Integrated math 3
- Analyzing structure word problem: pet store (1 of 2)
- Analyzing structure word problem: pet store (2 of 2)
- Rational equations word problem: combined rates
- Rational equations word problem: combined rates (example 2)
- Rational equations word problem: eliminating solutions
- Reasoning about unknown variables
- Reasoning about unknown variables: divisibility
- Structure in rational expression
Reasoning about unknown variables
Sal solves the following challenge: Given that a>0, b<0, and a/b>a*b, what further can we learn about the variables? Created by Sal Khan.
Want to join the conversation?
- at4:04why is b squared(4 votes)
a/b < ab You multiplied each side by b to get
a*b/b < ab*b and the b/b is 1 and disappears so you get
a < ab * b which is
a < a *b * b and b*b is b² so
a < ab²
That is where the b² came from.
Then when you divide both sides by a you get
1 < b²(14 votes)
- When you divide A out at4:20do you always replace the empty side with 1?(4 votes)
- When you divide any variable out of a equation, you always replace the number or variable with 1. 1 is like the benchmark in math, it always takes the place of the number if the initial number is divided out.(5 votes)
- At4:27, can you take the square root of both sides? I tried that and it gives tells me that 1<b and -1<b. 1<b cannot be possible due to the given constraints. That leaves us with -1<b which is not what Sal concluded. What did I do wrong?(1 vote)
- Yes, we can take the square root of both sides, we just have to be very careful.
First of all, we know that the square root function is strictly growing over the interval [0, ∞).
This means that since 𝑏² > 1, then √(𝑏²) > √1
And because √1 = 1, we can write √(𝑏²) > 1
√(𝑏²) is the positive number whose square is equal to 𝑏².
Since 𝑏 is negative then −𝑏 is positive, and (−𝑏)² = 𝑏².
Thus, √(𝑏²) = −𝑏, which gives us the inequality
−𝑏 > 1
Multiplying both sides by −1 we get
𝑏 < −1(4 votes)
- At3:52Sal changes the sign of the inequality saying that this is necessary, since we multiply both sides by a negative. Could anyone explain to me why this is the case? Like why do you need to change the sign and how do we know it, except from learning this as a rule?(2 votes)
- This is a often asked question in multiple videos.
So first way to think about it is finding values. If -x ≥ 3, put in values. If x=3, -3≥3 is false as would be any positive number. Try 0, 0≥3 still false. Try -3 and get -(-3)≥ 3 so 3≥3, finally true. The more negative we go, it would always stay true, so x ≤ -3.
Second way to show it, if -x ≥ 3, add x to both sides and subtract 3 on both sides to get -3 ≥ x, then flip the equation around to get x ≤ -3.(2 votes)
- at5:37how can b>1 when at the top of the screen it says b<0(0 votes)
- It can't.
And in the 30 seconds after5:37, Sal explains that b>1 cannot be part of the solution because b<0 is a restriction.(7 votes)
- Since we know b is negative, why don't we reverse the inequality when multiplying a/b * b?(2 votes)
- We did... at least, looking back at my notes I know I did. So, go ahead :)(1 vote)
- At3:50Sal mentioned that we knew b was negative so he changed the inequality. What do we do if we don't know for sure if b is negative or if it could be both?(2 votes)
- Can I use a condition |A| =/> |B| ?
