A formula for Pythagoreans with prime roots

[Vast3]

*The text is long and please don't leave rude comments if you don't have time to read it or don't have the patience to read it*
I will start from the beginning, well, we all know the Pythagorean formula, right? Well, if you want to use the Pythagorean formula for problems like: "Draw a point on the number line that shows the number 1 + square root of three, then if we use a geometric solution, it means an elongated triangle whose end points towards the numbers. is negative and draw two other triangles (which become smaller each time) with the sizes of the factors of the sides of the triangle above the triangle and attach it to the chord of the triangle; finally, if the vertex that is on the side that is perpendicular in the last triangle is half He drew a circle that coincidentally hits exactly on the point of the number axis that shows the number 1 + square root of three, And now, if we expand the axis of numbers to the two-dimensional coordinate plane, now by counting the size of each side of all triangles and simplifying and putting it in the Pythagorean formula, we will reach the chord of the primary plane on which all the sides of the triangle are drawn.
Well, we now have the geometric method, but what about the algebraic method, if we also use the Pythagorean formula, we will reach the number four in the end, but 1 + the square root of three is not the same as the number four, so what is the algebraic solution?
Well, I concluded that the best formula is this formula after a month of trial and error:
[math]a ^ 2 + (b ^ 2 - \frac{a ^ 2}{a}) = \sqrt{c}[/math]The letter a here represents the height of the right side of the triangle, b represents the length of the lower side of the triangle, and c is the hypotenuse of the triangle,
Now, let's use the same example of 1 + square root of three here since three is a prime number and we got to four by putting only its factors, one and three itself, now before progressing in solving this problem, we need to define some rules to reach a reasonable answer.
Rule 1: "a" is one if and only if "c" is prime
Rule 2: "a" changes concerning "c" when it is possible to find the solution by replacing "b" in the Pythagorean formula and equality is established
Rule 3: "b" is a complex number if a does not have a continuous ratio with "c".
Now it's time to start, we replace:
[math]1 ^ 2 + ( b ^ 2 - \frac{1 ^ 2}{1}) = \sqrt{3}[/math]Before jumping to conclusions, it is necessary to simplify as much as possible:
[math]1 + b ^ 2 - 1 = \sqrt{3}[/math]Now we remove the pesky 1's and arrive at the simplest form:
[math]b ^ 2 = \sqrt{3}[/math]Now we convert the quadratic equation to the standard form:
[math]b ^ 2 - \sqrt{3} = 0[/math]Now we have to define the constants and put them in the formula to solve the equation and simplify:
[math]a = 1, b = 0, c = -\sqrt{3} \rightarrow b = \frac{-0 \pm \sqrt{0 ^ 2 -4 \cdot 1 \cdot (-\sqrt{3})}}{2 \cdot 1}[/math]Now to simplify:
[math]b = \frac{\pm \sqrt{4 \sqrt{3}}}{2}[/math]Further simplification:
[math]b = \frac{\pm 2 \sqrt[4]{3}}{2}[/math]The last simplification and reaching the two roots of the equation:
[math]b_{1} = \sqrt[4]{3}, b_{2} = - \sqrt[4]{3}[/math]Now, if we use the positive root and let it be in the Pythagorean formula, we will reach the same acceptable answer as expected:
[math]1 ^ 2 + (\sqrt[4]{3}) ^ 2 = ?[/math]First, we simplify and arrive at the answer:
[math]1 + (\sqrt{3}) ^ 2 = 1 + \sqrt{3}[/math]You saw that we reached the answer and we were able to find the two sides, and now by substituting in my original formula, we can prove the equality. By placing the second factor, which is b, in my formula, we reach the square root of three, and this shows The reason is that Pythagoras is only able to solve the first three with the help of the geometric method. Still, without it, he is not able to solve threes like the factors of three, because he cannot show the point we want and only A chord solution leads to a completely different point appearing on the axis.
This formula can be attributed to a general formula for any Pythagorean multiple because it is possible to obtain the root of any number and develop the concept of prime numbers:
[math]a ^ n + (b ^ n - \frac{a ^ n}{a}) = \sqrt{c}[/math]

