully, and show that they contradict each other. If the assumption leads to something impossible, the assumption itself must be false.
That is proof by contradiction, and it is the first genuinely powerful argument in the whole of school algebra. It settles in six lines a question that no amount of calculation could settle.
The rest of this chapter is about the two families of numbers the proof separates.
- A rational number can be written as with and integers and
- An irrational number cannot, and its decimal expansion goes on for ever without ever repeating
Together they make up the real numbers, and every point on the number line is one or the other.
This page covers the first part of the ICSE Class 9 Mathematics chapter on rational and irrational numbers: classifying numbers, representing them on the number line by construction, the contradiction proofs, and converting decimals to form.
Which numbers are rational and which are irrational?
**A number is rational if it can be written as with integers and and ; otherwise it is irrational.
The real number system, from the inside out.
- Natural numbers**
- Whole numbers — the naturals with zero added
- Integers — the wholes with the negatives added
- Rational numbers — every integer, every terminating decimal and every recurring decimal
- Irrational numbers — everything else on the line
- Real numbers — the rationals and the irrationals together
Each set contains the one before it: .
What makes a number rational, in three equivalent ways. It can be written as ; or its decimal expansion terminates; or its decimal expansion recurs. Any one of the three implies the other two.
An irrational number's decimal expansion is non-terminating AND non-recurring. Both conditions are needed — a decimal that goes on for ever is not irrational if it repeats.
Worked classification.
- — rational
- — rational
- — rational
- — rational, because it terminates
- — rational, because it recurs
- — rational
- — rational
- — irrational
- — irrational
- — irrational
- — irrational, because the pattern grows and never repeats a fixed block
Now the two traps in that list.
A square root is not automatically irrational. and are both perfectly rational. The square root of a perfect square is rational, and the next section explains exactly why the proof of irrationality works for and fails for .
**And is not .** The fraction is rational, and is irrational, so they cannot be equal. , which recurs, while , which does not. They agree to two decimal places and part company at the third, and writing in a proof is an error, however useful the approximation is in a calculation.
The real number system, from the inside out.
- Natural numbers**
- Whole numbers — the naturals with zero added
- Integers — the wholes with the negatives added
- Rational numbers — every integer, every terminating decimal and every recurring decimal
- Irrational numbers — everything else on the line
- Real numbers — the rationals and the irrationals together
Each set contains the one before it: .
What makes a number rational, in three equivalent ways. It can be written as ; or its decimal expansion terminates; or its decimal expansion recurs. Any one of the three implies the other two.
An irrational number's decimal expansion is non-terminating AND non-recurring. Both conditions are needed — a decimal that goes on for ever is not irrational if it repeats.
Worked classification.
- — rational
- — rational
- — rational
- — rational, because it terminates
- — rational, because it recurs
- — rational
- — rational
- — irrational
- — irrational
- — irrational
- — irrational, because the pattern grows and never repeats a fixed block
Now the two traps in that list.
A square root is not automatically irrational. and are both perfectly rational. The square root of a perfect square is rational, and the next section explains exactly why the proof of irrationality works for and fails for .
**And is not .** The fraction is rational, and is irrational, so they cannot be equal. , which recurs, while , which does not. They agree to two decimal places and part company at the third, and writing in a proof is an error, however useful the approximation is in a calculation.
How do you mark root 2 and root 3 on the number line?
Build a right-angled triangle whose hypotenuse is the length you want, then swing that length onto the line with a compass.
A rational number first, because it needs no construction. To mark , divide the segment from to into five equal parts and take the third division point. Any rational number can be reached this way, because means "divide a unit into parts and take of them".
**Now by construction.**
- Draw the number line and mark at and at
- At , erect a perpendicular of length unit
- Join . By Pythagoras, , so
- With as centre and radius , draw an arc cutting the number line at
- Then , so is the point
**Then , using the point just found.**
- At (which is at ), erect a perpendicular of length
- Then , so
- Swing that radius onto the line to get the point
Repeating the step gives in turn, and the figure that results is called the square root spiral.
**And directly, which is quicker.** Note that . So erect a perpendicular of length at the point on the line; the hypotenuse from is . Swing it down as before.
A second method that works for any number, using a semicircle. To construct :
- On a straight line mark units and, continuing in the same direction, unit, so
- Draw a semicircle on as diameter
- At erect a perpendicular meeting the semicircle at
- Then
Why that works, since the step is not obvious. The perpendicular from the right angle of a triangle to its hypotenuse satisfies . Here that gives , so .
