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Number - Operations with Integers

Grade 6IB

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

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Integers are a set of whole numbers that include positive numbers, negative numbers, and zero: {...,βˆ’3,βˆ’2,βˆ’1,0,1,2,3,...}\{..., -3, -2, -1, 0, 1, 2, 3, ...\}. Visually, they are represented on a horizontal number line where zero is the origin, positive integers extend infinitely to the right, and negative integers extend infinitely to the left.

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The Absolute Value of an integer, denoted as ∣x∣|x|, is its distance from zero on a number line. Because it represents distance, it is always non-negative. For example, βˆ£βˆ’5∣=5|-5| = 5 and ∣5∣=5|5| = 5, meaning both numbers are exactly 55 units away from zero.

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Adding Integers with the Same Signs: To add integers with the same sign, add their absolute values and keep the common sign. Visually, adding two negative numbers like βˆ’2+(βˆ’3)-2 + (-3) is like starting at βˆ’2-2 on the number line and moving 33 units further to the left to land on βˆ’5-5.

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Adding Integers with Different Signs: Subtract the smaller absolute value from the larger absolute value and keep the sign of the number with the larger absolute value. Visually, this can be seen as 'Zero Pairs'; if you have 44 positive counters and 66 negative counters, 44 pairs cancel out to zero, leaving 22 negative counters, so 4+(βˆ’6)=βˆ’24 + (-6) = -2.

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Subtracting Integers: To subtract an integer, add its additive inverse (its opposite). The rule is aβˆ’b=a+(βˆ’b)a - b = a + (-b). Visually, subtracting a negative number like 2βˆ’(βˆ’3)2 - (-3) is the same as 2+32 + 3, moving 33 units to the right on the number line to reach 55.

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Multiplying and Dividing Integers with Same Signs: When multiplying or dividing two integers with the same sign (both positive or both negative), the result is always positive. For example, (βˆ’4)Γ—(βˆ’5)=20(-4) \times (-5) = 20 and (βˆ’10)Γ·(βˆ’2)=5(-10) \div (-2) = 5.

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Multiplying and Dividing Integers with Different Signs: When multiplying or dividing two integers with different signs (one positive and one negative), the result is always negative. For example, 8Γ—(βˆ’3)=βˆ’248 \times (-3) = -24 and (βˆ’15)Γ·3=βˆ’5(-15) \div 3 = -5.

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Order of Operations (BIDMAS/BODMAS): When an expression contains multiple operations, follow the order: Brackets, Indices, Division and Multiplication (from left to right), and then Addition and Subtraction (from left to right). This ensures consistent results when working with complex integer expressions.

πŸ“Formulae

a+(βˆ’b)=aβˆ’ba + (-b) = a - b

aβˆ’(βˆ’b)=a+ba - (-b) = a + b

(+)Γ—(+)=(+)(+) \times (+) = (+)

(βˆ’)Γ—(βˆ’)=(+)(-) \times (-) = (+)

(+)Γ—(βˆ’)=(βˆ’)(+) \times (-) = (-)

(βˆ’)Γ—(+)=(βˆ’)(-) \times (+) = (-)

(+)(+)=(+)\frac{(+)}{(+)} = (+)

(βˆ’)(βˆ’)=(+)\frac{(-)}{(-)} = (+)

(+)(βˆ’)=(βˆ’)\frac{(+)}{(-)} = (-)

∣x∣=xΒ ifΒ xβ‰₯0,Β and ∣x∣=βˆ’xΒ ifΒ x<0|x| = x \text{ if } x \geq 0, \text{ and } |x| = -x \text{ if } x < 0

πŸ’‘Examples

Problem 1:

Evaluate the expression: βˆ’12+(βˆ’8)βˆ’(βˆ’15)-12 + (-8) - (-15)

Solution:

Step 1: Solve the addition part: βˆ’12+(βˆ’8)=βˆ’20-12 + (-8) = -20. \ Step 2: Rewrite the subtraction as adding the opposite: βˆ’20βˆ’(βˆ’15)=βˆ’20+15-20 - (-15) = -20 + 15. \ Step 3: Solve the final addition: βˆ’20+15=βˆ’5-20 + 15 = -5.

Explanation:

We first combined the two negative numbers by adding their absolute values and keeping the negative sign. Then, we applied the rule that subtracting a negative is the same as adding a positive.

Problem 2:

Calculate: 18Γ·(βˆ’3)Γ—(βˆ’2)+418 \div (-3) \times (-2) + 4

Solution:

Step 1: Perform division (left to right): 18Γ·(βˆ’3)=βˆ’618 \div (-3) = -6. \ Step 2: Perform multiplication: (βˆ’6)Γ—(βˆ’2)=12(-6) \times (-2) = 12. \ Step 3: Perform addition: 12+4=1612 + 4 = 16.

Explanation:

Following the BIDMAS rule, we handle division and multiplication from left to right before performing the addition. A positive divided by a negative results in a negative, and a negative multiplied by a negative results in a positive.