This is a problem that commonly trips students when evaluating definite integrals. The bone of contention is what to do with the square-root of a square.
A great addition to this discussion would have been to graph y=sqrt(1-cos^2 x) and show that on the interval [0,2pi] the area bounded by the graph and the x-axis is definitely going to be a positive number. That way, when you evaluate the integral after doing it incorrectly (i.e. without the absolute value) and get 0 it's perfectly clear that something's gone wrong with your working. I know that effectively that's what you did when you graphed y=abs(sin x), but I think that graphing it in it's original form would have been better and would have shown why you get abs(sin x) instead of just sin x. Regardless, great video.
Muchas gracias por compartir tan interesante ejercicios de integrales definidas.😊. Mi hija y mi persona estamos muy agradecidos por la buena y clara explicación. 😊❤😊.
@2:20: I think you may have misspoke. You said "that only happens if this x is not a variable", meaning that sqr(x^2) = x. I think you meant to say that it only happens if x is never negative. It doesn't work for the constant -9, but does work for the function f(x) = sin(x) + 1. Sqr(f^2(x)) = f(x) since abs(f(x)) = f(x).
No , the point here is that since x is a variable, we may not tacitly presume that x is positive; x may be negative (and still produce a positive value x^2 under the square root as if x were positive). And that problem doesn't occur when the value under the square root is a constant: sqrt(9) = 3 , sqrt(-9) = 3i , never do we get something that is negative which must be converted into someting positive in order to get it right.
@yurenchu I would politely disagree. If x was the constant -9, x^2 would be 81 and the square root of 81 is 9, not the original x of -9, but rather abs(-9). It's the sign of x that matters, sqr(x^2) = x if and only if abs(x) = x. Whether x is constant or varying is inconsequential so long as for all values of x, abs(x) = x.
@@KMcCabe The problem that he wants to point out is that nobody makes the mistake of saying sqrt((-9)^2) = -9 (except for the few who confidently maintain that a square root can be negative; but that's a different story), but plenty do reduce sqrt(x^2) automatically and blindly to x because they overlook that x is a variable that may be negative. Furthermore, I think he is comparing the case of sqrt(x^2) versus the case of sqrt(81), not versus sqrt((-9)^2) . 81 is readily recognized as a square, hence we can automatically and easily reduce the square root expression sqrt(81) . x^2 is also readily recognized as a square, but we must be careful when reducing the square root expression sqrt(x^2), because there's a little hidden trickyness involved since x is a variable. I think that's what he's trying to say..
@@yurenchu I agree that the sqr((-9)^2) not equalling -9 is a trivial example. I get your point that it should be obvious when a constant number is involved. I've watched it a few times and what he said was that it only fails when it's a variable. My point is that he may have misspoke. It can fail to be true for a constant number and it can hold true for a variable function. I've given examples of both. The idea that you can just cancel an even exponent and root only holds when the base is non-negative. I think we can all agree with that. 🤓
The square root of x^2 is well defined for all values of x, but it has a subtle behavior when x is negative. If, for example, x = -3, then sqrt((-3)^2) = sqrt(9) = 3, so sqrt(x^2) is NOT equal to x if x = -3. Instead, if x = -a is negative (where a is positive), then sqrt(x^2) = sqrt((-a)^2) = sqrt(a^2) = a = -x = |x|. Thus, sqrt(x^2) = |x| is valid if x is any real number, positive or negative or 0.
You're great! Keep on with the videos! You are doing it with heart, and a good portion of pedagogic gift. You didn't really explain people why in case of a variable the sqrt of a square is the abs value. Because it can represent a negative number as well, and in that case the answer is positive at the end. You just mentioned that if it is a number it is always a positive.
As an engineering student, we were taught that integrals have an area with sign, so you can just integrate sinx (without abs) from 0 to pi, and then from pi to 2 pi, but changing signs. At the end it is the same but you showed a more “mathematical approach” to the problem, we just change the sign 😂
Since we know the integral goes from 0 to 2 pi which is a full sine, and that the absolute rectifies the sine graph, the two "humps" you're trying to find the area of are going to be the same. So could you just do 2 * integral from 0 to pi of sinx instead of splitting it into two integrals?
That's a shortcut that we can take (in this particular case), but he's showing us the basic proper way of how to handle problems like this, which we can rely on even when the graph doesn't show such convenient symmetries (for example, if instead of |sin(x)| the integrand were |sin(x) + ½| , or something like that).
