 So, the question says which of the following rational numbers have terminating, terminating decimal representation, right? So, terminating decimal representation that means the decimal representation has to end, right? It cannot be recurring or even non-repeating, okay? So, what type of rational numbers have terminal decimal representation? So, you know that a rational number is of the form of p by q, correct? And now we also know that if q is equal to 2 to the power m and 5 to the power n. If any denominator is of this form, then we know that the decimal representation is going to be terminating, okay? So, any denominator is of this form, powers of 2 and powers of 5 only, right? What does it mean? Let's say for example, 56 by 80. This is the number, okay? Now 56 by 80 is 56 divided by, don't touch the numerator, only denominator, prime factorize it. So, you will see this is nothing but 2 to the power 4 into 5 to the power 1. The only prime factors which are appearing in the denominator are either 2 or 5, nothing else. So, it can't have 3, it can't have 7 or 11, 13, nothing. They are not allowed. If only 2 and 5 are there in the denominator, then that p by q will be having a terminating decimal representation. It doesn't matter what is there in the numerator, right? Only you need to check is denominator. So, check the denominator, check the denominator, okay? Check the denominator, okay? And then prime, this is first one. Second, prime factorize, prime factorize the denominator, okay? Prime factorize it, right? If only 2 and 5 are the factors or rather are the prime factors guys, it should be prime. So, prime factors, then, then p by q has a terminating, terminating decimal, decimal representation, okay? Otherwise not. Otherwise, otherwise, it will have, it will have repeating, repeating, right? Non-terminating, repeating and non-terminating, non-terminating decimal representation, clear? Decimal representation, this is the idea behind how to find out without actually dividing it. So, let's check the options guys. So, what all the options are? So, here is 5. Clearly, only 5 is there. So, either 2, 5 or both of them should be there. So, either of them is also okay. So, this clearly will give you the terminating decimal representation. Here, 13 is a prime number. So, 2 and 5 are not only there. So, not allowed. 27 is 3 to the power 3. So, this is also not going to give you terminating decimal. This is also 7, apart from 2 and 5. So, these all are ruled out. Only, first one is there and you already know, this is 0.8. You could have just done the, you know, division process and figured out the option, but this is how it works. So, for example, tomorrow, if there is a bigger number given, let's say 57, 572 divided by 800, let's say 50, no, 850 will not give you. So, 800, okay, 800 only. So, then this will give you terminating decimal representation, isn't it? Or for that matter, 2796 into 4096, okay, or 24, yeah, 4096. If this is there, because this is just a power of 2, right? It is, I believe, 2 to the power 12. So, hence, this will also be having terminating decimal representation, both of them understood. But the moment there is something like 756 and 81. Clearly, without even division, you can say that it will have recurring decimal representation, because there's a factor 3 here. It's not 81 is 3 to the power 4. There is no 2, 5. And it should be only 2 and 5, even if there is 2 into 5 and some other number, then also it will not work. Only 2 and 5 should be there in denominator, or some powers of 2, or some powers of 5 only. Okay, I hope you understood this. Let's go to the next one. So, here is the next question, guys. It says, how many rational numbers can be found between two distinct rational numbers? And all of you know the answer. There are infinitely many, right, infinitely many, right, between two distinct, distinct meaning unique, different. So, let's say this is one, A and this is another one, B. Let's say A and B are rational numbers. So, you know that there are infinitely many rational numbers between A and B on the real line, right? So, there is no brainer in it. There are infinitely many. So, what is the option? Option is clearly 4, right? So, how many rational numbers can be found between two distinct rational numbers? Infinitely many. So, answer is 4. Okay, let me