Tuesday February 5
Today in math class we worked through some problems which were not in the common base 10 form. Instead of base 10, we used base 4, 6, 9... ect.
Some examples of these change of base problems are like so:
Time Problem:
3 hrs 15 min
-2 hrs 45 min
0 hrs 30 min = 12:30pm
The way that we got that answer was starting with the minutes first. 5 min - 5 min is obviously 0 min. Then we have to subtract 40 min from 10 min, since we dont have enough minutes anymore, we have move over to the hours place and borrow one hour. Since we are borrowing one hour and moving it to the minutes place, we give 60 min. So then we subtract 70-40 which is 30. Then we do the remaining 2 hrs-2 hrs. And end up with a time of 0 hrs 30 min. but since the hours are a base 12, the 0 hrs actually is supposed to be a 12. So we get a final time of 12:30.
Some simpiler problems are like so...
32
+23 (Base 4)
121
In a base 4 system, no numbers can ever be above a 4 ever, once a number reaches 4, it has to move to the next largest place value slot. So we start with the normal 2+3 which is 5, but since we are in a base 4 system fives dont exist, so we move a group of for to the tens place, and keep 1 in the ones place. Then we add up the 3+2 again, but we also have an extra 1 from the ones place that moved over. So we get 6, which is actually a group of 4 and two left over, so we move the group of four over to the hundreds place, and keep the 2 in the tens place. This results in a final answer of 121.
Some more problems are like so:
345
+266 (Base 7)
644
43
+25 (Base 6)
112
333
+101 (Base 4)
1100
Subtraction problems with a change of base use the same rules as normal subtraction, it just takes a little more effort and thinking...
63
-25 (Base 7)
35
We start with the ones and realize that we need to borrow a group of ones from the next place value over, but since this problem is in base seven, one group in the tens place is equivalent to seven individuals in the once place. So the 60 becomes a 50, and the 3 becomes a 10. This is because we borrowed 7 from the tens, and gave it to the ones, and 3+7=10. So then we move on with our subtraction problem with 10-5 which is 5. Then we move on over to the tens place and subtract 5-2 because we borrowed a group from the tens. We end up with an answer of 35.
Some more problems like this:
31
-12 (Base Four)
13
342
-253 (Base Six)
45
302
-103 (Base Four)
133
The next step of this process is to be given a problem, and then try to determine what base the problem is in:
231
+414
1200
So in this problem like most others you start from the right and work your way left. So we start with the 4+1. Since in this problem the 4+1 results in a 0 in the ones place, we can guess that this problem is in a base 5. We can assume this because there is a 0 in the ones place, and if this happens, a group of five moves to the tens place. Then you move to the tens place, and then do the 3+1+1 (the second one is being carried from the ones place). So we get an answer of 5 again, so we carry one over to the hundreds place. Then we do 4+2+1 which equals 7, but we are in base ten, so we move one group of five over to the thousands place, and keep 2 in the hundreds. Resulting in an answer of 1200.
More problems like this:
231
+414
1045
This problem is a base six because all the numbers are below a base 6 in the problem, and 400+200=600 which in this case would equal 1000 because its a base 6 problem.
344
+143
1042
This problem is in a base 5.
Subtraction Problems like this are done the exact same way:
523
-254
236
This problem is in the base of 7. Since we cant subtract 4 from three we have to borrow a group from the tens place. but we dont know how many we have to borrow since we dont know what base this problem is in, but we do know that whatever number we borrow, has to result in a subtraction problem that equals 6. so we ask ourselves, what plus three - 4 is six? The answer is 7. 7+3=10. 10-4=6. So from here we can assume the problem is in base 7, and finish the problem to double check and make sure.
More problems:
523
-254 (Base 8)
427
1020
-203 (Base 5)
312
Saturday, February 9, 2013
Saturday, February 2, 2013
Week Four.
Tuesday January 29th
Today in math class we did an activity out of our activity booklet. In this activity we were doing three diget addition and subtraction problems, at first with base ten blocks, then place value holders, then finally with paper and pencil. By using these three different ways to solve a problem, we were able to manipulate the ways that children go through the process of learning math problems.
First they need to have base ten blocks, where they can clearly see that 10 ones blocks, are equal to 1 tens block, and so on. The picture below illustrates the blocks that we used in the first activity.
In the first activity we rolled three dice to determine how many 'hundreds' we will have, then rolled two dice to find out how many 'tens' we would have, and so on. We had to do this for two three diget numbers.
Once we found out what numbers we were going to illustrate, we had to copy both numbers exactly with the base ten blocks. Making sure to not change 13 'tens' blocks into a 1 hundred block, and 3 tens blocks, that part comes later.
After you had both numbers laid out on the table, then you can manipulate the numbers to have the correct amount of numbers in each section, by not exceeding 10 blocks in one column.
