The closure property asks a simple question: If you take two real numbers and perform an arithmetic operation, will the result always be a real number?
For any two real numbers $a$ and $b$ ($a, b \in R$):
(i) Addition: $a + b \in R$ (Always a real number)
(ii) Subtraction: $a – b \in R$ (Always a real number)
(iii) Multiplication: $a \times b \in R$ (Always a real number)
(iv) Division: $\frac{a}{b}$ may or may not belong to $R$.
The Exception: Division is the “rebel” here. If you divide by zero, the result is undefined.
Example: $2$ and $0$ are real numbers, but $\frac{2}{0}$ is not a real number ($\frac{2}{0} \notin R$).
Summary: Real numbers are closed under addition, subtraction, and multiplication, but they are not closed under division.
The commutative property deals with order. Does it matter which number comes first?
For $a, b \in R$:
(i) Addition: $a + b = b + a$ (Order doesn’t matter)
(ii) Subtraction: $a – b \neq b – a$
Example: $3 – 2 \neq 2 – 3$ (because $1 \neq -1$)
(iii) Multiplication: $a \times b = b \times a$ (Order doesn’t matter)
(iv) Division: $a \div b \neq b \div a$
Example: $\frac{8}{4} \neq \frac{4}{8}$ (because $2 \neq 0.5$)
Summary: Real numbers are commutative under addition and multiplication, but not under subtraction or division.
The associative property concerns grouping. If you have three numbers, does it matter which two you group together first?
For $a, b, c \in R$:
(i) Addition: $(a + b) + c = a + (b + c)$
(ii) Subtraction: $(a – b) – c \neq a – (b – c)$
Example: $(5 – 3) – 2 \neq 5 – (3 – 2)$
Calculation: $2 – 2 = 0$, whereas $5 – 1 = 4$. Therefore, $0 \neq 4$.
(iii) Multiplication: $(a \times b) \times c = a \times (b \times c)$
(iv) Division: $(a \div b) \div c \neq a \div (b \div c)$
Identity properties involve numbers that, when used in an operation, leave the other number unchanged.
(i) Additive Identity: For any real number $a$, $a + 0 = 0 + a = a$. Here, 0 is the additive identity.
(ii) Multiplicative Identity: For any real number $a$, $a \times 1 = 1 \times a = a$. Here, 1 is the multiplicative identity.
Inverse properties describe how to “cancel out” a number to reach the identity element.
(i) Additive Inverse: For every $a$, there exists $-a$ such that $a + (-a) = (-a) + a = 0$. The value $-a$ is called the additive inverse of $a$.
(ii) Multiplicative Inverse: For any non-zero $a$, $a \times \frac{1}{a} = \frac{1}{a} \times a = 1$. The value $\frac{1}{a}$ is the multiplicative inverse (or reciprocal) of $a$.
Example: The multiplicative inverse of $2$ is $\frac{1}{2}$.
(iii) Product of Inverses: If $ab = 1$, then $b = \frac{1}{a}$ (where $a \neq 0$).
These properties highlight specific behaviors involving zero and one in various operations.
(i) Multiplication by Zero: $a \times 0 = 0 \times a = 0$. (Note: The source image contains a typo stating this equals $a$, but standard algebraic rules dictate the product is 0).
(ii) Subtraction with Zero: $a – 0 = a$.
(iii) Zero Minus a Number: $0 – a = -a$.
(iv) Identity Multiplication: $1 \cdot a = a$.
(v) Division by One: $\frac{a}{1} = a$.
(vi) Zero Product Property: If $a \times b = 0$, then either $a = 0$, $b = 0$, or both $a = b = 0$.
The distributive property is a powerful tool that links multiplication with addition and subtraction.
(i) Over Addition: $a \times (b + c) = (a \times b) + (a \times c)$, which simplifies to $a(b + c) = ab + ac$.
(ii) Over Subtraction: $a \times (b – c) = (a \times b) – (a \times c)$, which simplifies to $a(b – c) = ab – ac$.
(iii) Right Distribution: $(a + b)c = ab + ac$.
(iv) Right Distribution (Subtraction): $(a – b)c = ab – ac$.
(v) Binomial Distribution: $(a + b)(c + d) = ac + ad + bc + bd$.
Problem: Find the product of $(2 + \sqrt{3})(5 + \sqrt{3})$
First: Multiply the first terms: $2 \cdot 5 = \mathbf{10}$
Outer: Multiply the outer terms: $2 \cdot \sqrt{3} = \mathbf{2\sqrt{3}}$
Inner: Multiply the inner terms: $\sqrt{3} \cdot 5 = \mathbf{5\sqrt{3}}$
Last: Multiply the last terms: $\sqrt{3} \cdot \sqrt{3} = \mathbf{3}$
Combine them all:
Final Simplification:
Group the whole numbers ($10 + 3$) and the like radicals ($2\sqrt{3} + 5\sqrt{3}$):
Problem: Find the product of $(5 – \sqrt{2})(\sqrt{2} + 4)$
When subtraction is involved, be very careful with your signs!