@@yurenchu Right that does makes sense, use a simple example to explain a more complex solution so it's easier to demonstrate? In the wild however if presented with this particular problem, I think most would take the shortcut and give reasoning for it.
@@spudhead169 Yes, you are free to do so. But he wouldn't be a good teacher if he didn't show us how his method applies even when we don't use convenient shortcuts. I really don't understand why you are asking the question in your opening comment. It's clear that the two "humps" are equal, did you really need confirmation that in this case you could just multiply the integral from 0 to pi by the factor 2? Furthermore, it's not really " _instead_ of splitting it into two integrals", you too are essentially splitting the integral into two integrals, but you're immediately recognizing that the two integrals are equal and hence you take the shortcut of multiplying one integral by 2.
I knew it was absolute value but I was doing the math in my head and the negatives annoyed me LOL. But I did caught the square root of sin^2x was absolute value of sinx. Nice problem (Was doing this on my way to work - hence no paper lol)
I at first thought it would be zero, but then I remembered that the radical means that the graph would be forced above the x-axis. It's the same concept we use when finding the positive area between the x-axis and the curve.
At the beginning I thought you was wrong before I figured that we go from 0 to 2pi ! Thanks well done Another way consist using another variable such that : t = cosx !
Challenged by the thumbnail title: I came up with the answer 4 after roughly 45 seconds of integrating in my head. Seemed so easy, hope I got it right. 😂
Suppose we have an ANALOG voltmeter that measures a voltage of the form abs (sin x), for example u = U abs (sin x). If the frequency is high enough that the voltmeter needle cannot follow the voltage variations, can you tell what voltage the voltmeter will indicate ? This is a math problem and it is very interesting.
The question is faulty, though. 0 to 2pi is a complete revolution. Which is equal 0 to 0. Definite integrals with the same lower and upper limit should be 0, no?
I did 0 in my head at first, then realized my mistake and switched to 4. I think my problem was doing it in my head. If I had written it out, I might have caught it quicker.
I always hated the sqrt function. As per defination of function each element in domain has to have only unique mapping to the codomian But sqrt has 2 mappings. So sqrt is not a function, but everyone treats it as 1 by forcing the defination of sqrt to have a codomian of only positive real numbers. My issue is it's never mentioned what is the defination of sqrt and that annoys me to no end. I have seen exams where it's treated as a relationship in 1 problem and as a function in the next. Exams needs to understand that they have to give the exact defination before using such "functions"
The radical is always meant as the principal root for real numbers. For complex numbers, it does represent all solutions. But √4 is always 2, it is never negative -2 and if it is written as such, your teacher doesn't know what they're doing.
You guys are right, but here in India, to make problems harder they sometimes break rules or probably forget the rules. I have seena few times, where students were awarded marks because the question took negative values for sqrt. I guess it's not such a big issue in other countries.
Make a video showing all integers solutions for n which | (3n+16) / (4-n) | = k² | k € Q I made this question by myself and the solutions for n are n = -10, -5, -3, 0, 3, 32 I even put the equation on a C compiler and these are the solutions fr
Cool math puzzle! But I think those are not all solutions. What about n = 452 ? EDIT: Also n = 6304 and n = -1690 . I don't have proof, but i have a feeling that this problem has infinitely many solutions.
@@thomazsoares1316 I'm still intrigued by this puzzle. I also found: n = -122 n = -23530 n = 87812 It appears there are three sets of solutions: One finite set of solutions where (3n+16)/(4-n) > 0 : n ∈ {-5 , -3 , 0 , 3} Two infinite sets of solutions where (3n+16)/(4-n) < 0 : First define integer sequences tⱼ and sⱼ , as follows: t₀ = 0 t₁ = 1 tₘ₊₂ = 4*tₘ₊₁ - tₘ , for any non-negative integer m sₘ₊₁ = tₘ₊₁ - tₘ , for any non-negative integer m Then n ∈ { 4 + 28*(tⱼ)² | index j is a positive integer } ==> n ∈ {32 , 452 , 6304 , 87812 , 1223072 , ...} and n ∈ { 4 - 14*(sⱼ)² | index j is a positive integer } ==> n ∈ {-10 , -122 , -1690 , -23530 , -327722 , ...} appear to be solutions to the equation. I have confidence that each of these values is indeed a solution, and also that these include _all_ possible solutions (in other words, there are no other solutions); although I don't have mathematical proof for either claim.