 

[Dr.Peterson]

I will start from the beginning, well, we all know the Pythagorean formula, right? Well, if you want to use the Pythagorean formula for problems like: "Draw a point on the number line that shows the number 1 + square root of three, then if we use a geometric solution, it means an elongated triangle whose end points towards the numbers. is negative and draw two other triangles (which become smaller each time) with the sizes of the factors of the sides of the triangle above the triangle and attach it to the chord of the triangle; finally, if the vertex that is on the side that is perpendicular in the last triangle is half He drew a circle that coincidentally hits exactly on the point of the number axis that shows the number 1 + square root of three, And now, if we expand the axis of numbers to the two-dimensional coordinate plane, now by counting the size of each side of all triangles and simplifying and putting it in the Pythagorean formula, we will reach the chord of the primary plane on which all the sides of the triangle are drawn.

Please include a picture so we can see what you are saying here. It is very hard to follow, in part because your terminology is nonstandard.

Well, we now have the geometric method, but what about the algebraic method, if we also use the Pythagorean formula, we will reach the number four in the end, but 1 + the square root of three is not the same as the number four, so what is the algebraic solution?

Again, please show the work you are doing. I suspect you are making a basic mistake working with radicals.

 

[Vast3]

Again, please show the work you are doing. I suspect you are making a basic mistake working with radicals.

If there is a mistake, please tell me

 

[stapel]

If there is a mistake, please tell me

Once you "please show the work you are doing", the helpers can try to determine if you have indeed made "a basic mistake". Thank you!

 

[Vast3]

Once you "please show the work you are doing", the helpers can try to determine if you have indeed made "a basic mistake". Thank you!

Well, now the helpers have not mentioned exactly what the basic mistake is and where it happened so that the amount of error in the problem can be evaluated

 

[Vast3]

The whole thing I am saying is that Pythagoras is not capable of finding such problems because I have personally tried many ways to apply Pythagoras, but none of them worked, but this formula establishes equality. So, I think the mistake that happened is if we put the last answer of Pythagoras in the same triangle with the chord:
[math]\sqrt{3}[/math]In fact, the whole thing I am saying is that Pythagoras is not capable of finding such problems because I have personally tried many ways to apply Pythagoras, but none of them worked, but this formula establishes equality. So, I think the mistake that happened is if we write Pythagorean's last answer in the same triangle with chord [math]\sqrt{3}[/math] as if we write:
[math]c = \sqrt{a ^ 2 + b ^ 2}[/math]Now, if we enter the factors of the root number in the chord of the triangle, we will see that we reach the number four.
If there is still a mistake, please let me know.

 

[Vast3]

This is what I got from Geogebra to solve the given problem, and you can see the chord is: [math]\sqrt{3}[/math] and the b edge is also: [math]\sqrt[4]{3}[/math] This is exactly what we get algebraically for the problem through the formula. Now, by substituting the values in the Pythagorean formula, we get: [math]\sqrt{3} + 1[/math]

 

[BigBeachBanana]

Now, if we enter the factors of the root number in the chord of the triangle, we will see that we reach the number four.

How? please show how you get 4.

 

[Vast3]

How? please show how you get 4.

Oh, I see the problem, I was mistaken it takes you to answer the number square root of 10, I am very sorry for this mistake, but in any case, this may have been the only mistake in the use of radicals

 

[Dr.Peterson]

View attachment 36800This is what I got from Geogebra to solve the given problem, and you can see the chord is: [math]\sqrt{3}[/math] and the b edge is also: [math]\sqrt[4]{3}[/math] This is exactly what we get algebraically for the problem through the formula. Now, by substituting the values in the Pythagorean formula, we get: [math]\sqrt{3} + 1[/math]

What do you mean by "chord"? You have repeatedly mentioned "the chord of a triangle", but triangles don't have chords. My guess is that you mean "hypotenuse", namely BC; but that has length [imath]\sqrt{\sqrt{3}+1}\approx1.65289[/imath], not [imath]\sqrt{3}\approx1.732[/imath].

