Presentation on theme: "Warm-Up Make the numbers 1-10 using four 4s and operations and parentheses. Ex. Make the number 16 using four 4s and operations and parentheses: 16 = 4*4*(4/4)"— Presentation transcript:
Warm-Up Make the numbers 1-10 using four 4s and operations and parentheses. Ex. Make the number 16 using four 4s and operations and parentheses: 16 = 4*4*(4/4)
Agenda for today! Warm-Up Number systems--Egyptian, Babylonian, Mayan, and Roman Assign homework for week
Number System Numbers did not start out as 1, 2, 3, … The first number systems were often 1, 2, many. 0 wasn’t needed. More complex number systems arose out of need.
Number Systems Tally systems: |||| at a time. Babylonian--only two symbols Egyptian--probably earliest known Roman--most widespread, all over Europe from the Roman Empire Mayan--only three symbols Hindu-Arabic--what we use today
A brief look at Roman Numerals No zero. Symbols: I = 1, V = 5, X = 10, L = 50, C = 100, D = 500, and M = 1000 If the Roman Numerals are in order from greatest to least, then add: VII = 5 + 1 + 1 = 7; XVI = 10 + 5 + 1 = 16
Roman Numerals I = 1, V = 5, X = 10, L = 50, C = 100, D = 500, and M = 1000. If the Roman Numerals are NOT in order from greatest to least, then subtract where the order is wrong. IV = 5 - 1 = 4; IX = 10 - 1; XCII = 100 - 10 + 1 + 1 = 92
A brief look at Babylonian Numbers Initially, no zero. Later developed: Two symbols only: = 1; = 10. Additive when written from greatest to least: = 10+10+1+1+1 Use 60 as a base--there is a break after 60. means 60+60+60+10+1
A brief look at Mayan Numbers Used the concept of zero, but only for place holders Used three symbols: 150 Wrote their numbers vertically: is 3 + 5 = 8, --- is 5 + 5 = 10
Mayan Numbers New place value… left a vertical gap. is one 20, and 0 ones = 20. The new place value is vertically above the ones place value. is two ___ + 6
Key Points Roman: no zero, subtractive, calculations are cumbersome Babylonian: no zero initially but developed to indicate an empty place, two symbols (and zero), place value system (base 60) Egyptian: no zero, additive, symbols for powers of 10 but not a place value system (only additive!) Mayan: zero used to indicate empty space, symbols for 1 and 5, place value system (base 20)
Homework Due Monday: in Explorations, 2.8 fill out a rough draft of the Alphabitia table on p. 41 using the conventions we invented in class. Warm-Up: Make the numbers 1-10 using four 4s and operations and parentheses. (you must use all the 4s for each number!)
Homework Comments: Strategy/Justify +1 Strategy: You should be able to separate the method you used from what you did. This is your strategy. It should be clearly labeled as such. +1 Justify: The step-by-step why you did what you did. This can be done alongside or as a summary of the steps in your work.
Example #14 1080/4 = 270ft would be the side lengths if we had a square. The length needs to be 80ft longer than the width. The total difference between a length and width is 80 and because we must increase/decrease two sides by an equal amount we divide 80 by 2. (80/2 = 40). So, 270-40 = 230 = width and 270 + 40 = 310 = length. We can check and see that 310 + 310 + 230 + 230 = 1080!
Strategy: Answer the question “How did you do it?” NOT “What did you do?” Strategy: First we will assume the land is square, then if we increase a length side we must decrease a width side by the same amount to keep the perimeter constant. Our goal is to find out how much we must increase/decrease the sides by and then change the lengths of the sides to find the length and width of the rectangular piece of land. Note: We didn’t have to know anything about the numbers in the problem to talk about how we would solve the problem (or how we did solve the problem).
Extend Your Thinking The strategy could be used if the problem were changed is NOT extended thinking. How a problem might be changed is important! Your strategy may not actually work depending on how it is changed. Try… If the numbers (perimeter, and amount length is longer than the width) in the problem were changed, we could use this strategy. However, if the relationship between length and width changed, say, length is twice the width, this might be harder to do by “squaring the rectangle” and finding some amount to increase/decrease the lengths of the sides by.
For example: if perimeter were 1080 and length is twice the width, if we start with a square with side lengths = 270 then it isn’t as obvious how much I should increase each length side by and decrease a width side like it was before. (why...? Take a few guesses!) It might be easier to solve this type of problem using algebra.