Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
Even and Odd Definition: Suppose k is an integer. If there exists an integer r such that k=2r+1, then k is an odd number. If there exists an integer r such that k=2r, then k is an even number.
Explanation: as long as an integer can be divided by 2, it is an even number.
Zero is an even number.
Positive and Negative Definition: A positive number is a real number that is greater than zero. A negative number is a real number that is smaller than zero.
Zero is not positive, nor negative.
Last edited by HongHu on 09 Mar 2005, 14:58, edited 2 times in total.
Basic rules for inequalities: (in the example: a>b>0, c>d>0)
You need to flip signs when both side are multiplied by a negative number:
You need to flip signs when 1 is divided by both side:
You can only add or multiply them when their signs are in the same direction:
You can only apply substractions and divisions when their signs are in the opposite directions:
(You can't say a/c>b/c. It is WRONG)
Deal with negative numbers:
However the sign needs to be flipped one more time if you are doing multiplication or division (because you are multiplying/dividing a negative number):
If x<-4, y<-2, we know that xy>8, but we don't know how x/y compare to (-4)/(-2)=2 since you can only do division when their signs are in different directions
If x>-4 and y<-2 then x/y<2 but we don't know how xy is compared to 8 since we can only do multiplication when their signs are the same direction.
It is easier to do the derivation, though, if you first change them to postive. For example:
If x<-4, y<-2, then -x>4, -y>2, xy>8
If x<-4, y<2, then -x>4, y<2, -x/y>2, x/y<-2
Last edited by HongHu on 14 Mar 2005, 07:21, edited 3 times in total.
Cancelling out "common terms" on both sides [#permalink]
08 Mar 2005, 09:44
This post received KUDOS
This post was BOOKMARKED
Cancelling out "common terms" on both sides of an equation
You need to be very careful when you do algebra derivations. One of the common mistake is to divide both side by "a common term". Remember you can only do this safely if the "common term" is a constant. However you CAN't do it if it contains a variable.
You can't cancel out the x on both side and say x=3 is the solution. You must move the x on the right side to the left side.
The solutions are: x=0 and x=3
The reason why you can't divided both sides by x is that when x is zero, you can't divide anything by zero.
Equally important if not more, is that you CAN'T multiple or divide a "common term" that includes a variable from both side of an inequality. Not only it could be zero, but it could also be negative in which case you would need to flip the sign.
You CAN'T divided both sides by x and say x>1. What you have to do is to move the right side to the left:
Solution would be either both x and x-1 are greater than zero, or both x and x-1 are smaller than zero. So your solution is: x>1 or x<0
Again you CAN'T multiply both sides by x because you don't know if x is positive or negative. What you have to do is to move the right side to the left:
If x>0 then x^2-1>0 =>x>1
If x<0 then x^2-1<0 =>x>-1
Therefore your solution is x>1 or 0>x>-1.
You could also break the original question to two branches from the beginning:
if x>0 then x^2>1 =>x>1
if x<0 then x^2<1 => x>-1
Therefore your solution is x>1 or 0>x>-1.
Last edited by HongHu on 26 Mar 2005, 21:08, edited 3 times in total.
I really struggle with the number property stuff. i guess i didn't pay enough attention back in 9th grade. do you have any broader suggestions if you will. i always try plugging but a lot of times, you lead oyuirself down the wrong path.
You need to first make sure you are clear about the concept before you go happily plugging numbers in. What I'm trying to do in these topped thead is exactly this: to collect some basic concepts at which everybody needs to be crystal clear.
I'll add on some more over a period of time, but here are for starters:
Divisibility rules: - Know them well
Integer is divisible by:
2 - Even integer
3 - Sum of digits are divisible by 3
4 - Integer is divisible by 2 twice or Last 2 digits are divisible by 4
5 - Last digit is 0 or 5
6 - Integer is divisbile by 2 AND 3
8 - Integer is divisible by 2 three times
9 - Sum of digits is divisible by 9
10 - Last digit is 0
Some other things to note:
- If 2 numbers have the same factor, then the sum or difference of the two numbers will have the same facor.
(e.g. 4 is a factor of 20, 4 is also a factor of 80, then 4 will be a factor of 60 (difference) and also 120 (sum))
- Remember to include '1' if you're asked to count the number of factors a number has
Start with the units digit, add every other digit and remember this number. Form a new number by adding the digits that remain. If the difference between these two numbers is divisible by 11, then the original number is divisible by 11.
Is the number 824472 divisible by 11? Starting with the units digit, add every other number:2 + 4 + 2 = 8. Then add the remaining numbers: 7 + 4 + 8 = 19. Since the difference between these two sums is 11, which is divisible by 11, 824472 is divisible by 11.
Is the number 49137 divisible by 11? Starting with the units digit, add every other number:7 + 1 + 4 = 12. Then add the remaining numbers: 3 + 9 = 12. Since the difference between these two sums is 0, which is divisible by 11, 49137 is divisible by 11.
Building on a post by Antmavel about sum of a series...
A quick way to find the sum of the a series where each preceding term is incremented by the same number would be to find the middle term and multiply it by the number of terms.
The middle term can be found at taking the average of the first and last term. Or for the case of Antmavel's question where there are ten terms, you just need to work up to the 5th and 6th term then find the middle of the these two numbers.
E.g. Sum of 4,8,12,16,20
Middle term: 12
Number of terms: 5
Sum = 12*5=60
Yes, that's a good one. I've been long thinking of doing one about the absolute value questions but have forgot about it.
Absolute values The way to solve this kind of questions is to break the equation (inequality) into two parts, one is when the value is non negative, the other is when the value is negative.
You break it into two parts:
If x-4>=0, then x-4<9, solve for both you get x>=4, x<13. So your solution is 4<=x<13.
If x-4<0, then -(x-4)<9, ie x-4>-9. Solve for both you get x<4, x>-5. So your solution for this part is -5<x<4.
Combine the two solutions, you get -5<x<13 as your final solution.
If x+4>=0, then x+4>4. Solve for both you get x>=-4, x>0. So your solution is x>0.
If x+4<0, then -(x+4)>4, ie. x+4<-4. Solve for both you get x<-4, x<-8. So your solution is x<-8.
You final solution is x>0 or x<-8.
The same strategy can apply to square questions.
For example: (x+4)^2>4
You could solve it this way:
x>-2 or x<-6
Or you can solve it this way:
If x+4>=0 then x+4>2. Solve for them you get x>-2.
If x+4<0 then x+4<-2. Solve for them you get x<-6.
if y>=0, y+1>=0, y>y+1, no solution.
if y<0, y+1<0, -y>-(y+1), solution is y<-1
if y>=0, y+1<0, y>-(y+1), no solution.
if y<0, y+1>=0, -y>y+1, solution is -1<=y<-1/2
So your final solution is y<-1/2
You could also solve this question by going the square route.
Last edited by HongHu on 17 Aug 2005, 06:28, edited 1 time in total.