**Check it on **: take and , so the diameter is and the radius . Then and .
Notice what these constructions actually prove. Every one of these points is a definite place on the line — reached exactly, with a compass and a straight edge, not approximately.
So an irrational number is not a vague or unfinished number. Its decimal expansion never ends, and the number itself sits at one precise point. The endlessness is a fact about writing it down in base ten, not about the number — which is the single most useful thing to understand in this chapter.
A rational number first, because it needs no construction. To mark , divide the segment from to into five equal parts and take the third division point. Any rational number can be reached this way, because means "divide a unit into parts and take of them".
**Now by construction.**
- Draw the number line and mark at and at
- At , erect a perpendicular of length unit
- Join . By Pythagoras, , so
- With as centre and radius , draw an arc cutting the number line at
- Then , so is the point
**Then , using the point just found.**
- At (which is at ), erect a perpendicular of length
- Then , so
- Swing that radius onto the line to get the point
Repeating the step gives in turn, and the figure that results is called the square root spiral.
**And directly, which is quicker.** Note that . So erect a perpendicular of length at the point on the line; the hypotenuse from is . Swing it down as before.
A second method that works for any number, using a semicircle. To construct :
- On a straight line mark units and, continuing in the same direction, unit, so
- Draw a semicircle on as diameter
- At erect a perpendicular meeting the semicircle at
- Then
Why that works, since the step is not obvious. The perpendicular from the right angle of a triangle to its hypotenuse satisfies . Here that gives , so .
**Check it on **: take and , so the diameter is and the radius . Then and .
Notice what these constructions actually prove. Every one of these points is a definite place on the line — reached exactly, with a compass and a straight edge, not approximately.
So an irrational number is not a vague or unfinished number. Its decimal expansion never ends, and the number itself sits at one precise point. The endlessness is a fact about writing it down in base ten, not about the number — which is the single most useful thing to understand in this chapter.
How does the contradiction proof that root 2 is irrational work?
Assume it is rational, in lowest terms, and show that its numerator and denominator must both be even — which contradicts "lowest terms".
**Prove that is irrational.**
Suppose, for the sake of argument, that is rational. Then it can be written as
where and are integers, , and the fraction is in lowest terms — that is, and have no common factor other than . Every fraction can be reduced to lowest terms, so this costs nothing.
Square both sides:
So is even. And if is even then is even, because the square of an odd number is odd. Write for some integer :
So is even, and therefore is even too.
But now **both and are even**, so both are divisible by — which contradicts the assumption that they had no common factor.
The assumption has produced an impossibility, so the assumption is false. **Therefore is irrational.
The same argument for .** Suppose in lowest terms. Then , so divides , so divides . Write : then , so , so divides . Both and are divisible by — a contradiction. **So is irrational.
And for **, the identical argument with in place of .
Now the step that carries all the weight, and it is worth pausing on. The proof needs " divides , therefore divides ", and " divides , therefore divides ".
That is true **because , and are prime, and it is false for a composite number.
Test it on .** Take . Then , and divides — yet does not divide . So the step fails immediately.
**And that is exactly why is rational.** The proof does not merely fail to work for ; it cannot work, because really is a fraction. A proof that would have proved something false has to break somewhere, and the place it breaks tells you what the argument was relying on all along.
So the theorem is really this: if is a prime, is irrational. **A question asking you to prove irrational is asking for the same proof with substituted**, and a question about is asking whether you noticed that is not prime.
**Prove that is irrational.**
Suppose, for the sake of argument, that is rational. Then it can be written as
where and are integers, , and the fraction is in lowest terms — that is, and have no common factor other than . Every fraction can be reduced to lowest terms, so this costs nothing.
Square both sides:
So is even. And if is even then is even, because the square of an odd number is odd. Write for some integer :
So is even, and therefore is even too.
But now **both and are even**, so both are divisible by — which contradicts the assumption that they had no common factor.
The assumption has produced an impossibility, so the assumption is false. **Therefore is irrational.
The same argument for .** Suppose in lowest terms. Then , so divides , so divides . Write : then , so , so divides . Both and are divisible by — a contradiction. **So is irrational.
And for **, the identical argument with in place of .