Also, nothing here has length [imath]\sqrt{3}+1[/imath].

Now, if we enter the factors of the root number in the chord of the triangle, we will see that we reach the number four.
If there is still a mistake, please let me know.

Oh, I see the problem, I was mistaken it takes you to answer the number square root of 10, I am very sorry for this mistake, but in any case, this may have been the only mistake in the use of radicals

I see nothing here that is equal to either 4 or [imath]\sqrt{10}[/imath]. Please show exactly what you did, step by step, to obtain whatever you are talking about.

 

[Vast3]

What do you mean by "chord"? You have repeatedly mentioned "the chord of a triangle", but triangles don't have chords. My guess is that you mean "hypotenuse", namely BC; but that has length [imath]\sqrt{\sqrt{3}+1}\approx1.65289[/imath], not [imath]\sqrt{3}\approx1.732[/imath].

Also, nothing here has length [imath]\sqrt{3}+1[/imath].


I see nothing here that is equal to either 4 or [imath]\sqrt{10}[/imath]. Please show exactly what you did, step by step, to obtain whatever you are talking about.

This shows that Pythagoras can't give us equality because we wanted a triangle with a hypotenuse square root of three, but Pythagoras gave us a hypotenuse square root of 10. That's the whole point I was trying to make.

 

[Dr.Peterson]

This shows that Pythagoras can't give us equality because we wanted a triangle with a hypotenuse square root of three, but Pythagoras gave us a hypotenuse square root of 10. That's the whole point I was trying to make.

NO, IT DOES NOT!

There is no [imath]\sqrt{10}[/imath] here. That's why I've asked to show your work step by step, so we can see where you are going wrong.

My best guess is that you are confusing the square root with the square, so that you think that [math]\sqrt{\sqrt{3}+1}=\sqrt{9+1}=\sqrt{10}.[/math] But it does not. Rather, [math]\sqrt{{\color{Red}3^2}+1}=\sqrt{9+1}=\sqrt{10},[/math] and [math]\sqrt{\sqrt{3}+1}\approx\sqrt{{\color{Red}1.732}+1}\approx\sqrt{2.732}\approx1.65289.[/math]
If that's not what you did, please show us how you actually got [imath]\sqrt{10}[/imath].

Now, it has never been clear what your actual goal is. If it is to get a hypotenuse of [imath]\sqrt{3}[/imath], then you want the two legs to be 1 and [imath]\sqrt{2}[/imath]. Then the hypotenuse will be [math]\sqrt{1^2+\sqrt{2}^2}=\sqrt{1+2}=\sqrt{3}\approx1.732[/math]

 

[Vast3]

NO, IT DOES NOT!

There is no [imath]\sqrt{10}[/imath] here. That's why I've asked to show your work step by step, so we can see where you are going wrong.

My best guess is that you are confusing the square root with the square, so that you think that [math]\sqrt{\sqrt{3}+1}=\sqrt{9+1}=\sqrt{10}.[/math] But it does not. Rather, [math]\sqrt{{\color{Red}3^2}+1}=\sqrt{9+1}=\sqrt{10},[/math] and [math]\sqrt{\sqrt{3}+1}\approx\sqrt{{\color{Red}1.732}+1}\approx\sqrt{2.732}\approx1.65289.[/math]
If that's not what you did, please show us how you actually got [imath]\sqrt{10}[/imath].

Now, it has never been clear what your actual goal is. If it is to get a hypotenuse of [imath]\sqrt{3}[/imath], then you want the two legs to be 1 and [imath]\sqrt{2}[/imath]. Then the hypotenuse will be [math]\sqrt{1^2+\sqrt{2}^2}=\sqrt{1+2}=\sqrt{3}\approx1.732[/math]

