IB Mathematics Analysis and Approaches Topic SL…

Why Standard Form Exists

Some numbers are awkward to write: the mass of the Earth is about $5\,970\,000\,000\,000\,000\,000\,000\,000\,000$ kg, and the width of an atom is a tiny fraction of a metre.
Counting all those zeros is slow and error-prone, so we need a compact way to write very large and very small numbers.
Standard form (also called scientific notation) does exactly that by splitting a number into a simple decimal and a power of ten.

The Structure: A Number Between 1 And 10 Times A Power Of Ten

$$ a \times 10^{k}, \quad 1 \le a < 10, \; k \in \mathbb{Z} $$

The first part $a$ is the significand: it carries the digits of the number and must have exactly one non-zero digit before the decimal point, so $1 \le a < 10$.
The second part $10^{k}$ is the scale: the exponent $k$ is a whole number (an integer, $k \in \mathbb{Z}$) that says how many places the decimal point has moved.
A positive $k$ makes the number large and a negative $k$ makes it small, because multiplying by $10^{k}$ shifts the decimal point $k$ places.
The rule $1 \le a < 10$ matters because it makes the form unique: without it, the same value could be written many ways.

Example

$4.5 \times 10^{5}$, $45 \times 10^{4}$ and $0.45 \times 10^{6}$ are all the same number, but only the first obeys $1 \le a < 10$.

Converting Between Standard Form And Ordinary Numbers

Each factor of ten moves the decimal point one place, so the exponent is really just a place-counter.
To write an ordinary number in standard form, slide the decimal point until exactly one non-zero digit sits in front of it, then count how many places you moved.
1. Moving the point left (for a large number) makes $k$ positive; moving it right (for a small number) makes $k$ negative.
To go the other way, read the exponent and move the decimal point that many places, filling empty spots with zeros.

Example

Large number $450\,000$: slide the point left past 5 digits to reach $4.5$, so $k = 5$ and $450\,000 = 4.5 \times 10^{5}$.
Small number $0.000\,32$: slide the point right past 4 places to reach $3.2$, so $k = -4$ and $0.000\,32 = 3.2 \times 10^{-4}$.
Back the other way, $6.1 \times 10^{3}$ moves the point 3 places right to give $6100$.

Multiplying And Dividing In Standard Form

The trick is to deal with the two parts separately: handle the significands, then handle the powers of ten.
The powers of ten follow the laws of exponents from SL 1.5: to multiply you add the exponents, and to divide you subtract them, because $10^{m} \times 10^{n} = 10^{m+n}$. $$ (a \times 10^{m})(b \times 10^{n}) = (a \times b) \times 10^{m+n} $$ $$ \frac{a \times 10^{m}}{b \times 10^{n}} = \frac{a}{b} \times 10^{m-n} $$
After combining, the significand might no longer sit between 1 and 10, so you often need one more step.
That step is renormalising: rewrite the significand in standard form and carry the extra power of ten into the exponent.

Example

$$30 \times 10^{7} = (3 \times 10^{1}) \times 10^{7} = 3 \times 10^{8}$$

Example question

Evaluate, giving each answer in standard form:

(a) $(3 \times 10^{5})(4 \times 10^{-2})$
(b) $\dfrac{8 \times 10^{7}}{2 \times 10^{-3}}$

Solution

For (a):
- Significands: $3 \times 4 = 12$
- Powers: $5 + (-2) = 3$
- This gives $12 \times 10^{3}$
- Renormalise since $12 > 10$: $12 \times 10^{3} = 1.2 \times 10^{4}$
For (b):
- Significands: $8 \div 2 = 4$
- Powers: $7 - (-3) = 10$
- Answer $4 \times 10^{10}$, already in standard form.

Adding And Subtracting In Standard Form

Adding and subtracting is different: you can only combine the significands once the powers of ten are the same.
This is like needing a common denominator before adding fractions: the powers of ten have to match first.
Rewrite the term with the smaller exponent so it shares the larger exponent, then add or subtract the significands.

