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Assume that U is the universe of discourse Prove UA = 0Conclude that 4As U ACAis not necessarily true:...

Question

Assume that U is the universe of discourse Prove UA = 0Conclude that 4As U ACAis not necessarily true:

Assume that U is the universe of discourse Prove UA = 0 Conclude that 4 As U ACA is not necessarily true:



Answers

Suppose a and b are real numbers (0<b<a). Prove if n is a positive integer then a^n-b^n?na^n-1(a-b)

All right. So we're looking at we've got this interval of negative infinity to negative one. And so if we go to a number line, I forgot my bracket. If we go to a number line, then and we didn't negative infinity to negative one, What we end up within is coming from here, and we go all the way to negative one, and we have a close circle. That's what that looks like on a number line. The second part of this is that we have negative for to infinity. And so what does that look like on a number line? Well, here's negative for And if we have a bracket, that means we have a closed circle and we're going to infinity now because this has this symbol, this an intersection symbol. We're looking for this intersection of these two, and that indeed is going to be negative for to negative one. Um, and so in this case, this is a true statement

Okay. So I need to prove that A. Is greater than zero and there exists and such that one over less than a. This and then so we're going to use the property where A. Is greater than zero and B. Is greater than zero. Then for some integer N. We have that in times A. This period can be yes, it's a comedian happen. So first we'll say let's let A equal one and be equals A. Then there this property must exist and then such that and one times one. Which is it was a be greater than a. Okay in other words this this um and one description A. So now and say well let's let a little A. And this baby is going to equal on it. Okay, so by the same property there must be some other some different yeah such that A. It's greater than one over and okay. And this comes from did kind of skip a step. Um that's derived from if we'd written it out as and to a. Being greater than one the bible side. Yeah to it this thing so now we're just gonna let N. N. So the N. B. The maximum of these two introduce then what we have is that end it's greater than or equal to N. One and and it's greater than or equal to and to the only what happens to be more such ad A. Such that which that A. Is less than and one is less than or equal to then. Okay that's going to follow up the first one and yeah one over N. It's just an equal to one over and two less than A. That follows from that second round. We did okay so we just got to combine these two pieces, plug them together we get that A. Is greater than one over N. But less than and Okay and so and that is our solution.

So we need to prove that if A B, C and D are real numbers that satisfy this condition, then we need to prove the following implication, which is that? The measure of the union of the two intervals is equal to this expression here. If and only if danger of all stars this joint. So every time you see if you don't live, uh you have to remember they have to prove that two directions, right? So I've put the two directions down here. So let's start with the first direction. The first direction says, we assume that The measure of the Union of the two intervals is equal at this. And if so then that must imply the intervals are this joint. So I know to proceed. I have made some assumptions that are pretty standard. Well, the first one is that for any to measurable sets A and B. The measure of their union is equal to this expression here. So that's by definition. Also by definition, the measure of an interval from A to B is equal to the b minus A. And the third fact is that intervals are measurable sets That have non zero measure. So they're measure is positive. So with these three fundamental assumptions or axioms, we will proceed with the proof of the first implication. So we assume this hypothesis and we need to show that this sets that the intervals are destroyed. Well, here we have the measure of a union of two sets. So we will apply axiom number one here, an axiom number. One says that. Well, a common B. Mhm. Union. See comedy. The measure of these has to be equal to the sum of the measures of the individual said so A B blood measure of C. The minus the measure of their intersection. A B intersect. See the but we already made the assumption that this is equal to b minus a plus d minus E. So keep that in mind. So bye bye. Axiom # two. Well, we know that the measure of an interval is just the the difference of their of the end point. So this will be the minus A. Bless the minus C minus this expression that you see here. Mhm. Okay. And then finally, we already made the assumption that this is equal to these expression here. So we can only make this step in one way. And that way is if this expression here is zero. So If that expression here is zero what does that mean? Well, that means that the intersection. So let's look at it here. It means that the intersection or this expression that you see here means that this intersection is an all set. But we already established that intervals are measurable sets with non zero measure. So they cannot. They set here cannot be and all said other than the empty set, which is what we wanted to prove. Okay, and then we prove the other direction of the implication. We assume that the sets are this joint than this has to be the case. And this one follows this one is simpler because we can simply compute this expression here in this way, this is all still true. But now they're telling us, we are assuming our hypothesis is that A, and B, and C and D. Those two intervals are this joint, so so if they are destroying that means that they're equal to the empty set, and we know that the measure of the empty set is zero, which means that this expression here has to be b minus A plus B minus C, which proves our conclusion.

Zero is less than B is less than a. Let's consider the function A four X. Is equal to expire in and is it and teacher great and equal to do now there is a theory called mean value to you. This function is a polynomial. It's continuous and differentiable. In an interval beat way. It's continuous enclosing to Albert way and differential. An opening to a gateway. Right? Very function is continuous in a closed into a gateway and differential and opening to a gateway then like green just mean value to them like ranges mentality and says that F a B minus ffb divided by a minus B is equal to a dash of something between be any. A very powerful freedom in calculus. Lagrange is humanitarian. Let's use this. So what is my fook? Fook will be a parent minus B. Baron divided by a minus B. Is equal to a dash of. See what is the derivative of this function? It is an ex power in minus one. This is the radio off function. Experiment by power. So what will be a flash of C. N. C. Bar and minus one? All right now you can see here sees between B. And so that means he's less than me. You see is less than a C. Bar and minus one is less than a bar and minus one. Multiplying and pull tight N. C. Bar and minus one is less than in a bar in minus one. So this quantity is less than in a par and minus one. So that would imply a par n minus B part and divided by a minus B is less than any power and minus one for all and greater than or equal to two for one. What will happen is it will be exactly equal because a pas one minus people. One divided by a minus B. It's compared with one paper, one minus one. This is one and 180 is one. So for one when it is one, this particular inequality is actually an equal. But when you increase one, that is when you take higher than 12345 all natural numbers starting with two. This inequality will hold two and we included by Lagrange is mean value.


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