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## Real Numbers

Real numbers are numbers that can be found on the number line.

This includes both the rational and irrational numbers.

This includes both the rational and irrational numbers.

Concepts

9 min read

- Let's dig deep into the concepts of the chapter

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Real numbers are numbers that can be found on the number line.

This includes both the rational and irrational numbers.

This includes both the rational and irrational numbers.

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An **irrational number** is number that cannot be expressed as a ratio of integers, i.e. as a fraction.

Therefore, irrational numbers, when written as decimal numbers, do not terminate, nor do they repeat.

For example:- The number $2 =1.41421356.....$ does not end.

The same can be said for any irrational number.

Therefore, irrational numbers, when written as decimal numbers, do not terminate, nor do they repeat.

For example:- The number $2 =1.41421356.....$ does not end.

The same can be said for any irrational number.

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Let's see to how express any irrational number on number line

For example:- Take $2 $ as irrational number

1. Draw a number line, mark the origin and other integers.

2. Now, find distance between $0$ and $1$ using compass.

3. Draw a perpendicular to $1$ of the same length as between $0$ and $1$.

4. Join origin and other end of the new line i.e. perpendicular. You will get a figure as shown.

5. Name origin as $A$, $1$ on the number line as $B$ and end of perpendicular as $C$.

6. Distance between $A$ and $C$ is $2 $.

7. Use compass to find distance between $A$ and $C$.

8. Place compass at $A$ and mark the same length as between $A$ and $C$ on the number line.

9. And you get the point $2 $ at $D$.

Similarly, you can do it for other irrational numbers like $3 ,5 ,..$

For example:- Take $2 $ as irrational number

1. Draw a number line, mark the origin and other integers.

2. Now, find distance between $0$ and $1$ using compass.

3. Draw a perpendicular to $1$ of the same length as between $0$ and $1$.

4. Join origin and other end of the new line i.e. perpendicular. You will get a figure as shown.

5. Name origin as $A$, $1$ on the number line as $B$ and end of perpendicular as $C$.

6. Distance between $A$ and $C$ is $2 $.

7. Use compass to find distance between $A$ and $C$.

8. Place compass at $A$ and mark the same length as between $A$ and $C$ on the number line.

9. And you get the point $2 $ at $D$.

Similarly, you can do it for other irrational numbers like $3 ,5 ,..$

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A terminating decimal is a decimal that ends i.e. it has finite number of digits.

For a fraction in decimal form, while performing division after a certain number of steps, we get the remainder zero.

The quotient obtained as decimal is called the terminating decimal.

For eg:-

$83 =0.375$ has $3$ digits after decimal point

$45 =1.25$ has $2$ digits after decimal point

Hence, $0.375$ and $1.25$ are terminating decimals.

For a fraction in decimal form, while performing division after a certain number of steps, we get the remainder zero.

The quotient obtained as decimal is called the terminating decimal.

For eg:-

$83 =0.375$ has $3$ digits after decimal point

$45 =1.25$ has $2$ digits after decimal point

Hence, $0.375$ and $1.25$ are terminating decimals.

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While expressing a fraction in the decimal form, when we perform division we get some remainder.

If the division process does not end $i.e.$ we do not get the remainder equal to zero; then such decimal is known as non-terminating decimal.

In some cases, a digit or a block of digits repeats itself in the decimal part, then the decimal is non-terminating recurring decimal.

For eg:-

$1.6666....,0.141414...$

If the division process does not end $i.e.$ we do not get the remainder equal to zero; then such decimal is known as non-terminating decimal.

In some cases, a digit or a block of digits repeats itself in the decimal part, then the decimal is non-terminating recurring decimal.

For eg:-

$1.6666....,0.141414...$

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While expressing a fraction in the decimal form, when we perform division we get some remainder.

If the division process does not end i.e. we do not get the remainder equal to zero; then such decimal is known as non-terminating decimal.

And if a digit or a block of digits does not repeats itself in the decimal part, such decimals are called non-terminating and non-recurring decimals.

For eg:- $1.41421356.....$

If the division process does not end i.e. we do not get the remainder equal to zero; then such decimal is known as non-terminating decimal.

And if a digit or a block of digits does not repeats itself in the decimal part, such decimals are called non-terminating and non-recurring decimals.

For eg:- $1.41421356.....$

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Let $a$ and $b$ be any two positive integers. Then there exist unique integers $q$ and $r$ such that

$a=bq+r,$ where $0≤r<b$

If $b∣a$, then $r=0$.

Otherwise, $r$ satisfies the stronger inequality $0≤r<b$

$a=bq+r,$ where $0≤r<b$

If $b∣a$, then $r=0$.

Otherwise, $r$ satisfies the stronger inequality $0≤r<b$

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If $a$ and $b$ are positive integers such that $a=bq+r$, then every common divisor of $a$ and $b$ is a common divisor of $b$ and $r$, and vice-versa.

Example: Find HCF of $420$ and $130$.

Since $420>130$ we apply the division lemma to $420$ and $130$ to get ,Since $300$ , we apply the division lemma to $130$ and $30$ to get

$130=30×4+10$

$420=130×3+30$

Since $100$ , we apply the division lemma to $30$ and $10$ to get

$30=10×3+0$

The remainder has now become zero, so our procedure stops. Since the divisor at this step is $10$, the HCF of $420$ and $130$ is $10$.

Example: Find HCF of $420$ and $130$.

Since $420>130$ we apply the division lemma to $420$ and $130$ to get ,Since $300$ , we apply the division lemma to $130$ and $30$ to get

$130=30×4+10$

$420=130×3+30$

Since $100$ , we apply the division lemma to $30$ and $10$ to get

$30=10×3+0$

The remainder has now become zero, so our procedure stops. Since the divisor at this step is $10$, the HCF of $420$ and $130$ is $10$.

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Find the HCF of $6$ and $20$ by the prime factorization method.

We have : $6=2×3$

$20=2×2×5$

$20=2×2×5$.

Common factors of $6$ and $20$ are $2$ $1$ and $2$ $2$

So for HCF take the common number with lowest exponent.

HCF $=2×1=2$

We have : $6=2×3$

$20=2×2×5$

$20=2×2×5$.

Common factors of $6$ and $20$ are $2$ $1$ and $2$ $2$

So for HCF take the common number with lowest exponent.

HCF $=2×1=2$