Now the step that carries all the weight, and it is worth pausing on. The proof needs " divides , therefore divides ", and " divides , therefore divides ".
That is true **because , and are prime, and it is false for a composite number.
Test it on .** Take . Then , and divides — yet does not divide . So the step fails immediately.
**And that is exactly why is rational.** The proof does not merely fail to work for ; it cannot work, because really is a fraction. A proof that would have proved something false has to break somewhere, and the place it breaks tells you what the argument was relying on all along.
So the theorem is really this: if is a prime, is irrational. **A question asking you to prove irrational is asking for the same proof with substituted**, and a question about is asking whether you noticed that is not prime.
Formula
How do you slot numbers between two given numbers and turn a decimal into a fraction?
To insert rationals, take means or use a common denominator; to convert a recurring decimal, multiply by a power of ten and subtract.
**Inserting rational numbers between and .
Method 1 — the mean.** The average of two numbers lies between them:
Check it: and , so .
Method 2 — a common denominator, which gives many at once. Write both with denominator :
Then are nine rational numbers between them. Two of them simplify neatly: and .
**To insert rationals, choose a denominator big enough to leave gaps — which is why method 2 is the one to use when a question asks for a specific number of them.
Inserting irrational numbers between and .** Any square root of a non-square between and will do, since and :
All four are irrational and all lie between and . You can also simply write a non-terminating non-recurring decimal in the range, such as
**Converting a terminating decimal to .** Put the digits over the appropriate power of ten and cancel:
Converting a recurring decimal, which needs the subtraction trick.
Example 1 — a purely recurring decimal. Let There is one recurring digit, so multiply by :
Example 2 — two recurring digits. Let Multiply by :
Example 3 — a mixed recurring decimal, where some digits do not repeat. Let Here one digit does not recur and one does, so use both multipliers:
Example 4. Let Then , so
Why the subtraction works, which is the whole idea. Multiplying by the right power of ten shifts the decimal point so that the repeating tails line up exactly. Subtracting then cancels the infinite tail completely, and what is left is an ordinary equation in .
So the multiplier is chosen by counting the recurring digits: one recurring digit needs , two need , three need . If some digits do not recur, use two multipliers so that both shifts leave the same tail — that is the whole reason example 3 needs and rather than alone.
One last rule worth having. A rational number in lowest terms has a terminating decimal expansion exactly when has **no prime factor other than and **. So terminates because ; recurs because contains a .
**Inserting rational numbers between and .
Method 1 — the mean.** The average of two numbers lies between them:
Check it: and , so .
Method 2 — a common denominator, which gives many at once. Write both with denominator :
Then are nine rational numbers between them. Two of them simplify neatly: and .
**To insert rationals, choose a denominator big enough to leave gaps — which is why method 2 is the one to use when a question asks for a specific number of them.
Inserting irrational numbers between and .** Any square root of a non-square between and will do, since and :
All four are irrational and all lie between and . You can also simply write a non-terminating non-recurring decimal in the range, such as
**Converting a terminating decimal to .** Put the digits over the appropriate power of ten and cancel:
Converting a recurring decimal, which needs the subtraction trick.
Example 1 — a purely recurring decimal. Let There is one recurring digit, so multiply by :
Example 2 — two recurring digits. Let Multiply by :
Example 3 — a mixed recurring decimal, where some digits do not repeat. Let Here one digit does not recur and one does, so use both multipliers:
Example 4. Let Then , so
Why the subtraction works, which is the whole idea. Multiplying by the right power of ten shifts the decimal point so that the repeating tails line up exactly. Subtracting then cancels the infinite tail completely, and what is left is an ordinary equation in .
So the multiplier is chosen by counting the recurring digits: one recurring digit needs , two need , three need . If some digits do not recur, use two multipliers so that both shifts leave the same tail — that is the whole reason example 3 needs and rather than alone.
One last rule worth having. A rational number in lowest terms has a terminating decimal expansion exactly when has **no prime factor other than and **. So terminates because ; recurs because contains a .
Exam tip
Exam tip: state "lowest terms" in the proof and count the recurring digits
**Give the definition with the condition on ** — a rational number is with integers and .
An irrational decimal is non-terminating AND non-recurring. Both words, every time. A recurring decimal is rational however long it goes on.
Do not assume a square root is irrational. and are rational. Check for a perfect square first.
**.** The fraction is rational and recurs; does not.
In the construction, name Pythagoras explicitly and show the arithmetic: , so . Then say you swing the arc with centre to reach the line.