@@thomazsoares1316 I'm still intrigued by this puzzle. I also found: n = -122 n = -23530 n = 87812 It appears there are three sets of solutions: V₀ = {-5 , -3 , 0 , 3} V₁ = { 4 + 28*(tⱼ)² | index j is a positive integer } where t₀ = 0 , t₁ = 1 , and tₘ₊₂ = 4*tₘ₊₁ - tₘ (for any non-negative integer m) ==> V₁ = {32 , 452 , 6304 , 87812 , 1223072 , ...} V₂ = { 4 - 14*(sⱼ)² | index j is a positive integer } where sₘ₊₁ = tₘ₊₁ - tₘ (for any non-negative integer m) ==> V₂ = {-10 , -122 , -1690 , -23530 , -327722 , ...} So n ∈ { union of (V₀ , V₁ , V₂) }. I could be wrong, but I think these are all possible solutions.
@@thomazsoares1316 I think I have solved your math puzzle and I have posted two replies in this thread with the possible values of n , but for some reason youtube hides those replies unless you have the comment section sorted in "Newest first" order (instead of "Most popular" order).
He does have to do two integrals because he is teaching you the proper method of how to handle problems like this, which works even when the integrand doesn't have symmetry properties like it does here. You may choose to take a shortcut later, but at least he has done his job as a teacher.
Me: "oh, sure; the symmetry of the sine curve means you can equate the absolute value of sine to twice the output of the integral of sinx from 0 to π." Newton: do your absolute values properly; stop cutting corners
Never. -5 can be called _a_ second root or square root of 25, but not _the_ square root. When you use the sqrt() function or the radical symbol, it's *always* the non-negative root that's assumed. Note the tiny but important differences in notion.
"the beauitful thing is that you generated an answer incorrect" lmao
yikes. can't believe i forgot that lol. incredible how the result is transformed from 4 to 0 from such a simple error
This guy has impeccable blackboard etiquette. So neat and easy to read.
Never stop learning. Those who stopped learning have stopped living.
I keep seeing these when I go to listen to ASMR to go to sleep, ITS TOO INTERESTING EVERY NIGHT BEFORE BED I WATCH ONE OF THE VIDEOS......
This channel is high on my list of interesting enough to watch while awake and soothing enough to fall asleep to.
4;23 that is the most beautiful sine graph ive ever seen
Finally, one of your videos where I realized the 'gotcha' part even before watching it. I'm feeling quite proud of myself today!
A great addition to this discussion would have been to graph y=sqrt(1-cos^2 x) and show that on the interval [0,2pi] the area bounded by the graph and the x-axis is definitely going to be a positive number. That way, when you evaluate the integral after doing it incorrectly (i.e. without the absolute value) and get 0 it's perfectly clear that something's gone wrong with your working. I know that effectively that's what you did when you graphed y=abs(sin x), but I think that graphing it in it's original form would have been better and would have shown why you get abs(sin x) instead of just sin x. Regardless, great video.
Can do in your head if you spot the trick: 2 int[0 to pi] sin(x) dx = 2[cos(0) - cos(pi)] = 4
That's exactly what I did.
Yea, I definitely expected him to point out the symmetry in the video.
Muchas gracias por compartir tan interesante ejercicios de integrales definidas.😊. Mi hija y mi persona estamos muy agradecidos por la buena y clara explicación. 😊❤😊.
@2:20: I think you may have misspoke. You said "that only happens if this x is not a variable", meaning that sqr(x^2) = x. I think you meant to say that it only happens if x is never negative. It doesn't work for the constant -9, but does work for the function f(x) = sin(x) + 1. Sqr(f^2(x)) = f(x) since abs(f(x)) = f(x).
No , the point here is that since x is a variable, we may not tacitly presume that x is positive; x may be negative (and still produce a positive value x^2 under the square root as if x were positive). And that problem doesn't occur when the value under the square root is a constant: sqrt(9) = 3 , sqrt(-9) = 3i , never do we get something that is negative which must be converted into someting positive in order to get it right.
@yurenchu I would politely disagree. If x was the constant -9, x^2 would be 81 and the square root of 81 is 9, not the original x of -9, but rather abs(-9). It's the sign of x that matters, sqr(x^2) = x if and only if abs(x) = x. Whether x is constant or varying is inconsequential so long as for all values of x, abs(x) = x.