*absolute value of A has to be same or higher than absolute value of B(1 vote)
- How do you square a thing that isn't the bxh of a 2d object(2 votes)
- I completely follow the tutorials, they make perfect sense when I watch him manipulate the expressions, but as soon as I try to do the practice problems my brain shuts off.(1 vote)
- when you went from a/b>ab
why did u change the inequality symbol(1 vote)
- Because you are multiplying by a negative number, you switch the greater than/less than sign. We know that b < 0, which means that it can be any negative number. So when we multiply by 'b', you are multiplying by a negative number and you have to remember to swap out the sign.(2 votes)
You go for a job interview. And the first thing that your interviewer says is, look, you have great work experience. You seem like a nice young person. But what I really care about is your logical reasoning capabilities. So what she says is, just sit down. I'm going to ask you a question about some math expressions. And you say, sure, shoot away. Give me some questions about math expressions. And she says, all right. So you have two integers, integer a and integer b. And she tells you that integer a is greater than 0, and integer b is less than 0. Then she says that we also know that a/b is greater than a times b. So then she says, tell me some interesting things about a and b and a/b and a times b. And you say, OK, I'll give my best shot. And I encourage you to actually pause now and try to think about how these things might relate, or how b might be constrained, or a might be constrained, and then unpause it. So I assume you've unpaused it, so let's think about this a little bit. So the first thing, we know that a is positive, b is negative. So if I take a positive divided by a negative, what am I going to get? Well, this right over here, a positive divided by a negative is going to be a negative. What happens if I take a positive times a negative? Well, ab is also going to be a negative. A positive times a negative is also a negative. So really what we're saying is that we have two negative quantities, and this one right over here is greater than that one. So let's visualize that on a number line. So let's say that this right over here is 0. This right over here is the positive direction. This is the negative direction. We know both of these are negative, but a/b is larger. So a/b is going to be to the right of a times b. So we know that a/b, it has to be negative. It has to be the left of 0. a/b is going to be to the right of a times b. So one way to think about it is they're both negative, but a/b is going to be less negative. Or another way of thinking about it, it's going to have a smaller absolute value. Its distance from 0, which is another way of thinking about absolute value, its distance to the left of 0 is less than a times b's distance to the left of 0. But because they're both negative, and we're talking about distances from the left of 0, the one that has a smaller distance to the left of 0 is less negative, and is therefore a greater number. So, so far your interviewer seems impressed. So that was pretty good. You were able to get a lot of clues just out of this little bit of information that I gave you. But tell me more. Tell me more about what b has to be and whether we can constrain it in some way. And you say, well, OK. I've got some clues here, the fact that this absolute value right over here is going to be less than this absolute value. We know that the absolute value of a/b is going to be less than a times b, than the absolute value of a times b. Once again, it's going to be not as far left of 0 as a times b is. So that's how we can make that statement. But let's actually manipulate, algebraically manipulate this inequality here. We could multiply both sides times b. So let's do that. Let's multiply both sides times b. b Is less than the 0. So if you're multiplying both sides of an inequality by something less than 0, it swaps the inequality. That's why I got rid of it. I'm going to rewrite it. So b times a/b is going to negative. We know it's negative. So it's going to be less than ab times b. If we just multiply it out, these cancel out. We get a is less than a times b squared. Well, now if we want to simplify it, we can divide both sides by a. And since a is greater than 0, it does not change the inequality. So we divide both sides by a. This becomes 1, and this just becomes b squared. So we're left with 1 is less than b squared. Or another way we could think about this, we could say, well, if 1 is less than b squared, that means that b-- let's be careful here-- that means that the absolute value of b is going to be greater than 1. So this means that the absolute value of b is greater than 1. And you say, wait, Sal, how did you get from that to that right over there? Well, just think about this a little bit. If I square something, and if it's greater than 1-- so that means that either b is less than negative 1, because if it was negative 1, if you squared it, you would get 1. If it was greater than negative 1, or just directly greater than negative 1, if it was negative 0.99, then when you square it, it would be something less than 1. So that doesn't work. So b has to be less than negative 1 or b has to be greater than 1, because for the same exact logic. If it was exactly equal to 1, if you squared it, this would be equal. If it was 0.5, then squared it would be 0.25, which wouldn't be greater than. So you know both of these things are true. And this is another way of saying that the absolute value of b is greater than 1. Now, we have our other constraint, that b is less than 0. So since we know b is less than 0, we can take this out of the picture. And if the absolute value of b is greater than 1, and b is less than 0, then we know that b has to be less than negative 1. So wherever negative 1 is, b is going to be less than that. And your interviewer is very impressed and she says, look, you've done very good reasoning here. I think you deserve the job.