**For , use ** — a perpendicular of at the point — which is quicker than building the spiral.
In the contradiction proof, the phrase "in lowest terms" is where the mark is. Without it there is no contradiction to reach, because and being even proves nothing on its own.
Write the contradiction out in words at the end — " and are both even, so they have the common factor , contradicting our assumption" — and then conclude.
**The step " divides so divides " works because is PRIME.** It fails for , which is exactly why is rational.
To insert a stated number of rationals, use a common denominator, not repeated averaging.
For a recurring decimal, count the recurring digits to choose the multiplier — one digit means , two mean . If some digits do not recur, use two multipliers so the tails match.
And remember the terminating test: in lowest terms terminates exactly when 's only prime factors are ** and **.
An irrational decimal is non-terminating AND non-recurring. Both words, every time. A recurring decimal is rational however long it goes on.
Do not assume a square root is irrational. and are rational. Check for a perfect square first.
**.** The fraction is rational and recurs; does not.
In the construction, name Pythagoras explicitly and show the arithmetic: , so . Then say you swing the arc with centre to reach the line.
**For , use ** — a perpendicular of at the point — which is quicker than building the spiral.
In the contradiction proof, the phrase "in lowest terms" is where the mark is. Without it there is no contradiction to reach, because and being even proves nothing on its own.
Write the contradiction out in words at the end — " and are both even, so they have the common factor , contradicting our assumption" — and then conclude.
**The step " divides so divides " works because is PRIME.** It fails for , which is exactly why is rational.
To insert a stated number of rationals, use a common denominator, not repeated averaging.
For a recurring decimal, count the recurring digits to choose the multiplier — one digit means , two mean . If some digits do not recur, use two multipliers so the tails match.
And remember the terminating test: in lowest terms terminates exactly when 's only prime factors are ** and **.
Did you know
Why a number can be exact and its decimal endless at the same time
There is a persistent feeling that an irrational number is somehow unfinished — that is not quite a proper number because we can never write it all down.
The construction in this chapter settles that completely. Draw a square of side and its diagonal is , exactly, with nothing left over. You can measure it with a compass, swing it onto the number line and mark the point. The length is as definite as or .
So what is endless is not the number. It is the attempt to write it in base ten.
And base ten is an arbitrary choice. Look at what it does to perfectly ordinary fractions. One third is a simple, exact quantity — divide a rope into three and take a piece — and in base ten it is , endless. One seventh is worse: , six digits repeating for ever.
Nobody thinks one third is an unfinished number. Yet base ten cannot write it down either.
The rule from the last section explains why. A fraction terminates in base ten exactly when its denominator has no prime factor other than and — and and are the prime factors of . Change the base and the list changes with it. In base three, one third would be written as a single digit and finish immediately, while one half would become endless.
So "terminating" and "recurring" are not properties of a number at all. They are properties of the number together with the base you chose to write it in — and the only thing special about and is that we happen to have ten fingers.
What is genuinely a property of the number is whether it can be written as a fraction at all. That does not depend on the base, which is why the contradiction proof never mentions decimals anywhere in it. It works entirely with integers and — and that is what makes it a proof about rather than about our notation.
The construction in this chapter settles that completely. Draw a square of side and its diagonal is , exactly, with nothing left over. You can measure it with a compass, swing it onto the number line and mark the point. The length is as definite as or .
So what is endless is not the number. It is the attempt to write it in base ten.
And base ten is an arbitrary choice. Look at what it does to perfectly ordinary fractions. One third is a simple, exact quantity — divide a rope into three and take a piece — and in base ten it is , endless. One seventh is worse: , six digits repeating for ever.
Nobody thinks one third is an unfinished number. Yet base ten cannot write it down either.
The rule from the last section explains why. A fraction terminates in base ten exactly when its denominator has no prime factor other than and — and and are the prime factors of . Change the base and the list changes with it. In base three, one third would be written as a single digit and finish immediately, while one half would become endless.
So "terminating" and "recurring" are not properties of a number at all. They are properties of the number together with the base you chose to write it in — and the only thing special about and is that we happen to have ten fingers.
What is genuinely a property of the number is whether it can be written as a fraction at all. That does not depend on the base, which is why the contradiction proof never mentions decimals anywhere in it. It works entirely with integers and — and that is what makes it a proof about rather than about our notation.
