@NadiehFan is a legend in Math. He challenges every method put forward. Kudos to @NadiehFan . I LOVE your approach to Math Problems.

@SyberMath

14 күн бұрын

Absolutely! Thanks

I once got stumped in a trig. problem where the answer I got had a nested square root but the answer at the back of the book had two square roots. I eventually found out that the answers are the same by using a calculator and used another hour to prove it.

Interesting!

Hi FIRST AMAZING VIDEO

Syber cant younsokve WITHOUT using the conjugate since indont see angine thinking kf it at all..?

Don't let him cook again 💀💀

I've already commented on this video but since this comment is buried inside a thread were it is not likely to be seen or even visible for most viewers of this video I'll repost an edited and slightly expanded version of my comment here. If x and y are positive rational numbers with √y irrational and x² − y > 0, then √(x ± √y) is denestable, that is, two positive rational numbers p and q exist such that √(x ± √y) = √p ± √q _if and only if_ x² − y is the square of a rational number. If r is a positive rational number such that r² = x² − y then p = (x + r)/2 and q = (x − r)/2. This is a theorem I proved in a somewhat more general form in a comment on a previous video of SyberMath about the equation x⁴ − 12x = 17, see kzread.info/dash/bejne/ZJNm0paGh8yXYJs.html Regarding the method for denesting denestable nested square roots shown in the video, there is a perfectly good reason to add and subtract √(x + √y) and √(x − √y). If r is a positive rational number such that r² = x² − y then we have the identities (1a) √(x + √y) = √((x + r)/2)) + √((x − r)/2)) (1b) √(x − √y) = √((x + r)/2)) − √((x − r)/2)) which imply that we also have (2a) √(x + √y) + √(x − √y) = 2·√((x + r)/2)) (2b) √(x + √y) − √(x − √y) = 2·√((x − r)/2)) and therefore (3a) (√(x + √y) + √(x − √y))² = 2x + 2r (3b) (√(x + √y) − √(x − √y))² = 2x − 2r So, if √(x + √y) and √(x − √y) are denestable, then both the square of the sum √(x + √y) + √(x − √y) and the square of the difference √(x + √y) − √(x − √y) are rational. This provides an easy alternative way to denest denestable square roots if you don't remember (or don't care to remember) the identities (1a) and (1b) above with r = √(x² − y) which can be combined into (4) √(x ± √y) = √((x + √(x² − y))/2)) ± √((x − √(x² − y))/2)) As shown in the video, to denest √(7 + √13) we first confirm that 7² − (√13)² = 49 − 13 = 36 = 6² is the square of a rational number, which means that this nested square root can be denested. Using the identities (a + b)² = a² + 2ab + b² (a − b)² = a² − 2ab + b² (a + b)(a − b) = a² − b² and noting that the last of these identities implies √(7 + √13)·√(7 − √13) = √((7 + √13)·(7 − √13)) = √(49 − 13) = √36 = 6 we can easily find using the identities for (a + b)² and (a − b)² that we have (√(7 + √13) + √(7 − √13))² = 14 + 2·6 = 26 (√(7 + √13) − √(7 − √13))² = 14 − 2·6 = 2 and therefore √(7 + √13) + √(7 − √13) = √26 √(7 + √13) − √(7 − √13) = √2 Adding these two identities we get 2√(7 + √13) = √26 + √2 so we can conclude that √(7 + √13) = ¹⁄₂√26 + ¹⁄₂√2 I actually discussed this technique for denesting denestable nested square roots of binomial quadratic surds six months ago in my comment on this SyberMath Video: kzread.info/dash/bejne/aKV-yaScdJW6lLg.html Of course we can also subtract the second identity √(7 + √13) − √(7 − √13) = √2 from the first identity √(7 + √13) + √(7 − √13) = √26 which similarly allows us to conclude that √(7 − √13) = ¹⁄₂√26 − ¹⁄₂√2 Apart from applying the identity (4) directly or using the squares of the sum and difference of nested square roots of conjugate binomial quadratic surds we can also start from the theorem that guarantees the existence of two positive rational numbers p and q such that √(x ± √y) = √p ± √q if x and y are positive rational numbers with √y irrational if x² − y is the square of a rational number. Since 7² − (√13)² = 49 − 13 = 36 = 6² this theorem guarantees the existence of two positive rational numbers p and q such that √(7 + √13) = √p + √q Squaring both sides and noting that 2·√p·√q = √(4pq) we have 7 + √13 = (p + q) + √(4pq) and using the _theorem_ that says that if the sum of a rational and an irrational number is equal to another sum of a rational and an irrational number, then both the rational numbers and both the irrational numbers are equal this gives p + q = 7 4pq = 13 From this we get (p − q)² = (p + q)² − 4pq = 7² − 13 = 49 − 13 = 36 which implies p − q = 6 ⋁ p − q = −6 Since inverting the sign of p − q amounts to swapping the values of p and q we only need a single value for p − q. Selecting p − q = 6 we have 2p = (p + q) + (p − q) = 7 + 6 = 13 2q = (p + q) − (p − q) = 7 − 6 = 1 so we have p = ¹³⁄₂, q = ¹⁄₂ and therefore √(7 + √13) = √(¹³⁄₂) + √(¹⁄₂) Note that √(¹³⁄₂) = √13/√2 = (√13·√2)/(√2·√2) = (√26)/2 = ¹⁄₂√26 and √(¹⁄₂) = √1/√2 = (√1·√2)/(√2·√2) = (√2)/2 = ¹⁄₂√2 so this is indeed equivalent with √(7 + √13) = ¹⁄₂√26 + ¹⁄₂√2 So, we have seen three methods for denesting denestable square roots of binomial quadratic surds √(x ± √y) that do _not_ rely on trial and error: 1. Using the identity √(x ± √y) = √((x + √(x² − y))/2)) ± √((x − √(x² − y))/2)) directly. 2. Squaring both the sum and the difference of √(x + √y) and √(x − √y). 3. Setting √(x ± √y) equal to √p ± √q and squaring both sides.