@@KMcCabe The problem that he wants to point out is that nobody makes the mistake of saying sqrt((-9)^2) = -9 (except for the few who confidently maintain that a square root can be negative; but that's a different story), but plenty do reduce sqrt(x^2) automatically and blindly to x because they overlook that x is a variable that may be negative.
Furthermore, I think he is comparing the case of sqrt(x^2) versus the case of sqrt(81), not versus sqrt((-9)^2) . 81 is readily recognized as a square, hence we can automatically and easily reduce the square root expression sqrt(81) . x^2 is also readily recognized as a square, but we must be careful when reducing the square root expression sqrt(x^2), because there's a little hidden trickyness involved since x is a variable. I think that's what he's trying to say..
@@yurenchu I agree that the sqr((-9)^2) not equalling -9 is a trivial example. I get your point that it should be obvious when a constant number is involved. I've watched it a few times and what he said was that it only fails when it's a variable. My point is that he may have misspoke. It can fail to be true for a constant number and it can hold true for a variable function. I've given examples of both. The idea that you can just cancel an even exponent and root only holds when the base is non-negative. I think we can all agree with that. 🤓
The square root of x^2 is well defined for all values of x, but it has a subtle behavior when x is negative. If, for example, x = -3, then sqrt((-3)^2) = sqrt(9) = 3, so sqrt(x^2) is NOT equal to x if x = -3. Instead, if x = -a is negative (where a is positive), then sqrt(x^2) = sqrt((-a)^2) = sqrt(a^2) = a = -x = |x|. Thus, sqrt(x^2) = |x| is valid if x is any real number, positive or negative or 0.
"The race is not to the swift" is an appropriate epigram for this problem.
You're great! Keep on with the videos! You are doing it with heart, and a good portion of pedagogic gift.
You didn't really explain people why in case of a variable the sqrt of a square is the abs value. Because it can represent a negative number as well, and in that case the answer is positive at the end. You just mentioned that if it is a number it is always a positive.
im actually flabbergasted from the fact that many cannot comprehend the function co domain shits and just often end up getting these wrong
It's the co-domain shit in this case, lol. 😂
@Grecks75 true,i apologise for my terminology that i wrote while taking a shit at 3 in the morning,i didn't even realise i had a comment here lol
sir! I like your politeness in math and vibe
Great one! Absolute value integration was so neat! Thank you again for making it easy. Love to see more absolute value function integration.
@@Abby-hi4sf Wait till you see floor function integration! 😉
As an engineering student, we were taught that integrals have an area with sign, so you can just integrate sinx (without abs) from 0 to pi, and then from pi to 2 pi, but changing signs. At the end it is the same but you showed a more “mathematical approach” to the problem, we just change the sign 😂
Great sir❤
damn. I knew id missed something when i got 0 but didnt see it. thanks for this!
Thank you for another outstanding video.
Its the integral of |sin x| over the period [0,2π]
Thanks for the video brother and remember "Never stop spreading the Gospel,who stop spreading the Gospel stop living the Gospel"
Since we know the integral goes from 0 to 2 pi which is a full sine, and that the absolute rectifies the sine graph, the two "humps" you're trying to find the area of are going to be the same. So could you just do 2 * integral from 0 to pi of sinx instead of splitting it into two integrals?
That's a shortcut that we can take (in this particular case), but he's showing us the basic proper way of how to handle problems like this, which we can rely on even when the graph doesn't show such convenient symmetries (for example, if instead of |sin(x)| the integrand were |sin(x) + ½| , or something like that).
@@yurenchu Right that does makes sense, use a simple example to explain a more complex solution so it's easier to demonstrate? In the wild however if presented with this particular problem, I think most would take the shortcut and give reasoning for it.
@@spudhead169 Yes, you are free to do so. But he wouldn't be a good teacher if he didn't show us how his method applies even when we don't use convenient shortcuts.
I really don't understand why you are asking the question in your opening comment. It's clear that the two "humps" are equal, did you really need confirmation that in this case you could just multiply the integral from 0 to pi by the factor 2? Furthermore, it's not really " _instead_ of splitting it into two integrals", you too are essentially splitting the integral into two integrals, but you're immediately recognizing that the two integrals are equal and hence you take the shortcut of multiplying one integral by 2.
