 We've looked at points and vectors and you've seen it's really easy to do and especially if you use computer language like the Wolfram language, but now it's time for some exercises so the plan really is to do these exercises after each set of lectures. So easy enough now we're going to look at points and vectors and some example problems but we're going to start off with a very special type of problem and this problem is going to be different from what you've seen before. You see a bunch of problems in an example, you get taught these different steps to solve these problems but this one is slightly different. It's just going to show you this tiny little step towards high mathematics, tiny tiny tiny little step but just how to start thinking slightly differently, how to view things as part of something bigger. Example problems, number one, prove that for all vectors u and v in Rn so they both vectors in the same space which is the only way that we can add vectors remember that the commutative property holds that u plus v equals v plus u. So why on earth do we start example problems where they prove? Well in some courses you will be required to do proofs and you'll have one or two proofs in your exam and I want you to think clearly about these. No, not everyone likes proofs and some of them are difficult but just doing some proofs here in linear algebra I think is a very nice way to do it. By the way, in abstract algebra which I'm going to veer into slightly, we have algebraic structures or mathematical structures that allow us to look closely at proofs and it's a very nice way to be introduced to proofs. I think a very nice way or better at least than getting introduced to proofs in something like real analysis. So let's have a look at how we construct a proof like this because it should seem self-evident but there's a few things that we have to discuss and talk about and it really is a tiny little step in the direction of higher mathematics. How do you start thinking and hide the mathematics? And a tiny little step into higher mathematics that we're going to take is this idea of abstraction. We're just going to abstract to something more general than the very specific examples that we grow up with. We can all add 3 plus 5 and that equals 8. That addition is just a very specific type of addition. There are other types of addition and we've seen that with vectors. Certainly we use the term addition in both of them but that's a very different type of addition. 3 plus 5 and vector u plus vector v, those are different. You physically have to do different things and they mean different things. So you've already started this idea that there is more than one way, more than one type of addition. So what we have to start with in thinking about all of this is with a definition. How did we define vector addition? So let's just have a look. Define vector addition. Well we define that if we have vector u and that vector, let me grab my ruler here that I'll put on the screen. I like to do that for some bizarre reason. And vector u, remember, that was an R in space, so it's going to be u1, u2, all the way to un. And since I do have the ruler, let's put the square brackets around this column vector. And we had vector v and vector v was v1, v2, all the way to vn. And that was in R in. R specifies for us that each of these elements are real numbers. So let's put that here as well, where, let's say, u sub i and v sub i are elements of the real numbers. So let me give myself some more space here. So we defined u plus v, u plus v, that was going to be u sub 1 plus v sub 1 and u sub 2 plus v sub 2 all the way down to u sub n plus v sub n. That is how we defined, that's how we defined vector addition. And the second thing we have to define is, and I'm going to write it out here, is how did we define that two vectors are the same vector? So let's say two vectors are the same vector, if and only if, so there's double ifs, u sub i equals v sub i. So we have this idea that u sub 1 equals v sub 1, and u sub 2 equals v sub 2, all the way down to u sub n equals v sub n, and they've got to be in R in space. And that's the exact same vector. So we're not in the same, we don't think about these things in linear algebra as we do in physics. So in physics we'll have this two dimensional space, and I'll have a vector there, and I'll have a vector there. And let's say for instance I drew them in the exact same direction and with the exact same magnitude, they can physically be applied in different areas on a plane or in three space, but for us we're talking about positional vectors here, so that we bring them back down with their tails to the origin, and then they are coincident, same direction, same length, and they are the same vector. So for us where we use vectors or we express vectors as this ordered tuple of numbers, these are exactly the same things. And now you can say well if they're all real numbers, of course if u plus v, that must be the same as v plus u, 3 plus 5 is 5 plus 3, so of course that is so. But we're taking for granted some of the properties are real numbers that we just grew up with, and somehow we have to formalize those, and there's a good reason to formalize these, so to generalize them. So I'm going to scroll all the way up here, and we're going to have a look at this. So in abstract algebra we have this idea of algebraic structures, mathematical structures, and there are the common ones, groups, you might have heard of those in group theory, we have rings, we have fields, that didn't come out well, fields. And why why do this? The idea is that if we just concentrate on fields for a moment, there are an infinite number of infinite fields, there are finite fields, and the beauty about them is if I have an example or member of a field and another member, one of them might be very difficult to study, but because they're both fields they're going to share some commonalities, and I might gain new insights into this new field. I might understand this new field better or be able to deal with it in an easier way or different way, because