Example

To subtract $7.2 \times 10^{6} - 5 \times 10^{5}$:

Match powers: $5 \times 10^{5} = 0.5 \times 10^{6}$
Now subtract the significands: $$7.2 - 0.5 = 6.7$, giving $6.7 \times 10^{6}$

Exam technique

A calculator shows standard form in a shorthand called E-notation, for example displaying $5.2 \times 10^{30}$ as $5.2\text{E}30$.
That shorthand is not accepted in written answers; you must copy it out in full as $5.2 \times 10^{30}$.
Standard form is assumed prior knowledge, so it has no entry in the formula booklet; you are expected to know the $a \times 10^{k}$ rule by heart.

Common Mistake

The most common slip is forgetting to renormalise, leaving an answer like $12 \times 10^{3}$.
The next most common is flipping the sign of $k$ when converting a small number.
Always finish by checking that $1 \le a < 10$.

Active recall

State the two conditions a number in the form $a \times 10^{k}$ must satisfy.
Why does the rule $1 \le a < 10$ make standard form unique?
Convert $0.000\,067$ to standard form, explaining the sign of $k$.
Evaluate $(2 \times 10^{6})(4 \times 10^{-9})$ and renormalise if needed.
Why must the powers of ten match before you add two numbers in standard form?

Why Standard Form Exists

Some numbers are awkward to write: the mass of the Earth is about $5\,970\,000\,000\,000\,000\,000\,000\,000\,000$ kg, and the width of an atom is a tiny fraction of a metre.
Counting all those zeros is slow and error-prone, so we need a compact way to write very large and very small numbers.
Standard form (also called scientific notation) does exactly that by splitting a number into a simple decimal and a power of ten.

The Structure: A Number Between 1 And 10 Times A Power Of Ten

$$ a \times 10^{k}, \quad 1 \le a < 10, \; k \in \mathbb{Z} $$

The first part $a$ is the significand: it carries the digits of the number and must have exactly one non-zero digit before the decimal point, so $1 \le a < 10$.
The second part $10^{k}$ is the scale: the exponent $k$ is a whole number (an integer, $k \in \mathbb{Z}$) that says how many places the decimal point has moved.
A positive $k$ makes the number large and a negative $k$ makes it small, because multiplying by $10^{k}$ shifts the decimal point $k$ places.
The rule $1 \le a < 10$ matters because it makes the form unique: without it, the same value could be written many ways.

Example

$4.5 \times 10^{5}$, $45 \times 10^{4}$ and $0.45 \times 10^{6}$ are all the same number, but only the first obeys $1 \le a < 10$.

Converting Between Standard Form And Ordinary Numbers

Each factor of ten moves the decimal point one place, so the exponent is really just a place-counter.
To write an ordinary number in standard form, slide the decimal point until exactly one non-zero digit sits in front of it, then count how many places you moved.
1. Moving the point left (for a large number) makes $k$ positive; moving it right (for a small number) makes $k$ negative.
To go the other way, read the exponent and move the decimal point that many places, filling empty spots with zeros.

Example

Large number $450\,000$: slide the point left past 5 digits to reach $4.5$, so $k = 5$ and $450\,000 = 4.5 \times 10^{5}$.
Small number $0.000\,32$: slide the point right past 4 places to reach $3.2$, so $k = -4$ and $0.000\,32 = 3.2 \times 10^{-4}$.
Back the other way, $6.1 \times 10^{3}$ moves the point 3 places right to give $6100$.

Multiplying And Dividing In Standard Form

The trick is to deal with the two parts separately: handle the significands, then handle the powers of ten.
The powers of ten follow the laws of exponents from SL 1.5: to multiply you add the exponents, and to divide you subtract them, because $10^{m} \times 10^{n} = 10^{m+n}$. $$ (a \times 10^{m})(b \times 10^{n}) = (a \times b) \times 10^{m+n} $$ $$ \frac{a \times 10^{m}}{b \times 10^{n}} = \frac{a}{b} \times 10^{m-n} $$
After combining, the significand might no longer sit between 1 and 10, so you often need one more step.
That step is renormalising: rewrite the significand in standard form and carry the extra power of ten into the exponent.