Yes, this answer is correct, but I did not say that: [math]\sqrt{\sqrt{3} + 1 } = \sqrt{9 + 1} =\sqrt{10}[/math] No, I think I caused some confusion and I'm sorry for that, but while I say: [math]\sqrt{3 ^ 2 + 1 ^ 2} = \sqrt{9 + 1} = \sqrt{10}[/math] While we wanted to have a triangle with hypotenuse square root of 3 and we also know that 9 and 1 and even 1 and square root of 2 are not factors of 3 because we know that they must be two numbers that if they reach the power of two and together Now let me say this while you mentioned this at the end of your message: [math]c ^ 2 = 1 ^ 2 + (\sqrt[4]{3}) ^ 2[/math] The answer we get is: [math]\sqrt{1 + \sqrt{3}}[/math] Now we have reached the second factor, which is the square root of 3, and we know that the square root of 3 is not a perfect square root, but we also know that when we take the square root of a number, we are actually reaching a number that, if it reaches a power, becomes a sub-radical number. The formula is the same as 3, but from the calculation above, we also realized that the first factor is not one, because if the first factor was one, we would not have gotten a number other than the square root of 3, and now we are faced with this paradox that mathematics says that prime numbers Apart from 1, they do not have any other factors, while in this calculation we have concluded that the first factor is not one and why is it not one, because if this was a factor of 3, then we would have reached the square root of 3, the reasons why I used the fourth root of 3 for Pythagoras because it would establish an equality. What I am trying to say is that when we have two numbers that are going to produce a specific number that is obviously a prime number, we know that the number we are producing is 3 primes and only the numbers 1 and 3 produce the number 3, in this case, those two numbers cannot produce 3, but they do in Pythagoras, but not in normal arithmetic, so we have this paradox that creates the problem of why Pythagoras still Factoring does not give us three numbers. There are two possible answers: the numbers 1 and 3 are not factors of 3, or Pythagoras has a problem, which is more likely, of course, I do not intend to cause confusion.

 

[BigBeachBanana]

we know that the square root of 3 is not a perfect square root, but we also know that when we take the square root of a number, we are actually reaching a number that, if it reaches a power, becomes a sub-radical number.

What is "power" and "sub-radical"?

The formula is the same as 3, but from the calculation above, we also realized that the first factor is not one, because if the first factor was one, we would not have gotten a number other than the square root of 3, and now we are faced with this paradox that mathematics says that prime numbers Apart from 1, they do not have any other factors, while in this calculation we have concluded that the first factor is not one and why is it not one, because if this was a factor of 3, then we would have reached the square root of 3, the reasons why I used the fourth root of 3 for Pythagoras because it would establish an equality.

Can't parse the rest of this.
I don't think you know what a factor is. One is a factor of [imath]3[/imath] because [imath]3= 1 \times 3[/imath]
One is not a prime number.

What I am trying to say is that when we have two numbers that are going to produce a specific number that is obviously a prime number, we know that the number we are producing is 3 primes and only the numbers 1 and 3 produce the number 3, in this case, those two numbers cannot produce 3, but they do in Pythagoras, but not in normal arithmetic, so we have this paradox that creates the problem of why Pythagoras still Factoring does not give us three numbers. There are two possible answers: the numbers 1 and 3 are not factors of 3, or Pythagoras has a problem, which is more likely, of course, I do not intend to cause confusion.

Are you claiming that you can only produce the hypotenuse of 3 if and only if the legs are 1 and 3?

 

[Vast3]

What is "power" and "sub-radical"?


Can't parse the rest of this.
I don't think you know what a factor is. One is a factor of [imath]3[/imath] because [imath]3= 1 \times 3[/imath]
One is not a prime number.