lose. So, answer is 4. Okay, infinite, infinite. Clear? Let's go to the next question then. So, here is our next question, guys. It says the value of 2 plus root 3 and 2 minus root 3. It's nothing but you have to simply multiply it. It is a compound third 2 plus root 3, 2 minus root 3. You can just open the brackets and solve it. So, 2 plus root 3, use the distributive property and you will be able to do this. This is nothing but if I write the steps, write proper steps and in clear handwriting, you will never get errors. So, root 3, 2 minus root 3. Simply, yep. 2 plus 2 times 2 minus root 3 and root 3 times 2 minus root 3. So, now again use the distributive property. So, this will be 2 into 2. This is not decimal, guys. 2 into 2 minus 2 times root 3 plus root 3 times 2 minus root 3 square. This is one way of doing it, right? If you don't know the identities, there is another way. I'll tell you that also. So, if you see, this is getting cancelled out. These are same. 2 root 3 and 3 root 3 times 2 is same. So, cancelled out. So, what is left for minus root 3 square is 3 since 1, right? Alternatively, you could have directly used the identity a square minus b square. So, if you see, a minus b and a plus b results into a square minus b square. So, clearly you can see here 2 plus root 3 into 2 minus root 3. So, a plus b a minus b. So, it is nothing but 2 square minus root 3 square, which is 4 minus 3 is equal to 1. So, both ways it is coming out to be 1. So, answer is again a. Okay. Clear. So, let's go to the next question, guys. Next question. What is there in the next question? Now, the question here is 27 to the power minus 2 by root 3 is equal to what? Very easy question again. So, how to solve this one? So, 27 to the power minus 2 by root 2 by 3 can be written as 27 to the power 1 by 3 and this is minus 2. I can write this. Why? Write the reason also in the bracket. So, a to the power m to the power n is equal to a m n. Correct? So, hence I can write this, this expression like this, right? So, then this is equal to so, you can clearly see 27 cube root of 27 is 3 and now minus 2, 3 to the power minus 2 is nothing but 1 upon 3 square. Hence, it is 1 upon 9, right? How do I know that this is 1 upon 3 square? Then write the rule because a to the power minus n is equal to simply 1 upon a to the power n. Correct? So, using this 1 upon 9. So, hence my answer is 2 1 upon 1 upon 9. This is the right answer. Correct? It is always good to check, you know, if others are, you know, any possibility of others to be the right answer, but clearly you could do the calculations you figured out that 1 by 9 is the answer. Let us go to the next question. So, here is the next question it says, simplify this, simplify 3rd root of 2 and then 4th root of 3, right? So, what is this? This question is nothing but making the surge of the same order, okay? So, this will be written as like that, okay? So, 3rd root of 2 is 2 to the power 1 by 3, 4th root of 3 is 3 to the power 1 by 4. Now, how do I multiply? So, we will not be able to multiply until and unless we have similar, similar surge or not similar surge, we must have same order surge, same order surge. So, now here order is 3, here is order 4, I cannot multiply different order surge. So, let us try to make the same order. So, what should I do? I should take the LCM of 3 and 4. So, I am also writing the steps LCM of 3 and 4, which happens to be 12, okay? Now, let us do something so that 3 becomes 12. How do I do that? So, I can write this as multiply the numerator. So, 1 by 3, I have to make this numerator 12. So, I will multiply by 4 at this denominator 12. So, multiply by 4. So, similarly you have to multiply by 4 on the numerator also to compensate it. And then here next is 3. Again, I have to make this denominator of this power as 12. So, I have to multiply by 3. Similarly, you have to multiply by 3, okay? So, this operation can be done. Now, this is nothing but 2 to the power 4 upon 12. And this one is 3 to the power 3 upon 12. Okay? Now, 2 to the power 4 by 12 and 3 to the power 3 by 12. Now, the order looks same. Why? Because you can write this as 2 to the power 4, 1 upon 12. And this one, 3 to the power 3 and 1 upon 12. Am I right? Perfect. Now, what is 2 to the power 4? You all know this is 16. Do not make any mistakes in calculation. So, 2 to the power 4 is 16. 