Exam relevance
Why does JEE Main keep coming back to irrational numbers?
Because proof by contradiction is the method Class 11 uses for its hardest results, and the rational-irrational distinction decides what a whole class of answers may look like.
This is the foundation for Class 11 Mathematics Sets, Complex Numbers and Mathematical Reasoning, examined in JEE Main. Mathematical Reasoning names proof by contradiction as a method alongside direct proof and proof by contrapositive, and asks you to identify or apply it. **The proof is the standard worked example of the method, and questions on the validity of a statement and its negation use exactly this shape of argument.
The real number system becomes set notation.** Class 11 Sets treats , , , and as sets with the chain of inclusions used here, defines intervals on the real line, and asks about , the set of irrationals. Questions on subsets, complements and intervals of the real line are standard in JEE Main, and the nesting on this page is what they assume.
The prime-divisibility step reappears in number theory. The step " prime and divides implies divides " is the fundamental theorem of arithmetic at work, and it is used in Class 11 and 12 for divisibility problems and in principle of mathematical induction questions.
Irrationality controls the form of an answer. Once trigonometry and coordinate geometry begin, answers routinely come out as , or , and they must be left in surd form rather than replaced by a decimal. **A JEE Main answer given as where was required is simply wrong, and the reason is the distinction drawn on this page.
The construction reappears in coordinate geometry.** Plotting by Pythagoras is the distance formula in its simplest form, and the semicircle method is the geometric mean — which returns in Class 11 Sequences and Series as the GM of two numbers, and in the AM-GM inequality. So the construction on this page is an inequality result in disguise, and the fact that the perpendicular is the geometric mean of and is examined in that chapter.
Decimal-to-fraction conversion becomes an infinite series. Class 11 Sequences and Series shows that a recurring decimal is an infinite geometric series and sums it — so . The subtraction trick used here is the elementary version of summing that series, and the series method is what JEE Main examines.
What the questions look like. For board work, expect classify given numbers as rational or irrational, **represent , or on the number line by construction, prove that or is irrational, insert a stated number of rationals between two numbers, and express a recurring decimal in form. The proof must contain the phrase "in lowest terms". For JEE Main, expect set and interval questions, identification of proof methods, geometric-series conversion and surd-form answers.
How board and competitive emphasis differ. A board paper rewards the full written proof and the labelled construction. A competitive paper assumes both and asks which proof technique applies, or sums a recurring decimal as a series.
The single trap that costs the most marks.** Omitting "in lowest terms" from the proof. Without that assumption, discovering that and are both even is not a contradiction at all — plenty of fractions have an even numerator and an even denominator, and is one of them. The contradiction exists only because we assumed at the start that no common factor remained. The defence is to write that assumption as a numbered step and then point back to it by number when the contradiction arrives, because an examiner is looking for exactly that reference and a proof without it earns very little however correct the algebra.
This is the foundation for Class 11 Mathematics Sets, Complex Numbers and Mathematical Reasoning, examined in JEE Main. Mathematical Reasoning names proof by contradiction as a method alongside direct proof and proof by contrapositive, and asks you to identify or apply it. **The proof is the standard worked example of the method, and questions on the validity of a statement and its negation use exactly this shape of argument.
The real number system becomes set notation.** Class 11 Sets treats , , , and as sets with the chain of inclusions used here, defines intervals on the real line, and asks about , the set of irrationals. Questions on subsets, complements and intervals of the real line are standard in JEE Main, and the nesting on this page is what they assume.
The prime-divisibility step reappears in number theory. The step " prime and divides implies divides " is the fundamental theorem of arithmetic at work, and it is used in Class 11 and 12 for divisibility problems and in principle of mathematical induction questions.
Irrationality controls the form of an answer. Once trigonometry and coordinate geometry begin, answers routinely come out as , or , and they must be left in surd form rather than replaced by a decimal. **A JEE Main answer given as where was required is simply wrong, and the reason is the distinction drawn on this page.
The construction reappears in coordinate geometry.** Plotting by Pythagoras is the distance formula in its simplest form, and the semicircle method is the geometric mean — which returns in Class 11 Sequences and Series as the GM of two numbers, and in the AM-GM inequality. So the construction on this page is an inequality result in disguise, and the fact that the perpendicular is the geometric mean of and is examined in that chapter.