First: $5 \cdot \sqrt{2} = \mathbf{5\sqrt{2}}$
Outer: $5 \cdot 4 = \mathbf{20}$
Inner: $-\sqrt{2} \cdot \sqrt{2} = \mathbf{-2}$
Combine and Simplify:
Following the logic in the image:
Combine the integers ($20 – 2$) and the radicals ($5\sqrt{2} – 4\sqrt{2}$):
Have you ever looked at a long math equation with division, multiplication, and brackets all mixed together and wondered, “Where do I even start?”
If you solve operations in the wrong order, you’ll end up with the wrong answer. That is why mathematicians use a specific hierarchy called the VBODMAS Rule. This rule acts as a roadmap, telling you exactly which operation to carry out first to simplify an arithmetic expression correctly.
VBODMAS is an acronym where each letter stands for a specific mathematical operation. To get the correct result, you must follow the letters in order from top to bottom:
V – Vinculum: This is a horizontal line drawn over a group of terms (similar to a bar). If you see this, solve the numbers under the bar first.
B – Brackets: Remove brackets in this specific order: circular brackets ( ), then curly braces { }, and finally square brackets [ ].
O – Order: This refers to powers, roots (like $\sqrt{x}$), or exponents ($x^2$). In some contexts, ‘O’ can also stand for “Of” (which implies multiplication).
D – Division ($\div$): Perform any division in the expression.
M – Multiplication ($\times$): Perform any multiplication.
A – Addition ($+$): Perform any addition.
S – Subtraction ($-$): Finally, perform any subtraction.
Let’s look at an example to see how the rule works in real-time.
Example 1: Simplify $17 – 24 \div 6 \times 4 + 8$
When we look at this expression, we have subtraction, division, multiplication, and addition. Following the VBODMAS order, we must handle the Division first.
Locate the division: $24 \div 6$.
$24 \div 6 = 4$.
Now our expression looks like this:
$> 17 – 4 \times 4 + 8$
Next in the rule is multiplication: $4 \times 4$.
$4 \times 4 = 16$.
Now our expression looks like this:
$> 17 – 16 + 8$
Now we handle the addition. To make it easier, group the positive numbers: $17 + 8 = 25$.
Now our expression looks like this:
$> 25 – 16$
Finally, perform the subtraction:
$25 – 16 = 9$.
Final Answer: 9
Problem: Simplify $[25 – 3(6 + 1)] \div 4 + 9$
The Step-by-Step Solution:
Inner Parentheses: Solve $(6 + 1)$ first.
$[25 – 3(7)] \div 4 + 9$
Multiply inside the bracket: $3 \times 7 = 21$.
$[25 – 21] \div 4 + 9$
Finalize the Square Bracket: $25 – 21 = 4$.
$(4 \div 4) + 9$
Divide and Add: $4 \div 4 = 1$, then add $9$.
Final Answer: 10
Some expressions include a “vinculum” (a bar over numbers). This bar acts as the highest priority bracket.
Problem: Find the value of $60 + [7 \div (6 + (1 – \overline{5 – 3}))] \text{ of } \frac{10}{7}$
The Step-by-Step Solution:
Solve the Vinculum: $\overline{5 – 3} = 2$.
$60 + [7 \div (6 + (1 – 2))] \times \frac{10}{7}$
Inner Brackets: Solve $(1 – 2) = -1$.
$60 + [7 \div (6 + (-1))] \times \frac{10}{7}$
Parentheses: Solve $(6 – 1) = 5$.
$60 + [7 \div 5] \times \frac{10}{7}$
Simplify Fractions: Notice that the $7$ in the bracket and the $7$ in the denominator cancel out, and $10 \div 5 = 2$.
$60 + 2$
Final Answer: 62
When an entire expression is under a square root, simplify the terms inside completely before attempting to find the root.
Problem: $\sqrt{5^2 \times 41 \times 5 – 17^2 – 75} = ?$
The Step-by-Step Solution:
Calculate Powers: $5^2 = 25$ and $17^2 = 289$.
$\sqrt{25 \times 41 \times 5 – 289 – 75}$
Multiply: $25 \times 41 \times 5 = 5125$.
$\sqrt{5125 – 289 – 75}$
Subtract: Perform the subtractions inside the radical.
$\sqrt{4761}$
Find the Root: Identify the number that, when multiplied by itself, equals $4761$.
Final Answer: 69
Let’s see how these rules work in practice using the number 123,456.
Rounding to the nearest Tens: * Target digit: 5 (in the tens place).
Digit to the right: 6 (more than 5).
Result: 123,460
Rounding to the nearest Hundreds: * Target digit: 4 (in the hundreds place).
Digit to the right: 5 (equal to 5).
Result: 123,500
Rounding to the nearest Thousands: * Target digit: 3 (in the thousands place).
Digit to the right: 4 (less than 5).
Result: 123,000
Rounding to the nearest Ten Thousands: * Target digit: 2 (in the ten thousands place).
Digit to the right: 3 (less than 5).
Result: 120,000
Rounding works the same way for decimals. Let’s look at two more examples:
Example 1: Rounding to Decimals What is the approximate value of 2.23607 correct to two decimal places?
Target digit: 3 (the second decimal place).
Digit to the right: 6 (more than 5).
Result: 2.24
Example 2: Rounding Whole Numbers with Decimals What is the number 76.0684 when rounded to the nearest 10?
Target digit: 7 (in the tens place).
Digit to the right: 6 (more than 5).
Result: 80