Example

$$30 \times 10^{7} = (3 \times 10^{1}) \times 10^{7} = 3 \times 10^{8}$$

Example question

Evaluate, giving each answer in standard form:

(a) $(3 \times 10^{5})(4 \times 10^{-2})$
(b) $\dfrac{8 \times 10^{7}}{2 \times 10^{-3}}$

Solution

For (a):
- Significands: $3 \times 4 = 12$
- Powers: $5 + (-2) = 3$
- This gives $12 \times 10^{3}$
- Renormalise since $12 > 10$: $12 \times 10^{3} = 1.2 \times 10^{4}$
For (b):
- Significands: $8 \div 2 = 4$
- Powers: $7 - (-3) = 10$
- Answer $4 \times 10^{10}$, already in standard form.

Adding And Subtracting In Standard Form

Adding and subtracting is different: you can only combine the significands once the powers of ten are the same.
This is like needing a common denominator before adding fractions: the powers of ten have to match first.
Rewrite the term with the smaller exponent so it shares the larger exponent, then add or subtract the significands.

Example

To subtract $7.2 \times 10^{6} - 5 \times 10^{5}$:

Match powers: $5 \times 10^{5} = 0.5 \times 10^{6}$
Now subtract the significands: $$7.2 - 0.5 = 6.7$, giving $6.7 \times 10^{6}$

Exam technique

A calculator shows standard form in a shorthand called E-notation, for example displaying $5.2 \times 10^{30}$ as $5.2\text{E}30$.
That shorthand is not accepted in written answers; you must copy it out in full as $5.2 \times 10^{30}$.
Standard form is assumed prior knowledge, so it has no entry in the formula booklet; you are expected to know the $a \times 10^{k}$ rule by heart.

Common Mistake

The most common slip is forgetting to renormalise, leaving an answer like $12 \times 10^{3}$.
The next most common is flipping the sign of $k$ when converting a small number.
Always finish by checking that $1 \le a < 10$.

Active recall

State the two conditions a number in the form $a \times 10^{k}$ must satisfy.
Why does the rule $1 \le a < 10$ make standard form unique?
Convert $0.000\,067$ to standard form, explaining the sign of $k$.
Evaluate $(2 \times 10^{6})(4 \times 10^{-9})$ and renormalise if needed.
Why must the powers of ten match before you add two numbers in standard form?

IB Mathematics Analysis and Approaches SL 1.1—using Standard Form Notes

IB Mathematics Analysis and Approaches SL 1.1—using Standard Form Notes

Why Standard Form Exists

The Structure: A Number Between 1 And 10 Times A Power Of Ten

Converting Between Standard Form And Ordinary Numbers

Multiplying And Dividing In Standard Form

Adding And Subtracting In Standard Form

Why Standard Form Exists

The Structure: A Number Between 1 And 10 Times A Power Of Ten

Converting Between Standard Form And Ordinary Numbers

Multiplying And Dividing In Standard Form

Adding And Subtracting In Standard Form

Number and Algebra

Functions

Geometry & Trigonometry

Statistics & Probability

Calculus

IB Mathematics Analysis and Approaches SL 1.1—using Standard Form Notes

IB Mathematics Analysis and Approaches SL 1.1—using Standard Form Notes

Number and Algebra

Functions

Geometry & Trigonometry

Statistics & Probability

Calculus

Why Standard Form Exists

The Structure: A Number Between 1 And 10 Times A Power Of Ten

Converting Between Standard Form And Ordinary Numbers

Multiplying And Dividing In Standard Form

Adding And Subtracting In Standard Form

Why Standard Form Exists

The Structure: A Number Between 1 And 10 Times A Power Of Ten

Converting Between Standard Form And Ordinary Numbers

Multiplying And Dividing In Standard Form

Adding And Subtracting In Standard Form