Then it is a simple addition or subtraction problem from here, where you either add or subtract the needed manipulative. Doing addition or subtraction this way makes it easier for children to see what is actually happening with the numbers, instead of just having numbers written on a piece of paper.
This is the final answer to the addition problem above, 571
The second activity that we did in class was the same type of activity only a step above the method of using the base ten blocks. In this activity instead of being able to see the relationship that 10 ones is the same as 1 ten, the manipulative that you are using are the exact same size for every place value in the problem.
Similar to the problem before, you pick two numbers then add or subtract them from one another. Only this time it is easier to lose track of your place values because you cant clearly see their relationship. You have three sections labeled ones, tens and hundreds, then place the two numbers that you are adding together in those slots by using the place value holder chips.
In this problem, the answer is 595.
The last activity that we did in class was one more step above using the place value holders. It is all done on paper with pencil, it is the traditional way of doing an addition or subtraction problem...
634 634
+120 +120
754 700---- 600+100
50----30+20
4----4+0
754
Thursday January 31st
Thursday in class we spent all class period going over the different ways that you can solve both addition and subtraction problems.
We learned that there are 5 ways to solve an addition problem. Those ways are illustrated as follows:
Traditional:
The traditional way of doing a subtraction problem is just what you would think, you stack the numbers and add them together up and down.
59
+83
142
Decomposing:
The decomposing method is illustrated on a number line, where you 'hop' your way from one number to the other, using friendly numbers to figure out the answer.
Instructional/Partial Sums:
The instructional or partial sums method is a way of doing an addition problem by breaking up the numbers into ones and tens, and even hundreds if needed. By doing this the problem seems much smaller, and its easier to see how you got your answer.
59
+83
130---50+80
12---9+3
142
Compensating:
The compensating method is a way of solving a problem by rounding a single number in the problem to make it more friendly, then at the end, adding or taking away the amount that made it a friendly number to get your answer.
59+83= --------60+83= 143 143-1= 142
since we added one more then the problem asked to the number 59 to make it 60, we had to take one away from the answer that we got to set the problem back into balance.
Give and Take:
This method is giving part of one number to another to make it a more friendly number to work with.
59+83=
59+(1+82)=
(59+1)+82= (60) + 82= 142
We also learned that there are 6 ways of solving a subtraction problem:
Traditional:
75
-42
33
Partial Differences:
75
-42
30---70-40
3----5-2
33
Decomposing:
Difference/Distance:
The difference between the decomposing method and the difference method is that the difference method can be seen as an addition problem that is done on the number line, where you start with the smaller number, and add up to your larger number finding the difference of the two which ends us being your answer.
Compensating:
75-42=? 75-45=30 30+3=33
Holy Shift:
This method is similar to the give and take method for addition.
75-42=
(75-2)-40=
73-40=33
In this method we are just shifting the problem to make it easier to solve. the numbers are different, but since we are trying to find the difference between the two numbers, we can move the numbers up or down and ultimately get the same answer.
Today in math class we did an activity out of our activity booklet. In this activity we were doing three diget addition and subtraction problems, at first with base ten blocks, then place value holders, then finally with paper and pencil. By using these three different ways to solve a problem, we were able to manipulate the ways that children go through the process of learning math problems.
First they need to have base ten blocks, where they can clearly see that 10 ones blocks, are equal to 1 tens block, and so on. The picture below illustrates the blocks that we used in the first activity.
In the first activity we rolled three dice to determine how many 'hundreds' we will have, then rolled two dice to find out how many 'tens' we would have, and so on. We had to do this for two three diget numbers.
Once we found out what numbers we were going to illustrate, we had to copy both numbers exactly with the base ten blocks. Making sure to not change 13 'tens' blocks into a 1 hundred block, and 3 tens blocks, that part comes later.
After you had both numbers laid out on the table, then you can manipulate the numbers to have the correct amount of numbers in each section, by not exceeding 10 blocks in one column.
Then it is a simple addition or subtraction problem from here, where you either add or subtract the needed manipulative. Doing addition or subtraction this way makes it easier for children to see what is actually happening with the numbers, instead of just having numbers written on a piece of paper.
This is the final answer to the addition problem above, 571The second activity that we did in class was the same type of activity only a step above the method of using the base ten blocks. In this activity instead of being able to see the relationship that 10 ones is the same as 1 ten, the manipulative that you are using are the exact same size for every place value in the problem.
Similar to the problem before, you pick two numbers then add or subtract them from one another. Only this time it is easier to lose track of your place values because you cant clearly see their relationship. You have three sections labeled ones, tens and hundreds, then place the two numbers that you are adding together in those slots by using the place value holder chips.
In this problem, the answer is 595.