@SyberMath

24 күн бұрын

Thank you for your contributions! 😍

It looks not possible😅

Let sqrt[x+sqrt(y)]=sqrt(u)+sqrt(v) Squaring: x+sqrt(y)=u+v+2sqrt(uv) u+v=x 2sqrt(uv)=sqrt(y) --> 4uv=y 4u(x-u)=y u²-xu+¼y=0 u=½[x+sqrt(x²-y)] v=½[x-sqrt(x²-y)] u=½[x-sqrt(x²-y)] v=½[x+sqrt(x²-y)]

Still looks nested.

@SyberMath

24 күн бұрын

When x^2-y is the square of a rational number, then it's denested.

@mcwulf25

24 күн бұрын

​@SyberMath true.

Sorry, I don't buy none of your explanations ... 1. In the first line you write sqrt(x-sqrt(y)), but x-sqrt(y) could be negative (x=-1 & y=4) 2. I don't understand your request about perfect squares. 3. The last equation at 03:00 is completly false ! For the second method, you claim that the conjugate of sqrt(x+sqrt(y)) is sqrt(x-sqrt(y)), without proof and wich seems false. I don't say the result is wrong, but the proof is not correct.

@FisicTrapella

28 күн бұрын

1. And what's the problem with that?? 2. If it is not a perfect square you'll keep the radical 3. Why?? 🤔

@bsmith6276

28 күн бұрын

Sybermath does tend to be weak about rigor. First, it is implied that x and y are positive. Then when he mentioned conjugate he probably should have more specific and say he is taking the conjugate of the radicand when introducing sqrt[x-sqrt(y)] as related to sqrt[x+sqrt(y)]. Also he glossed over a very important part: x^2-y>=0 is a necessary condition for this denesting to apply over the Reals; furthermore for integer x and y we want the expression x^2-y to be the square of an integer for the denesting to actually work out.

@FisicTrapella

28 күн бұрын

OK, but I don't know if the denesting works for complex numbers, which would allow (x-sqr(y)) to be negative...

@XJWill1

28 күн бұрын

@@bsmith6276 Actually, x^2 - y does not need to be the square of an integer. It needs to be the square of a rational number. In other words, sqrt(x^2 - y) must be rational, but it need not be an integer.

@NadiehFan

28 күн бұрын

@@XJWill1 Exactly. If x and y are positive rational numbers, √y irrational and x² − y > 0 then √(x ± √y) is denestable _if and only if_ x² − y is the square of a rational number. I proved this in a comment on a previous video of SyberMath about the equation x⁴ − 12x = 17: kzread.info/dash/bejne/ZJNm0paGh8yXYJs.html