Math Detours · Grade 6 · Sharpen · 4

Below zero

Negative numbers stop being confusing the moment you treat them as moves on a line. Then the famous rule, minus times minus is plus, turns out to be forced on us.

Name ____________________Date ____________

Tonight's idea

Think of every number as a position on a line, and every addition or subtraction as a move. Adding 5 means step 5 to the right. Adding \(-5\) means step 5 to the left. Subtracting is the opposite move: subtracting 5 is 5 left, and subtracting \(-5\) is the opposite of 5 left, which is 5 right. That is the whole reason \(3 - (-5) = 8\).

−6−5−4−3−2−10123456783 − (−5) = 8: the opposite of 5 left
Subtracting a negative is the opposite of a leftward move.

Absolute value, written \(|x|\), is distance from zero, ignoring direction. \(|-7| = 7\). More useful: \(|a - b|\) is the distance between \(a\) and \(b\). Distance is never negative.

Why is \((-1) \times (-2) = +2\)? Look at the pattern: \(3 \times (-2) = -6\), \(2 \times (-2) = -4\), \(1 \times (-2) = -2\), \(0 \times (-2) = 0\). Each time the first number drops by 1, the answer rises by 2. Continue: \((-1) \times (-2)\) must be \(2\), and \((-2) \times (-2)\) must be \(4\). Any other answer would break the pattern. The rule is forced.

Watch one

Compute \(-3 - (-7)\). Subtracting \(-7\) is the opposite of moving 7 left, so move 7 right from \(-3\): land on 4. Another view: \(a - b\) is "how far, and which way, from \(b\) to \(a\)". From \(-7\) up to \(-3\) is 4 steps right, so \(+4\).

Compute \((-2) \times 3 \times (-5) \times (-1)\). Three negative signs, an odd number, so the answer is negative. Size: \(2 \times 3 \times 5 \times 1 = 30\). Answer \(-30\).

Your turn

  1. Compute: \(-5 + 8\), \(\;-5 - 8\), \(\;6 - (-4)\), \(\;-2 - (-9)\), \(\;-7 + (-7)\).
  2. Compute \(|{-7}|\), \(|3 - 10|\), \(|{-4}| - |4|\). Then find both numbers that are at distance 5 from \(-2\).
  3. At night it is \(-8\)°C. By noon it has risen 13 degrees. By midnight it has fallen 19 degrees from noon. Give the noon and midnight temperatures.
  4. Compute: \((-3)(-4)\), \(\;(-2)^3\), \(\;(-1)^{10}\), \(\;(-1)^{11}\), \(\;(-2)(3)(-5)(-1)\). State a rule for the sign of a product with \(n\) negative factors.
  5. Plot \(A(-3, 2)\), \(B(4, 2)\), \(C(4, -4)\). Name the quadrant of each point (quadrants are numbered I to IV anticlockwise, starting top right). Find the lengths \(AB\) and \(BC\) by counting, then find the fourth corner \(D\) that makes \(ABCD\) a rectangle and its area.
  6. An elevator starts at floor \(-2\) (second basement). It goes up 7, down 11, up 3. Which floor is it on? Write it as one sum.

Stretch: the straight-line distance

In question 5 you found \(AB = 7\) and \(BC = 6\). The straight line from \(A\) to \(C\) is the diagonal of the rectangle. It is clearly longer than 7 and shorter than \(7 + 6 = 13\). Can you pin it down better?

A fact you will prove on another sheet: the diagonal \(d\) satisfies \(d^2 = 7^2 + 6^2\). Work out \(7^2 + 6^2\). Which two whole numbers is \(d\) between? Is it closer to the lower or upper one? (Compare \(d^2\) with \(9^2\) and \(10^2\).)

Answers

  1. 3, \(-13\), 10, 7, \(-14\).
  2. 7, 7, 0. Distance 5 from \(-2\): \(3\) and \(-7\).
  3. Noon: \(-8 + 13 = 5\)°C. Midnight: \(5 - 19 = -14\)°C.
  4. 12, \(-8\), 1, \(-1\), \(-30\). An even number of negative factors gives a positive product; an odd number gives a negative product.
  5. \(A\) is in quadrant II (top left), \(B\) in quadrant I, \(C\) in quadrant IV. \(AB = 7\) (from \(-3\) to 4), \(BC = 6\) (from 2 down to \(-4\)). \(D = (-3, -4)\), in quadrant III. Area \(7 \times 6 = 42\).
  6. \(-2 + 7 - 11 + 3 = -3\). Third basement.

Stretch. \(7^2 + 6^2 = 49 + 36 = 85\). Since \(81 < 85 < 100\), \(d\) is between 9 and 10, and because 85 is much closer to 81 than to 100, \(d\) is a little over 9 (about 9.22). You have just used Pythagoras's theorem; the proof is waiting in "Around the corner".

← PreviousNext: Multiply like a magician →