it is also a field. So let's have a look at this idea of this notion of an algebraic structure, and as I say we're veering off a little bit from linear algebra, but I think it's such a nice way to be introduced to this this more generalized thinking. So we have this notion of a set and we're not going to quibble about how we define a set, we're just going to use the mellow frankle set theory, everything is nice and neat. So let's just do that. So we have the set, a set of elements, and again there of course we're going to deal with numbers, but we want to get away from this idea that numbers are just there to quantify things. For us, for in many cases, they are just a quantification. We just use numbers as a quantification in as much as they can enumerate something for us, but numbers exist differently than that. We have a symbol for one, the number one, we write down one, but naturally it just quantifies something for us, but that symbol that we write down, it has a more general meaning. And so we have the set, and I suppose in the terms that we are familiar with we can think of the rational numbers, we can think of real numbers, those are sets of elements. And then we're also going to be dealing with binary operations. So let's do the binary operations there. Binary operations between ordered pairs of a set. So ordered pairs is I take any two, so if I think of the real numbers I take three and five, that's different from five and three, there's some order to those. And I apply a binary operation to them, and a binary operation means I combine them in some way. So if I bake a cake I would add some ingredients and we bake the cake. So that's some way of combining these things. And we're going to think only of two operations, and that's all we really need to do, and that's the binary operations of addition there and of multiplication. And again I urge you to get away from this notion of addition and multiplication in the very specific instance we did with real numbers or integers that we grow up with. It is much more general than that. Again you've already seen that with vector addition. If we add two vectors that's a very different way from adding two real numbers, it's a different process. So it's a much more general thing, so much so that we're going to use different symbols. So for addition we're going to use this symbol with plus with a circle around it, and for multiplication a dot with a circle. Just to signify the fact that we're talking about some general way of adding things and some general way of multiplying things, not the same as we did with scalars, with just real numbers. So we're going to have this idea of closure. You see closure there, and you're going to see closure there again. And the eleven steps that you see here are the axioms of fields. Axioms or properties are something that we decided on. We said these are the things that we put down, and if you bring me a set of elements and you define two binary operations on them called addition and multiplication, and they obey these axioms, then they are a member of this algebraic structure. Now there's a better way to define that we define fields in, and that's by using groups in the definition, but I wanted to write all these out just so that we can get familiar with this. And you see the two closures there. Not all textbooks will have that in because some would say that the binary operation of addition and multiplication, they are operations that map the set to itself. So what we're going to do with fields, we're going to call our set F. I think most people would use F. And these two binary operations map F to F. In other words, if I combine this ordered pair, this arbitrary pair of elements, then the result that I do get is also an element of that set. So some define it that way, or if we write it out explicitly, there'll be closure. So closure under addition would mean the following. So I would say for all a, b elements of F. So I'm taking any two arbitrary elements in F, this field. And because it's two arbitrary ones, it means all of them. It doesn't matter which two ones I take. So if all a be an element of F, it follows that. And I'm going to use this right arrow for follows that. A, and this idea of combining them under addition is also going to be an element of F. I don't get outside of it. I don't get a result that's not in F. And as I say, you can clearly see there if we defined initially the two binary operations as a mapping of F to F, I wouldn't have to put this closure. But here for closure under multiplication, it's the same thing. For all a, b elements of F, it follows that a and its binary operation of multiplication B is also going to be an element of F. So don't get out of the element of F. Then we have this commutativity. So that would be simple for all a, b elements of F. It has to follow that a and its binary operation with B, that must be equal to B, binary operation of addition with A. Any two elements that I take. And the same here with the commutative property under multiplication or axiom of commutativity under multiplication and seven here, we're going to say for all a, b elements in F, it's going to follow that a and this binary operation of multiplication with B is the same as B with A. Exactly the same thing. We're also going to have the associative axiom that we have if we have three elements a, b and c elements of this field, it's going to follow that if I do this binary operation first, and then the result of that with C, that's going to be the exactly the same as doing B and C first. Same here with multiplication. So if I have a, b and c elements of the field, it is going to follow that a and b. If I then do c, that's going to be the same as doing a and then b with C. And then we have this idea of an additive identity. So we're going to say not for all, we're going to say there exists. So this backwards e there exists. I'm going to say e element of F. And so that's a very specific one, there exists an e element of F, such that we use this straight vertical line for such that for all a element of F, it follows that