Split interval into quadrants
Simplify integrand to sin(x)
In fact splitting interval to [0,pi] and [pi,2pi]
is enough here
I knew it was absolute value but I was doing the math in my head and the negatives annoyed me LOL. But I did caught the square root of sin^2x was absolute value of sinx. Nice problem (Was doing this on my way to work - hence no paper lol)
Useful for competitive exam, sir. S Chitrai Kani
this is so cool teacher, tks alot
I at first thought it would be zero, but then I remembered that the radical means that the graph would be forced above the x-axis. It's the same concept we use when finding the positive area between the x-axis and the curve.
The precision he draws sine curves scares me.
Thank you for a great explanation.
I like your explanation
I like your video
At the beginning I thought you was wrong before I figured that we go from 0 to 2pi ! Thanks well done
Another way consist using another variable such that : t = cosx !
Thanks Sir 👍
Got me ! Good concept. Will not repeat these kind of mistakes again 😅
Thank you for the crystal clear explanation sir. Won't do that mistake again :D
Integrate[Sqrt[1-(Cos[x])^2],{x,0,2π}]=4 It’s in my head.
Me too
Thank you for your explanation for the modulus |sinx|.
Good luck, you make a great job.
Here we use only sinx. Read my NOTE.
First Reaction: " Something isn't adding up. "
Challenged by the thumbnail title: I came up with the answer 4 after roughly 45 seconds of integrating in my head. Seemed so easy, hope I got it right. 😂
Love from India 🇮🇳
Got that! sqrt((sin x)^2) = |sin x|
Here we have to use sinx only. Read my NOTE.
@@Misha-g3bProof?
I solve it mentally 💫✍️🐺
Suppose we have an ANALOG voltmeter that measures a voltage of the form abs (sin x), for example u = U abs (sin x). If the frequency is high enough that the voltmeter needle cannot follow the voltage variations, can you tell what voltage the voltmeter will indicate ? This is a math problem and it is very interesting.
the race is indeed not for the swift
Very fitting in this context. 😊
I think schools need to teach this more
Much simple: Integral from 0 to 2.pi square root 1-cos^2dx=2×Integral from 0 to pi sin x.dx= --2.cos x evaluated from 0 to pi.
Noice one but please bring back the og olympiad qns😊
Sir how to e mail u a problem
Genius
The question is faulty, though. 0 to 2pi is a complete revolution. Which is equal 0 to 0. Definite integrals with the same lower and upper limit should be 0, no?
No
Draw the graph of y = sqrt( 1 - (cos(x))^2 ) and you'll see that the area under the graph from x=0 to x=2pi isn't equal to 0 .
This is like watching paint dry. Excruciating.
I did 0 in my head at first, then realized my mistake and switched to 4. I think my problem was doing it in my head. If I had written it out, I might have caught it quicker.
those who cant do it they need to look in the mirror
Could you just integrate sin(x) from 0 to pi and the multiply that answer by 2?
So long as you're allowed to make an argument from symmetry or the periodicity of sine along with it being an odd function, that's fine.
yes if you are wanting the total area under the curve sinx from 0 to 2pi...
Clearly yes, provided you have figured out for yourself what abs(sin(x)) looks like (on the given interval).
4? ( i didnt watch the vdo yet just tried in my head )
I always hated the sqrt function.
As per defination of function each element in domain has to have only unique mapping to the codomian
But sqrt has 2 mappings.
So sqrt is not a function, but everyone treats it as 1 by forcing the defination of sqrt to have a codomian of only positive real numbers.
My issue is it's never mentioned what is the defination of sqrt and that annoys me to no end. I have seen exams where it's treated as a relationship in 1 problem and as a function in the next.
Exams needs to understand that they have to give the exact defination before using such "functions"
The radical is always meant as the principal root for real numbers. For complex numbers, it does represent all solutions. But √4 is always 2, it is never negative -2 and if it is written as such, your teacher doesn't know what they're doing.
Which is why under the reals you'll often find that √(x²) is written as |x| because it is true.
No, sqrt refers strictly to the positive square root.
You guys are right, but here in India, to make problems harder they sometimes break rules or probably forget the rules.
I have seena few times, where students were awarded marks because the question took negative values for sqrt.
I guess it's not such a big issue in other countries.
@@flightyavian the teachers know, but the text book author and exam creators don't. They often forget and it gets really annoying.
Solved by watching thumbnail only without even pen🇮🇳🇮🇳
Make a video showing all integers solutions for n which | (3n+16) / (4-n) | = k² | k € Q
I made this question by myself and the solutions for n are n = -10, -5, -3, 0, 3, 32
I even put the equation on a C compiler and these are the solutions fr
Cool math puzzle! But I think those are not all solutions. What about n = 452 ?