16 to the power 1 by 12. And this is 27 to the power 1 by 12, isn't it? 3 to the power 3 is 27. So, this will be nothing but 16 into 27 to the power 1 by 12. Okay? So, if you now multiply this, 16 into 27 is 112, isn't it? So, 112, 11 is the carry. So, 16 times 2 is 32 plus 11, 4, 32. So, 4, 32 to the power 1 by 12 is the answer. Or 12th root, 4, 32. This is the answer. Right? So, what do we learn in this question? So, whenever there are different order, order means what is the nth root you are trying to find out. So, in this case, order was 3. In this case, order is 4. So, take the LCM, make the same order and then do some compensation to show that the orders are same. So, that's what I did. And then take the common powers and do the, apply the exponential rules. What exponential rules we applied? a, b to the power n is a to the power n, b to the power n. Correct? And the second one which we used is a to the power m to the power n is equal to a to the power m, n. These are the two rules which we applied. Okay? So, let's go to the next question. So, here is another question which says find two rational numbers between half and 1 by 3. So, if you notice 1 by 3 is less than half. Okay? So, for our rough idea, if you see this is my 1 upon 3 and this is 1 upon 2. So, there are two methods to do. Actually, you know 1 by 3 is somewhere 0.33 bar. So, this is 1 by 3 and half is 0.5. So, any number between these will work. So, for example, even 0.4 will do, 0.4 is here and 0.0, 0.4 will be not there exactly. In fact, less than the midpoint. So, yeah. So, somewhere here is 0.4. Okay? So, 4 upon n. These are, these are one numbers and then you can find infinite numbers between these decimal numbers. But the formal way of doing it is let A is equal to 1 by 3 and B is equal to 1 by 2. It's always advisable to take smaller number first. So, one number, one number or one rational number basically between A and B is equal to simply A plus B upon 2. Okay? So, let's find out A is 1 by 3. B is half divided by 2. So, if you take the LCM in the numerator, 6. So, 2 plus 3 divided by 2. So, hence it is 5 upon 12. This is one number between 1 by 3 and 1 by 2. But they are asking 2. Is it a two rational numbers? So, what you can do is now the moment you know what is 5 upon 12. So, what you can do is find, take this number as another one and another set is this and then try to find out a number here. Okay? So, how to do it? So, now another number, another number between, between 1 by 3 and 5 by 12. So, whatever is the number between 1 by 3 and 5 by 12 will also be between 1 by 3 and 1 by 2. Isn't it? This number here is between 1 by 3 and 1 by 2. So, let's find this out. It's nothing but 1 by 3 plus 5 upon 12 upon 2. So, LCM of 3 and 12, clearly it is 12. So, 3 times 4. So, here it will be 4 plus simply 5 and divided by 2. Correct? So, this is nothing but 9 upon 12 into 2. Correct? Now, 3 times 3 and 3 times 4. Right? 3 times 3 and 3 times 4. So, this number is 3 upon 8. Isn't it? So, this is also between 1 by 3 and 1 by 2. You can also check this by long division method. What is 3 upon 8? So, it will be simply 3 and 8, 3 star 24. So, 60. So, 8, 7, 5, 0.375. Right? Clearly, 0.375 is between 0.33 and 0.5. So, this is to check. This is to check. And 5 upon 12 if you calculate by long division method, it will be 0.416 something something like that. You can go like that. So, this is also between 0.33 and 0.5. So, that means our answers are correct. Okay? Let's go to the next one. Now, here is another question which simply says multiply these two serves. Right? So, multiplication you know, you will apply distributive law. Not a big deal, not a difficult problem at all. Only thing is be careful with signs and multiplications. Right? So, apply distributive law. You will get 3 times 6 plus root 2. And write as many steps as possible. This will reduce your error percentage. Okay? So, 3 times 6 plus root 2 and minus root 5 times 6 plus root 2. Let's open it now. So, 3 times 6 is 18 plus 3 times root 2. Okay? And this one is 6 times root 5. Right? And minus root 10. 