e, this binary operation of addition with a is going to equal a. And if you think about the real numbers, that would just be zero, for instance, but in vectors, that would be the zero vector. And those are two different things. And that's why we use this plus for the circle around because it's this general idea of addition. So there exists this e element of F such that for all a element of F, it follows that e plus this binary operation of addition a equals a. So it doesn't change that element in any way. And I could also suppose right here that we have a and e, but we don't usually write that because we already have commutativity right there. We already define this, we already have this axiom of commutativity. So I don't have to really do that there, but you can if you want. And then there's this unique multiplicative identity. So we're going to say there exists, let's write it this way around, there exists. And let's make it z element of F, such that for all a element of F, it follows that z and then binary operation of multiplication with a, that is just going to leave me with a. So in the case of real numbers, that's going to be one, of course, we multiply one by any number, any real number, we're going to get one. And then we have this idea of an additive inverse here, number five, an additive inverse for every element. So we're going to say, for all a element of F, there exists a minus a element of F. Now hesitate to put that minus because that immediately just leads us back to real numbers. And that's not what this additive inverse means, because the additive inverse of a vector is going to be something different than the additive inverse of real number. But anyway, for all a elements of F, there exists an a inverse, this inverse additive inverse of F, such that if I do this with the additive inverse, I'm just going to get the additive identity. Then we also have this idea of the unique multiplicative inverse for every element. So we're going to say there exists, there exists this, and we're going to call it a like that, element of F, such that for all a element of F, it's going to follow that if I have this a and multiply by its inverse, I'm going to get z back, the multiplicative identity. And again, I hesitate to use this a with a power negative one because it just reminds us just of real numbers, but other types of sets with these binary operations will also have their own unique inverse. And the inverse of a matrix is not the same as, we can talk about matrices, of course, it's not the same as the inverse of a real number. But we have this notion that if we do this binary operation of multiplication, that we get back the multiplicative identity, which is one. And in some cases, if we look say just at the real numbers, of course, we've got to exclude one of the elements. We've got to exclude zero in as much as we cannot divide by zero. So zero doesn't have a multiplicative inverse. And then we have this distributive property of multiplication over addition. And this says for all a, b and c elements of F, we have it follows that if we have this notion of a, and then b plus c, so this will be lift, the lift distributive property. And that is just going to equal, that's a multiplied by b, and then a plus a plus a multiplied by c. And you can also have it on the right side. So we're going to say that these are the axioms, these are the axioms of fields. And now we can have a look and say, well, can you give me a set? And you please define how you add and multiply these things. And let's see if that's a field. So the one obvious one that we're going to start off with is the real numbers. The real numbers are a set of elements, they're all unique, infinite number of them. And we have to, we have to find the binary operations of addition on them, and of multiplication on them. That's something we naturally learn even before school how to add and multiply numbers. So would they, do they obey these properties? And by the way that we've set them up, if you set them up as something that enumerates something, if you've defined them in that way, of course, there are some people who feel there is a problem with that, because if you have an infinite decimal point, can you really define how addition is going to work? Because you might have to carry over an element to the element, if you have elements right towards infinity there, that's going to affect all sorts of numbers as you work, work your way from addition from right to left as we do addition. But let's just not think of that for the moment. We have this idea of adding real numbers, and we have this idea of multiplying real numbers, we've defined them, and if you run through all the properties here, you see that they obey all of these 11 axioms. So we can say that the reals, they really are, they are a field. So we can show that they obey all these 11 properties and they are a field. Now, after all of that thinking, and you can think, can you find other common sets and how we define addition and multiplication on them if they are a field as well? The rational number sure would be a field, but the integers definitely are not a field. We don't have multiplicative inverses for those, so the integers are not a field. So now we get back to our first example problem here, we want to prove that for all vectors u and v in Rn, that u plus v equals v plus u. So these are vectors over the field of real numbers. So let me write that down, because that's important, that's an important way to talk about it. These are vectors over the field of real numbers. That means the elements of which they are made up inherit the properties of the specific field called the real numbers. And now we can say that that's how we define that, how do we define v plus u? So v plus u, well that's going to be v1 plus u1, v2 plus u2, all the way down to vn plus un. That's how we define them. And because we've defined, oh that was loud, because we've defined how two vectors are similar, we have to have this notion of u1 plus v1. Let's make it ui plus vi just to generalize over all of them, must be equal to