EDIT: Also n = 6304 and n = -1690 . I don't have proof, but i have a feeling that this problem has infinitely many solutions.
@yurenchu that is true, i think the same, thanks for telling me
@@thomazsoares1316 I'm still intrigued by this puzzle. I also found:
n = -122
n = -23530
n = 87812
It appears there are three sets of solutions:
One finite set of solutions where (3n+16)/(4-n) > 0 :
n ∈ {-5 , -3 , 0 , 3}
Two infinite sets of solutions where (3n+16)/(4-n) < 0 :
First define integer sequences tⱼ and sⱼ , as follows:
t₀ = 0
t₁ = 1
tₘ₊₂ = 4*tₘ₊₁ - tₘ , for any non-negative integer m
sₘ₊₁ = tₘ₊₁ - tₘ , for any non-negative integer m
Then
n ∈ { 4 + 28*(tⱼ)² | index j is a positive integer }
==> n ∈ {32 , 452 , 6304 , 87812 , 1223072 , ...}
and
n ∈ { 4 - 14*(sⱼ)² | index j is a positive integer }
==> n ∈ {-10 , -122 , -1690 , -23530 , -327722 , ...}
appear to be solutions to the equation.
I have confidence that each of these values is indeed a solution, and also that these include _all_ possible solutions (in other words, there are no other solutions); although I don't have mathematical proof for either claim.
@@thomazsoares1316 I'm still intrigued by this puzzle. I also found:
n = -122
n = -23530
n = 87812
It appears there are three sets of solutions:
V₀ = {-5 , -3 , 0 , 3}
V₁ = { 4 + 28*(tⱼ)² | index j is a positive integer }
where t₀ = 0 , t₁ = 1 , and tₘ₊₂ = 4*tₘ₊₁ - tₘ (for any non-negative integer m)
==> V₁ = {32 , 452 , 6304 , 87812 , 1223072 , ...}
V₂ = { 4 - 14*(sⱼ)² | index j is a positive integer }
where sₘ₊₁ = tₘ₊₁ - tₘ (for any non-negative integer m)
==> V₂ = {-10 , -122 , -1690 , -23530 , -327722 , ...}
So n ∈ { union of (V₀ , V₁ , V₂) }.
I could be wrong, but I think these are all possible solutions.
@@thomazsoares1316 I think I have solved your math puzzle and I have posted two replies in this thread with the possible values of n , but for some reason youtube hides those replies unless you have the comment section sorted in "Newest first" order (instead of "Most popular" order).
You do not have to do 2 integrals. Just double it. 2 times the integral from pi to 0.
He does have to do two integrals because he is teaching you the proper method of how to handle problems like this, which works even when the integrand doesn't have symmetry properties like it does here. You may choose to take a shortcut later, but at least he has done his job as a teacher.
@@yurenchu Then, he should teach both.
I know someone by your last name. Hope you're related to her.
@@kevinstreeter6943 Why? You still need to be shown how to multiply something by 2?
It's a part of basic mathematics, that you have used to solve..
not particularly clear explanation of square root of square
4
Pwend! ☺
Who the f can't solve this problem 🤷🏻
Easy
|sin²x|
Me: "oh, sure; the symmetry of the sine curve means you can equate the absolute value of sine to twice the output of the integral of sinx from 0 to π."
Newton: do your absolute values properly; stop cutting corners
Being Indian, we solved it in 12th standard, when we were mere 18 years!
That isn't exactly an accomplishment. In China and Eastern Europe everyone would have solved this earlier on.
Dawg… that’s what people in the states do also.
@@ElVerdaderoAbejorro I agree. 13 year old bulgarian on the keyboard.
Indian math education may be great or even superior to other countries', but this is not an example, lol.
@@ElVerdaderoAbejorroYou're right but I wouldn't say the same about Western Europe these fine days. 😢
Total useless! 😅
square root of 25 can also be -5
No. (-5)² = 25, but √25 only outputs a value in R^+ U {0}. It is defined as a single valued function
Never. -5 can be called _a_ second root or square root of 25, but not _the_ square root. When you use the sqrt() function or the radical symbol, it's *always* the non-negative root that's assumed. Note the tiny but important differences in notion.
In my head I worked out, it should be zero.
The calculator said, it should be four 🫣🙃 (I understood the mistake)
4