Decimal-to-fraction conversion becomes an infinite series. Class 11 Sequences and Series shows that a recurring decimal is an infinite geometric series and sums it — so . The subtraction trick used here is the elementary version of summing that series, and the series method is what JEE Main examines.
What the questions look like. For board work, expect classify given numbers as rational or irrational, **represent , or on the number line by construction, prove that or is irrational, insert a stated number of rationals between two numbers, and express a recurring decimal in form. The proof must contain the phrase "in lowest terms". For JEE Main, expect set and interval questions, identification of proof methods, geometric-series conversion and surd-form answers.
How board and competitive emphasis differ. A board paper rewards the full written proof and the labelled construction. A competitive paper assumes both and asks which proof technique applies, or sums a recurring decimal as a series.
The single trap that costs the most marks.** Omitting "in lowest terms" from the proof. Without that assumption, discovering that and are both even is not a contradiction at all — plenty of fractions have an even numerator and an even denominator, and is one of them. The contradiction exists only because we assumed at the start that no common factor remained. The defence is to write that assumption as a numbered step and then point back to it by number when the contradiction arrives, because an examiner is looking for exactly that reference and a proof without it earns very little however correct the algebra.
Key takeaways
Rational and irrational numbers: quick revision
- A rational number is with integers and . An irrational number cannot be written so.
- The sets nest: , and is the rationals together with the irrationals.
- Three equivalent tests for rational: expressible as ; a terminating decimal; a recurring decimal.
- An irrational decimal is non-terminating AND non-recurring.
- Rational: , , , , , , . Irrational: , , , ,
- A square root of a perfect square is rational — do not assume every root is irrational.
- **** — the fraction recurs and does not.
- A rational number on the line: to mark , divide to into 5 parts and take the 3rd.
- ** by construction**: perpendicular of at the point ; ; swing the arc with centre .
- ****: perpendicular of at the point ; . Repeating gives the square root spiral.
- ** directly**: , so a perpendicular of at the point .
- **Semicircle method for **: mark and , draw a semicircle on , erect a perpendicular at meeting it at ; then , so .
- **Proof that is irrational**: assume in lowest terms; then , so is even; put to get , so is even; both even contradicts lowest terms. Hence irrational.
- ** and ** follow identically with and .
- **The proof relies on , and being PRIME.** It fails for : divides but not — and that is exactly why is rational.
- Inserting rationals: the mean ; or a common denominator — and give nine numbers between them.
- **Inserting irrationals between and **: , since .
- **Terminating to **: ; .
- **Recurring to **: gives , . gives , . needs both and , giving , . gives .
- Count the recurring digits to pick the multiplier, and use two multipliers when some digits do not recur.
- ** in lowest terms terminates exactly when has no prime factor other than and .**
Write out the proof from memory and check that the words "in lowest terms" appear in it — then try the same proof on and find the line where it breaks.
- The sets nest: , and is the rationals together with the irrationals.
- Three equivalent tests for rational: expressible as ; a terminating decimal; a recurring decimal.
- An irrational decimal is non-terminating AND non-recurring.
- Rational: , , , , , , . Irrational: , , , ,
- A square root of a perfect square is rational — do not assume every root is irrational.
- **** — the fraction recurs and does not.
- A rational number on the line: to mark , divide to into 5 parts and take the 3rd.
- ** by construction**: perpendicular of at the point ; ; swing the arc with centre .
- ****: perpendicular of at the point ; . Repeating gives the square root spiral.
- ** directly**: , so a perpendicular of at the point .
- **Semicircle method for **: mark and , draw a semicircle on , erect a perpendicular at meeting it at ; then , so .
- **Proof that is irrational**: assume in lowest terms; then , so is even; put to get , so is even; both even contradicts lowest terms. Hence irrational.
- ** and ** follow identically with and .
- **The proof relies on , and being PRIME.** It fails for : divides but not — and that is exactly why is rational.
- Inserting rationals: the mean ; or a common denominator — and give nine numbers between them.
- **Inserting irrationals between and **: , since .
- **Terminating to **: ; .
- **Recurring to **: gives , . gives , . needs both and , giving , . gives .
- Count the recurring digits to pick the multiplier, and use two multipliers when some digits do not recur.
- ** in lowest terms terminates exactly when has no prime factor other than and .**
Write out the proof from memory and check that the words "in lowest terms" appear in it — then try the same proof on and find the line where it breaks.