vi plus ui. But because this is a vector over the field of real numbers, we inherit all the properties of a field and commutativity is one of them. In other words, these two things are equal to each other. And finally, our thought processes have allowed us to prove the fact that there's commutativity with the vectors over a field of real numbers. It's really as simple as that. So if you have to do any of the things, the vector arithmetic that we've done, if you have, if in your exam you have to prove that those properties hold, think yourself these are vectors over field of real numbers, might be dealing with complex numbers in your course. But this whole notion of they are part of a mathematical structure of a algebraic structure. And that means they have certain properties. And now we can make use of those properties in our proof. And it really is as simple as that. Problem number two, we asked to find x, y and z if the following two vectors are equal to each other. We've already seen how we define when vectors are equal to each other. So we have the fact that for these two vectors, if they were u and v, that ui must equal vi. For all the i's, one, two, three in this instance, it's an r3. So we're going to have the fact that x must equal four. Two must equal y plus one. And if two equals y plus one, from that follows that y equals one. Simple algebra there. And then z equals x minus one. And if z equals x minus one, that means z equals, we know what x is, it's four. That's four minus one, which is three. So we have our solutions there. x equals four, y equals one, and z equals three. Clearly because of the way that we define equality between two vectors. Let's do the following problem. We asked to add these two vectors, vector one four with elements one four seven and minus one four eight. And how we define vector addition, that's just going to be adding the specific elements. We have one plus minus one, that equals zero, four plus zero is four, zero plus four is also four, and seven plus eight is 15. So we see the result there. So what I want to do here is just bring this, I've got a second monitor here. I just want to bring in the Wolfram language for us there. This is increase the size a little bit. And there's our problem there. This was problem number five, add these two vectors. And what we can do in the Wolfram language is just to check. Remember, we represent vectors as lists. Lists go inside of curly braces. And each element is separated by a comma. And then I'm just adding these two. And the result, I want it displayed as a column vector. And this is what this function does. It has nothing to do with how the arithmetic is taking place. That's a function that will just determine how things are displayed to the screen. And because it is a function, it has a set of opening and closing square brackets, and the arguments go inside there. And if we run this, we see the result that we saw before, zero, four, four, 15. So it's easy in the Wolfram language just to check whether we are doing the right thing. We see question six here. Add these two vectors. Of course, that answer is not defined. Because this vector is, the first vector is in four space, and the second vector is in three space. And we cannot add, we don't have a definition for the addition of those. And we can't do that. Let's look here at this problem. Calculate the scalar vector multiplication in seven, plot and explain what the scalar multiplication does to the vector. So I'm just going to move it away. There we have problem number four. This is scalar vector multiplication. So we know how we defined scalar vector multiplication. The way that we defined it is if we have some scalar c, and we multiply it by u sub one, u sub two, all the way down to u sub n. Trying to keep these square brackets going with my ruler here. The way that we defined that was c times u sub, u one c, u sub two, all the way down to c, u sub n. And this is again over the field of real numbers. So we inherit the property of the multiplication of real numbers. So that makes this very easy problem. We're going to have three, negative three, and three. And we asked in the Wolfram language problem, let me just bring that up for you. So what does that mean for the physicality of vectors? If I just bring this in from the left hand side, we've got this three times and in the Wolfram language, if I leave a space in between there, that's the same as multiplication. I could also do a star. It's also multiplication, but it's easy just to put a little space and that means multiplication. And then we're just going to draw that. So we're going to use the graphics 3d. We've got the error game from zero zero zero, the origin to one negative one and one. And then from the origin to three, negative three, three. And if we run that again, we're using the matrix form function there, just to print this to the screen, we can check our result. Of course, it's correct. And then if we were to just plot this, we see the difference there. What does vector scalar multiplication, scalar vector multiplication do? The little red one there in the beginning, that's our first one. And then this one is just three times as long. So three times as long as the original was. And that is what the scalar multiplication is going to do. It's going to increase the magnitude of our vector. Let's do number five. Next, we ask to find three times u minus two times v plus two times w as vectors. And we see the vectors there in three space. And this problem, it is vector, we're just going to do vector, a scalar vector multiplication. And then we're going to use vector addition. So three times u is just going to be three times one, two, three. This ruler, as I said before, this is a physical ruler. And to that, we're going to add, we're going to add this idea of negative two, we're going to add that to four, two, two. And then we're going to add twice, one negative one, two. And doing this the long way around, that's going to be three, that's going to be six, that's going to be nine. And to that, we're going to add negative