Are you claiming that you can only produce the hypotenuse of 3 if and only if the legs are 1 and 3?

sub-radical is a typo I was the mean number under the radical and I know what a factor is, and the word "power" is the same as "exponent", for example, if we want to say 2 to the power of 2, we don't say 2 "exponent" to 2, now let's go back to the issue of the sides of the triangle. There is a confusion here and I'm sorry for that. It is better to say that I said in the text that the paradox is that prime numbers are only divisible by one and themselves. Still, in our calculations, we came to the opposite of this, as the calculations showed that the number one is not a factor of the number 3, because The calculations should have reached an acceptable number, which is the square root of 3, but it was not. I am not saying that we only reach the square root of 3 in the hypotenuse when the other sides are 1 and 3. c let's put a and b in, so at the stage where c is ^, then we know that the numbers a and b must be a factor of three. The reason for this problem that a and b must be a factor of c is known in the Pythagorean geometrical proof itself, but the simpler explanation is that "if there are two numbers that produce a prime number, then those two numbers must be factors of that number because Only the numbers 1 and 3 produce the number 3, so we know that if a and b have a relationship with c and the number c is the square root of a prime number, then naturally those two numbers must be its factors, but if the factors in Pythagoras enter, they do not give the root of three.
It has been said from the beginning that why Pythagoras gives you the chord of 1 + the root of three is that, Pythagoras is not able to find it.

 

[Dr.Peterson]

While we wanted to have a triangle with hypotenuse square root of 3 and we also know that 9 and 1 and even 1 and square root of 2 are not factors of 3 because we know that they must be two numbers that if they reach the power of two and together

The formula is the same as 3, but from the calculation above, we also realized that the first factor is not one, because if the first factor was one, we would not have gotten a number other than the square root of 3, and now we are faced with this paradox that mathematics says that prime numbers Apart from 1, they do not have any other factors, while in this calculation we have concluded that the first factor is not one and why is it not one, because if this was a factor of 3, then we would have reached the square root of 3, the reasons why I used the fourth root of 3 for Pythagoras because it would establish an equality.

The reason for this problem that a and b must be a factor of c is known in the Pythagorean geometrical proof itself, but the simpler explanation is that "if there are two numbers that produce a prime number, then those two numbers must be factors of that number

There is something fundamental that you are either misunderstanding, or failing to say clearly.

The Pythagorean theorem has nothing to do with factors, or with prime numbers. Whatever you are claiming, it makes no sense to me (and hasn't from the start).

 

[Vast3]

There is something fundamental that you are either misunderstanding, or failing to say clearly.

The Pythagorean theorem has nothing to do with factors, or with prime numbers. Whatever you are claiming, it makes no sense to me (and hasn't from the start).

The only thing I want to say is that this problem has a second solution, there is no need to use 1 and the square root of 2 because using the formula establishes equality. Now, I'm sorry if I can't express my meaning well, but Anyway, there is this problem. If you want to see if you can try other numbers on the formula that was presented, you can try a = 2 and c = 6 and see if equality is established. I don't want to. I may be annoying or rude, but I want you to see that the formula can be solved. I think the mistake and misunderstanding that I had was that I did not see that equality was established with 1 and the square root of 2 and unfortunately I wrote the messages quickly. Forgive me.
Although I am only providing alternative solutions, it is very difficult to prove such things

 

[fresh_42]

The only thing I want to say is that this problem ...

Which problem exactly?

... has a second solution ...

to what the first solution is?

Although I am only providing alternative solutions

To what? Pythagoras is a formula, not a problem. It is a quadratic formula with three quantities. Those quantities are usually associated with a positive length. But as a quadratic formula, there are always negative solutions, too.

 

[Dr.Peterson]

Anyway, there is this problem. If you want to see if you can try other numbers on the formula that was presented, you can try a = 2 and c = 6 and see if equality is established.

I want you to see that the formula can be solved.

The trouble is that you have never clearly stated what "problem" it is that you think you are "solving". If you ever manage to do that, there will be something to discuss. As it is, there is nothing more to say.

 

[Vast3]

The trouble is that you have never clearly stated what "problem" it is that you think you are "solving". If you ever manage to do that, there will be something to discuss. As it is, there is nothing more to say.

The problem was explained at the beginning of the text, which is: "How to use a right-angled triangle to find a point on the number line"
The problem is that if we use a triangle, we have to draw an arc from the highest vertex of the triangle, which we have to open from a caliper with an opening the size of the hypothesis. This is because I say that the sides must have numerical relationships with each other because If the sides have no numerical relationship with anything and we put this, we will not reach the answer that the "problem" wants.