5 into 10 root means the order is same. So, you can simply multiply the numbers within the root 5 into 2, 10. Right? So, check the signs. There will be two negative signs here. Clear? There are two positive signs. Right? So, this is the final result. You don't need to go further. This is quite sufficient. If you want, you wish you can order them, put it in order, but not required. This is good enough. So, this is very, very simple question. And you can, if there are like terms, then do go for simplifying it. But there here, everything is different. Root 2 is different from root 5, root 10. You cannot really simplify it further. Okay? So, this is the answer. 18 plus 3 root 2 minus 6 root 5 minus root 10. Okay? Let's go to the next one. Here is a typical question of number system. Express 0.888888 repeating in the form of p by q. And you know the answer already how to approach such. In shortcut, I will tell you if a number is 0.8 bar. Okay? There is one digit repeating after decimal. And there is no non-repeating digits. Right? So, every number should be repeating. So, in this case, after decimal, 8 is repeating. Right? So, then this is directly written as 8 upon 9. Right? So, what we need to do is remove all the decimals and divide it by simply 1, 9. How many 9s will be there in the denominator? As many as the repeating digits are. Right? So, if it was 0.21 bar, it will be simply 21 upon 99. Okay? 0.31 bar is 31 upon 99. 0.413 bar is 413 by 999. Right? But if it is 0.012 bar, then you have to be careful. Then you can't write 1, 2 by 999. There is one fixed number and rest of them are repeating. So, hence the methodology will be little different here. But there is another elaborative way of doing it. And what is that? Let x be. So, this is the common method. Let x equals to 0.8 bar. Right? Then we multiply this by 10. So, 10x is equal to 8.8 bar. And why do we do this? Why? Because our next step is going to be a subtraction step. There, if the number after decimal is same, then subtraction becomes very easy. Isn't it? So, if I subtract. Right? So, 1 and 2, you write subtracting. Subtracting 1 from 2. Subtracting 1 from 2. What will I get? 10x minus x is equal to 8.8 bar minus 0.8 bar. So, hence 9x is equal to 8. So, x is equal to 8 upon 9. Right? This is in the form of p-work you always make sure. So, there is an error zone. So, make sure reduce the fraction or reduce the rational number to most simplified form. Most simplified form. Right? That means there should not be any common factor between numerator and denominator. Right? You should cancel all the common terms. Right? So, if let's say your answer is coming out to be, for example, in this case, 21 by 99 can be written as 7 upon 33. So, this should be the final answer. Reduced form. Okay? So, this is how you have to find out 0.8 bar or anything else in the form of p by q. Let's go to the next question now. So, in this question it is asking simplify. Simplify it should be. Simplify by rationalizing. So, sorry. It should be simplify. Simplify by rationalizing denominator. So, let's rationalize the denominator. So, you know how to do it. You have to multiply it by conjugate. So, this is my 7 plus 3 root 5 divided by 7 minus 3 root 5. Here what you need to do is multiply and divide the numerator and denominator by the conjugate of the denominator. So, it is asking denominator. So, I have to rationalize the denominator. If it asks to rationalize the numerator, then just do same things with the numerator. So, but be careful. Usually it is given denominator. So, there is a general tendency of all the students to ignore this term. But be careful. Error zone in this case would be. What is error zone in this case? Check what is to be rationalized. What is being asked to rationalize. Many people just blindly go and rationalize the denominator. But many a times there could be a trap laid for you where they are asking to rationalize the numerator. So, be careful. Okay. And how to rationalize? First of all, find the rationalizing factor. So, rationalizing factor in this case will be. Rationalizing factor is nothing but the conjugate of denominator. What is conjugate? You just change the sign of the irrational part. Conjugate of denominator. Correct. So, change the sign of the irrational part. So, 7 minus 3 root 5 will become 7 plus 3 plus 3 root 5. That will be the conjugate of it. So, let us write. This was the original number given 7 minus 3 root 5 and then multiply and divide by the conjugate. 7 plus 3 root 5. I am sorry. I have to write 3 root 5 like that. And in the numerator also same thing. 