eight, negative four, negative four. And to that, we're going to add two minus two and four, as I just multiply this out by each other. Remember, that's how we define that. Scalar vector multiplication. And then it's just going to be going to be component wise addition. So three plus minus eight minus five or negative five plus two, that's going to be a negative three. And we're going to have six minus four is two, minus another two is zero, and nine minus four is five, plus another four is nine. And the only problem with these things, of course, is making very simple little arithmetical errors. And that creeps in all the time. So if you do these, check yourself two at a second time. Better yet, let's drag in the warframe language. Remember, you can do all of this free of charge. If you just open a warframe cloud account online, just start a new notebook. And there we go. Three times one, two, three, minus two times four, two, two, plus two times one, negative one and two. And we're going to wrap that inside the matrix form. So we see our result. And we see, and check ourselves, negative three, zero, nine, negative three, zero, nine. Absolutely correct. So here's our next problem. We asked to write w as a linear combination of u and v. A linear combination is exactly what we saw up here. Let's just do this in gene. Let's define this as a linear combination. So we're combining a couple of vectors, and we're doing it in such a way that it's linear. In other words, each of them are just multiplied by the scalar. So we need this idea of something times u. So let's write this down. Let's back to blue. We're going to have some scalar x, element of the real numbers, times one and two. And that is going to be to that we're going to add a scalar multiple of one and three. And that is going to have to give us five and 12. So let's just do this. And this is going to become so critical in the section where we start talking about matrices, this idea of linear combination. Now what you see here are actually, it's actually a way to write two equations, because what we have here is x and 2x, and we add to that y and 3y, and that's going to equal five and 12. So it's this idea of these vectors. But remember this is how we defined addition, that was just component y's. So we have to have the fact that x plus y equals five, and 2x plus 3y is going to equal 12. So we have, it's a way to represent a linear system, a system of linear equations. I have two equations, and I can look at that, I can write them as the addition of vectors. That is a very interesting thing. So from this it follows that y, I can write y as five minus x, and from this it then follows that I can write this as 2x plus three times five, three times five minus x, and that's going to equal 12. So 2x plus 15 minus 3x is going to equal 12. In other words minus x is going to equal negative three. So x equals three, and that means y equals five minus three, and that is two. So I have this linear combination, I can say x is three. So three times u plus two times y equals a w. I can write w as a linear combination of two other vectors, but the beauty of all of this is that I can use the vectors as we've done here to solve the linear systems, and we're going to have quite a good look at that. Problem number seven, compute the dot product of these two vectors in, there are two vectors in three space, and remember how we do this, the dot product, let's call this one u, and let's call this one v, then if we have the dot product of u and v, that is just going to be the sum, and this instance i equals one to three, that's three of u, i, v, i. So I just have to multiply each of them and do that addition. So we're going to have two multiplied by negative one, so that leaves us with a negative two, to that we add the following, four times two, that's eight, to that we add the following, four times two is eight, eight and eight is 16, 16 minus two is 14, and there we go, we've gone from the definition, I like to have definitions here in green, so we went from the definition of the dot product or inner product or scalar product, and we can just easily do this multiplication, so it's element wise multiplication, and then we just add all of those elements to each other, so there's my first element, my second element, my third, and we just add those and we get to the result of 14. So let's bring in the Wolfram language again, we like to check that we've done our example problems correctly, so there was our problem, compute the dot product between the vectors in 11, 2, 4, 4, negative 1, 2 and 2, so let's just squish this up a little bit, so it's 2, 4, 4, negative 1, 2, 2, yes, we've got the same problem there, and I'm going to use the dot function, and it's a function, so it goes inside of a set of square brackets, and I list my two vectors there, remember vectors are lists of elements, lists go inside of curly braces, so I've got those two there, and I pass all of this, the dot product function with its opening and closing square brackets, I pass that to the matrix form function as an argument, so the matrix form function also has its set of, you see the green there, the open and closing, its open and closing set of square brackets, and let's execute that in lo and behold, the answer is 14, easy enough for us to check our results. It's a good problem, find u and v, u plus v multiplied by w as vectors, and we ask is this the same as u multiplied by w plus v multiplied by w, well we've seen the distributive property, we know that it holds, let's just use it as an example, just to get some exercises on our bell, so these things just become second nature, so u plus v, we're going to do that first, that's component wise, so the one plus the two, that equals three, and two plus and negative one, the two disappeared, they are me, two plus negative one, this is one, and then three plus negative two, that's also one, so I have my first vector there as my addition, and then we're going to do the dot product of that with one, one, two, so what are we going to get, we're going to get a scalar, so this becomes