The last activity that we did in class was one more step above using the place value holders. It is all done on paper with pencil, it is the traditional way of doing an addition or subtraction problem...
634 634
+120 +120
754 700---- 600+100
50----30+20
4----4+0
754
Thursday January 31st
Thursday in class we spent all class period going over the different ways that you can solve both addition and subtraction problems.
We learned that there are 5 ways to solve an addition problem. Those ways are illustrated as follows:
Traditional:
The traditional way of doing a subtraction problem is just what you would think, you stack the numbers and add them together up and down.
59
+83
142
Decomposing:
The decomposing method is illustrated on a number line, where you 'hop' your way from one number to the other, using friendly numbers to figure out the answer.
Instructional/Partial Sums:
The instructional or partial sums method is a way of doing an addition problem by breaking up the numbers into ones and tens, and even hundreds if needed. By doing this the problem seems much smaller, and its easier to see how you got your answer.
59
+83
130---50+80
12---9+3
142
Compensating:
The compensating method is a way of solving a problem by rounding a single number in the problem to make it more friendly, then at the end, adding or taking away the amount that made it a friendly number to get your answer.
59+83= --------60+83= 143 143-1= 142
since we added one more then the problem asked to the number 59 to make it 60, we had to take one away from the answer that we got to set the problem back into balance.
Give and Take:
This method is giving part of one number to another to make it a more friendly number to work with.
59+83=
59+(1+82)=
(59+1)+82= (60) + 82= 142
Traditional:
75
-42
33
Partial Differences:
75
-42
30---70-40
3----5-2
33
Difference/Distance:
The difference between the decomposing method and the difference method is that the difference method can be seen as an addition problem that is done on the number line, where you start with the smaller number, and add up to your larger number finding the difference of the two which ends us being your answer.
Compensating:
75-42=? 75-45=30 30+3=33
Holy Shift:
This method is similar to the give and take method for addition.
75-42=
(75-2)-40=
73-40=33
In this method we are just shifting the problem to make it easier to solve. the numbers are different, but since we are trying to find the difference between the two numbers, we can move the numbers up or down and ultimately get the same answer.
Saturday, January 19, 2013
Week Two
Tuesday, January 15th
When I walked into the classroom on Tuesday there was immediately a problem on the board that we all had to solve independently. The problem was like so... If there is a dartboard with three sections labeled with the numbers 1, 5, and 10, how many different sums can you obtain if you hit the board using three darts.
The first way that I tried to tackle this problem was to make a list of all the possible outcomes starting with the number one.
After I did this, I moved onto the number 5, and then onto the number 10.
In this list I had a section labeled what the sum of the numbers added together was, to make sure that I had no repeats.
The list ended up looking like so....
Then after we all figured out the problem using our own methods, the class discussed the importance of using clear tables when solving problems.
The next problem that we did in class was a word problem. The word problem had five different types of cars, all finishing a race at different times. We were given clues on how to figure out the order in which the cars finished the race.
In this problem I tried to solve it in my head and was thinking to myself that there had to be a better way to solve this problem. After I talked to the girls that were sitting at the table with me, I saw the best way to solve this problem. Using a number line...
Using a number line is a great way to solve problems that have a time factor in them, because they can also act as a time line, using the tick marks to indicate seconds.
The last problem that my class did on Tuesday was one that I was more familiar with, It was an algebra problem.
There were four hexagons attached together side to side, and we needed to figure out a way to calculate the perimeter of the hexagons, without counting the sides.
The way that I solved the problem was by counting the sides of the beginning and ending hexagons, since there were five sides showing on both of them, and two end pieces, I mulpiplied 5x2. Then I counted the middle pieces, and how many sides they had showing, and multplied 2x4.
Then I added together (5x2)+(4x2) and got an answer of 18.
Then we talked as a class, and I herd of other ways that people solved this problem. Some people started out with knowing that each hexagon had to have at least four sides, and then added the sides necessary from there. Like so....
There were four total hexagons so they started out with 4x4. Then they knew that the two end hexagons had one extra side, and added two to their original total.
(4x4) + 2 = 18
And then someone else did the same process, but started out with six sides on each hexagon and subtracted the necessary sides. The way that they figured out how many to subtract was by how many sides the middle hexagons shared, then multiplied that number by two, because there were two sides being shared.
(6x4) - (3x2) = 18
Thursday, January 17th
On Thursday in class we started to learn about sets, and we started off with Venn Diagrams.
We learned some Venn Diagram lingo first to better understand how to group items together.
A u B = (A union B) is the set of everything in set A or B, or Both.
A n B = (A intersect B) is the set of what set A and B both share, what they have in common.
The term Complement means opposite, and is expressed by a line over the set.