7 plus 3 root 5. Okay. So, then what happens? If you see this is nothing but 7 plus 3 root 5 whole squared divided by 7 square minus 3 root 5 whole squared. How? How? So, a square minus b square is equal to a minus b a plus b. This is what we have used here. Now, clearly top one in you know using the identity a plus b whole square. Open it. So, it will be 7 square plus 2 times 7 times 3 root 5. Right. Plus 3 root 5 squared. Correct. Divided by what? 7 square is 49. Minus 3 square. 9 times 45. 9 times 5 is 45. Okay. I hope this is clear. 3 root 5 whole square is 45. So, do not get confused. Now, this is 49 plus this will be 45 again. So, I am writing it here next to 49 because I have to add them. And this is 2 times 7, 14 times 342. Correct. So, 42 root 5 and divided by 4. Okay, guys. So, hence what is it now? Let us add. So, 49 plus 45 is 94. Right. 9 plus 5, 14, 14, 5, 49. Yes. And 42 root 5 divided by 4 we can cancel 1, 2 from numerator and denominator to get 47 plus 21 root 5 by 2. Right. So, 47 plus 21 root 5 by 2 is the answer. Okay. Let us go to the next question. So, here is the next question, guys. What is it? It says simplify this big expression. Okay. So, do not get intimidated by such things. These are very, very simple actually. So, in such cases, there are two ways. Either you can do this first, then this and then that or use the exponential law. So, hence, can't I say this as 625, right, to the power of 1 minus 1 by 2 times minus 1 by 4 times 2. All will be multiplied together. Why? Because a to the power m to the power n is simply a to the power mn. And then if you go for one more round, it will again get multiplied in the exponent. Okay. So, minus half minus 1 by 4 into 2, be careful with the signs. So, this is 625 and this 2 and this 2 will disappear and this is simply 1 upon 4. Right. Negative, negative will become positive and this 2 and this 2 will get cancelled. So, fourth root of 625, which is simply 5. Right. It was simple. Another way of doing it is this. You first do 625 to the power minus half. Okay. So, this will become 1 by 6 to the power half. First step is this, then minus 1 by 4 and then here there is a 2. Okay. So, now let's evaluate this first. So, square root of 625 is 25. So, hence, it is now 1 by 25 to the power minus 1 by 4. Okay. Squired. So, which is nothing but 1 by 25. And again, using this particular law, I can write minus 1 by 4 into 2, which is nothing but 25 to the power minus half 1 by 25. Sorry. So, it is nothing but 1 by 25 to the power minus 1 by 2. So, which is nothing but 25 to the power 1 by 2, which is nothing but 5 again. Very, very simple. Square root of 25 is 5. So, hence, this is how you should do it. So, let's go to the next one. Now, here is visualize 3.76 on the number line using successive magnification. Now, in most likelihood, it is not there in your syllabus this year. But for, you know, let's say if I have to do it, what to do? Visualize 3.76 on the number line using successive magnification. So, what you need to do is 3.76 is between 3 and 4. And it is somewhere on third quarter, right? So, 3.76. Now, what you need to do is go to 3.7. So, this is 3.7. And this is 3.8. Is it? So, this has to be magnified. And here, it will come like that. Okay. So, you magnify it. And here, it is 3.7. And at this point, it is 3.8. So, 3.76 is going to lie somewhere here. Okay. So, between 3 and 4, you magnified around 3.75. So, this is 3.7. This is 3.75 in the middle of it, right? So, where is 3.76? 3.76 is here. Okay. So, you have to divide it into 10 parts, let's say. So, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, like that. Okay. So, you have to use a ruler and scale. So, 1, 2, 3, 4, 5, 6. This is 3.76. Okay. So, this is how you have to do it. I have done very, very crudely here. So, you have to take a ruler, mark point like 3 and 4. And then here is something you draw a small circle. And again, draw like that. And in this line, you demarcate 10 equal points. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Let's say. So, let me, again, you have to do very, very. So, 10, this thing you draw. And this 3, you connect this 3 to the first one. And 4 to the last one. Okay. So, this is 3. Yep, 3. And, yeah, sorry. So, this one is 3.7. Sorry, my bad. I have to draw these, not these from here. Because this is being magnified. So, this is 3.7. And this one is 3.8. Like that. So, I am now magnifying between 3.7 and 3.8. So, this is 3.71, 72. This is 3.73. This is 3.74, 3.75, 3.76, 3.77. Like that. So, this number is what you are asking for. Okay. So, it's always advisable to label it A, B, C, D. So, this is 3.76 by successive magnification. Okay. Let's go to the next problem. So, here is another very interesting problem. We have to prove that this expression is equal to 1. Okay. So, let's try proving it. So, what is 1 upon? Let's start from LHS. So, you write LHS is equal to 1 upon 1 plus x to the power B minus A can be written as this. And this one will be x to the power C upon A. Right. Plus next one is 1 plus x to the power A divided by x to the power B. Sorry, this is x to the power A guys. Okay. Then, this one is x to the power C divided by x to the power B. And plus 1 upon 1 plus x to the power A by x to the power C plus x to the power B divided by x to the power C. And why am I writing this? Because A to the power M by A to the power N is simply A to the power M minus M. This is the reason. Exponential law. Now what? I can take LCM or common denominator. So, this x multiply and divide the numerator and denominator by x to the power A. Right. So, if you see, this is going to become x to the power A into 1 divided by x to the power A 1 plus x to the power B by x to the power A plus x to the power C by x raised to A. Why am I doing this? Why? Because you will see this denominator x to the power A will just get eliminated. So, hence, in this case, you multiply by whatever is the denominator here. So, x to the power B multiplied by 1. And this one will be x to the power B 1 plus x to the power A, x to the power B, x to the power C, x to the power B. Like that. Then again, the third term multiply it by x to the power C and divide also by x to the power C. So, that it is having no impact whatsoever in the result or in the value of the expression. Correct. But it is going to give us a very ordered expression. Okay. So, now let's do multiplication in the denominator. So, numerator becomes simply x to the power A and denominator will become x to the power A plus when this x A to the power A. Oh, sorry. Sorry, guys. This is extra power A. So, when this is also x to the power B here, sorry my bad. So, when this x to the power A and x to the power A gets multiplied, you will simply get x to the power B here. And similarly, you will get x to the power C. What about this one? x to the power B divided by x to the power B plus x to the power A plus x to the power C. And the final one is x to the power C divided by x to the power C plus x to the power A plus x to the power B. Now, if you look carefully, all the denominators are same. Their order is different, but they are same. So, hence I can take common denominator as x to the power A plus x to the power B plus x to the power C. And in the numerator, what will you get? x to the power A plus x to the power B plus x to the power C, isn't it? So, hence, if you see numerator and denominator is same, so this is simply one and which is what you needed to prove RHS. So, proving problems in a way are easy, why? Because you know the end result. Now, the catch was since it is an exponent form, you have to express this in this form B minus A and that was the catch. And then make the denominator common in all of them by simply multiplying by requisite factors or expressions here. So, x to the power A, x to the power A I multiplied. I multiplied purposefully with x to the power A. Why? Because the denominator is going to be cancelled. This x to the power A in the denominator's place, this will be cancelled. So, hence, and then it was also giving me a common denominator across the three terms and hence I could prove this problem. Okay. So, this is how you should be writing. So, what are the take aways? Take aways are write as many steps as possible, write as neatly as possible, wherever required write the reasons why it is happening. Okay. So, I keep saying that the more you make the life of an examiner easier, she will make your life easier. Okay. So, hence, try and explain every bit of it, try and show them, show the examiner your sincerity, your effort through your presentation. So, presentation also matters a lot. Okay, guys. So, see you in the next worksheet. Till then, bye-bye. Take care and all the best.