component wise multiplication, so those two are multiplied by each other, three times one, that is the three, and then one times one, that's this one, so we add the one, and the last one we have one times two, you multiply that that's two, so we get here three plus one is four plus two is six, so we get a solution there of six, let's do it the other way around, let's do u multiplied by w, so in this instance we're going to get one, two, and three as our first vector here, and we're going to do the dot product with w, and w is one, one, two, so let's do that one on the side here, so that's going to be one times one is one, plus two times one is two, and three times two is six, so I have that on this side, let's add to that v and w, so that's two minus one minus two, and one, one, two, those are my vectors that say the only way to make a mistake here is this silly arithmetical errors, and they happen all the time, don't obviously feel bad about those, so again it's two times one, and that's two, we're going to have here and another one, we have negative one times one, so that's minus one there, and we have negative two times two, so we're going to add to that a negative four, so two minus one is one, one minus three is negative three, so on this side I have a negative three, and on this side I have still the one plus two plus six, so one plus two three plus six nine, and nine minus three is six, and lo and behold the two are exactly the same, because we know that this distributive property is going to hold by the way that we've defined these things, we can go back to say that this is over the field of reals, and if you had to prove this it would be quite simple, you're going to write out the definition of how we define addition, and then multiplication, vector addition, vector multiplication, we're going to say it's over the field of reals, so we inherit all the properties of the field of real numbers, and lo and behold the proof would be quite easy. In the next problem we're asked to calculate the norm of u being one two three, the vector one two column vector one two three, and we also have to calculate the square root of u dot u, so let's do the norm of that vector, remember how we define that, let's just write it out in green, we have this definition of the norm of a vector, that's going to be the square root of this sum of i equals one to n, how many they are, and it's just going to be ui squared, so I just square all of those values, and you can clearly see why they asked us to do this one on this side, because these two things are exactly the same, the dot product squares, if it's the same thing it's just going to square all those values, and here we're going to square because the norm, the norm of vector u is going to be the square root of, we just square all these things, one squared plus two squared plus three squared, so if we do that that's just the square root of one plus four plus nine, and that's the square root of, let's put it here, the square root of fourteen, but what happens if we do u dot u, u dot u with itself, so that's going to be one two three and one two three, let's make some noise with the ruler, the dot product of that, and it's one times one is one, one times one is one, and two times two is four, and three times three is nine, and that's fourteen, but we're not asked to do that, we're asked to get the square root of u dot u, so that's going to be the square root of fourteen, these two things are exactly the same, because that is another way to calculate the norm, there's another definition for the norm of a vector, so let's write that in green here, so we have this vector u, and we're taking its norm, we're taking its norm, that's exactly that's that, so that's exactly the same as this, it's the square root of u dot u, the cross product of the vector with itself, the dot, listen to me cross product, the dot product of the vector with itself, next problem we have to normalize u or express it as a unit vector, this would be the same as, let's say express u as u hat, so what do we have to do, how do we define, let's say seven in green, so we remember that u hat is this u divided by the norm of u, so we're just going to divide each component by the length, the magnitude, so let's do that, what is the norm of u, well we've just done that just now, we saw that u, the norm of u was the square root of fourteen, so u hat is just going to be one divided by square root fourteen, I mean two divided by square root fourteen and three divided by square root fourteen, remember the way that we set that up, the norm of this unit vector is going to be just one, if I square all of those add or take the square root I'm just going to get one, in some cases remember you're asked just to express it in a different way, so this is going to be square root fourteen over fourteen and so you just multiply the numerator and denominator by the denominator, so this is going to be two square root fourteen, that's ugly over square root, over fourteen itself, over fourteen and then three square root of fourteen divided by fourteen and we can just clean that up as the square root of fourteen over fourteen and this is going to be the square root of fourteen over, we can divide fourteen by two and that gives us seven and then this is going to remain exactly the same fourteen over there, so you might just be asked just to clean things up and not to have square roots in your denominator but those things are exactly the same, you can just leave it as the first one, let's bring in the Wolfram language and we can see the the matrix form of this vector divided by its norm and we can see the results, they are exactly the same except that the second one is expressed as the square root of two over seven and if I just take that away remember if we have two square root of fourteen that will be exactly the same as the square root of two over the square root of fourteen that's the square root of four over fourteen and that's the square root of two over seven, so there's various ways that we can express each one of these real numbers no problem