A u B = (Complement of everything in A and B) What is not in the set of A u B
From this information we broke into small groups and began to play a game using different sizes and shapes.There were two teams, and a Venn Diagram. We had to pick a card to express the set that was going to be placed in each circle of the diagram.
Then when we put the two sets together we got a complete Venn Diagram that looked like so...
Then from there each team picked a card that had a task on it, such as "what is the complement of the other teams set" and from there you had to pick the shapes that applied to the card. You got points for how many shapes applied to the card drawn.
When I walked into the classroom on Tuesday there was immediately a problem on the board that we all had to solve independently. The problem was like so... If there is a dartboard with three sections labeled with the numbers 1, 5, and 10, how many different sums can you obtain if you hit the board using three darts.
The first way that I tried to tackle this problem was to make a list of all the possible outcomes starting with the number one.
After I did this, I moved onto the number 5, and then onto the number 10.
In this list I had a section labeled what the sum of the numbers added together was, to make sure that I had no repeats.
The list ended up looking like so....
Then after we all figured out the problem using our own methods, the class discussed the importance of using clear tables when solving problems.
The next problem that we did in class was a word problem. The word problem had five different types of cars, all finishing a race at different times. We were given clues on how to figure out the order in which the cars finished the race.
In this problem I tried to solve it in my head and was thinking to myself that there had to be a better way to solve this problem. After I talked to the girls that were sitting at the table with me, I saw the best way to solve this problem. Using a number line...
Using a number line is a great way to solve problems that have a time factor in them, because they can also act as a time line, using the tick marks to indicate seconds.
The last problem that my class did on Tuesday was one that I was more familiar with, It was an algebra problem.
There were four hexagons attached together side to side, and we needed to figure out a way to calculate the perimeter of the hexagons, without counting the sides.
The way that I solved the problem was by counting the sides of the beginning and ending hexagons, since there were five sides showing on both of them, and two end pieces, I mulpiplied 5x2. Then I counted the middle pieces, and how many sides they had showing, and multplied 2x4.
Then I added together (5x2)+(4x2) and got an answer of 18.
Then we talked as a class, and I herd of other ways that people solved this problem. Some people started out with knowing that each hexagon had to have at least four sides, and then added the sides necessary from there. Like so....
There were four total hexagons so they started out with 4x4. Then they knew that the two end hexagons had one extra side, and added two to their original total.
(4x4) + 2 = 18
And then someone else did the same process, but started out with six sides on each hexagon and subtracted the necessary sides. The way that they figured out how many to subtract was by how many sides the middle hexagons shared, then multiplied that number by two, because there were two sides being shared.
(6x4) - (3x2) = 18
Thursday, January 17th
On Thursday in class we started to learn about sets, and we started off with Venn Diagrams.
We learned some Venn Diagram lingo first to better understand how to group items together.
A u B = (A union B) is the set of everything in set A or B, or Both.
A n B = (A intersect B) is the set of what set A and B both share, what they have in common.
The term Complement means opposite, and is expressed by a line over the set.
From this information we broke into small groups and began to play a game using different sizes and shapes.There were two teams, and a Venn Diagram. We had to pick a card to express the set that was going to be placed in each circle of the diagram.
Saturday, January 12, 2013
Week One.
Everyone in the class had to make a name tag and think of something mathematical about themselves. (At the time I had no idea what that meant, along with many other people in the class). When it was my turn to present, a tingling sensation overtook my hands and feet. My throat felt very dry and I was incredibly nervous.
After the presentations Christina, the instructor for the class, began to go over the syllabus and relay her expectations for the class.
Then she put a simple math problem on the board, and asked us to solve it in our heads.
The problem was:
66-29=?
So naturally I tried to solve this problem the way most of us were originally taught. I stacked the numbers and then proceeded to borrow, to solve the problem. I messed up a few times but eventually got the right answer of 37.
Then we split into different groups according to the way that we solved the problem. I was absolutely shocked to see how many different ways you could solve one simple subtraction problem. Total there were 7 different ways that people in the class solved the problem.
One person from each group had to illustrate how their group solved the problem on a chalk board, and explain it to the rest of the class. The different strategies varied from rounding one number, to both numbers, to using a number line, and the traditional way that I used.
Before I knew of these other ways to solve subtraction problems, I always had to pull out the calculator on my phone to get the right answer.
I no longer have to use a calculator to do subtraction problems! I can do them in my head now.
On Thursdays class we started off with a few more simple math problems to do in our head. This time I tried to use a strategy that I learned in Tuesdays class, and with no surprise to me, the other strategy worked wonders better.
I cant believe that I was never aware of these other ways to solve math problems, especially since they are mush easier. And the probability of obtaining the right answer using these new strategies are much higher.
I have already learned so much in this class and it has only been one week. I cant wait to come to Tuesdays class and learn about other strategies or material that will make myself a better student, and help me become a better teacher in the future.
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