# Coded Data Proofs: Mean & Standard Deviation, y=kx+C

Coded Data Proofs (3):

Say y=kx+C and also that:

x={p, q} and y={kp+C, kq+C}

This would mean that:

$\frac { \Sigma y }{ n } =\frac { kp+C+\left\{ kq+C \right\} }{ n } \\ \\ =\frac { kp+kq+nC }{ n } \\ \\ =\frac { k\left( p+q \right) +nC }{ n } \\ \\ =\frac { k\left( p+q \right) }{ n } +\frac { nC }{ n } \\ \\ =k\cdot \frac { \Sigma x }{ n } +C$

And if the above is true:

$\frac { \Sigma { y }^{ 2 } }{ n } =\frac { { \left( kp+C \right) }^{ 2 }+{ \left( kq+C \right) }^{ 2 } }{ n } \\ \\ =\frac { \left( kp+C \right) \left( kp+C \right) +\left( kq+C \right) \left( kq+C \right) }{ n } \\ \\ =\frac { { k }^{ 2 }{ p }^{ 2 }+nkpC+{ C }^{ 2 }+\left\{ { k }^{ 2 }{ q }^{ 2 }+nkqC+{ C }^{ 2 } \right\} }{ n } \\ \\ =\frac { { k }^{ 2 }{ p }^{ 2 }+{ k }^{ 2 }{ q }^{ 2 }+nkC\left( p+q \right) +n{ C }^{ 2 } }{ n } \\ \\ =\frac { { k }^{ 2 }\left( { p }^{ 2 }+{ q }^{ 2 } \right) +nkC\left( p+q \right) +n{ C }^{ 2 } }{ n } \\ \\ =\frac { { k }^{ 2 }\left( { p }^{ 2 }+{ q }^{ 2 } \right) }{ n } +\frac { nkC\left( p+q \right) }{ n } +\frac { n{ C }^{ 2 } }{ n } \\ \\ ={ k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } +kC\cdot \Sigma x+{ C }^{ 2 }\\ \\ ={ k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } +C\left( k\cdot \Sigma x+C \right)$

Therefore:

${ \sigma }_{ y }=\sqrt { \frac { \Sigma { y }^{ 2 } }{ n } -{ \left( \frac { \Sigma y }{ n } \right) }^{ 2 } } \\ \\ =\sqrt { { k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } +C\left( k\cdot \Sigma x+C \right) -{ \left( k\cdot \frac { \Sigma x }{ n } +C \right) }^{ 2 } } \\ \\ =\sqrt { { k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } +C\left( k\cdot \Sigma x+C \right) -\left( k\cdot \frac { \Sigma x }{ n } +C \right) \left( k\cdot \frac { \Sigma x }{ n } +C \right) } \\ \\ =\sqrt { { k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } +C\left( k\cdot \Sigma x+C \right) -\left\{ { k }^{ 2 }\cdot { \left( \frac { \Sigma x }{ n } \right) }^{ 2 }+nkC\cdot \frac { \Sigma x }{ n } +{ C }^{ 2 } \right\} } \\ \\ =\sqrt { { k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } +C\left( k\cdot \Sigma x+C \right) -{ k }^{ 2 }\cdot { \left( \frac { \Sigma x }{ n } \right) }^{ 2 }-nkC\cdot \frac { \Sigma x }{ n } -{ C }^{ 2 } } \\ \\ =\sqrt { { k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } +C\left( k\cdot \Sigma x+C \right) -{ k }^{ 2 }\cdot { \left( \frac { \Sigma x }{ n } \right) }^{ 2 }-kC\cdot \Sigma x-{ C }^{ 2 } } \\ \\ =\sqrt { { k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } +C\left( k\cdot \Sigma x+C \right) -{ k }^{ 2 }\cdot { \left( \frac { \Sigma x }{ n } \right) }^{ 2 }-C\left( k\cdot \Sigma x+C \right) } \\ \\ =\sqrt { { k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } -{ k }^{ 2 }\cdot { \left( \frac { \Sigma x }{ n } \right) }^{ 2 } } \\ \\ =\sqrt { { k }^{ 2 }\left\{ \frac { \Sigma { x }^{ 2 } }{ n } -{ \left( \frac { \Sigma x }{ n } \right) }^{ 2 } \right\} } \\ \\ =\sqrt { { k }^{ 2 } } \cdot \sqrt { \frac { \Sigma { x }^{ 2 } }{ n } -{ \left( \frac { \Sigma x }{ n } \right) }^{ 2 } } \\ \\ =k\cdot { \sigma }_{ x }$

# Coded Data Proofs: Mean & Standard Deviation, y=x/k

Coded Data Poofs (2):

Say y=x/k and that: x={p, q}, y={p/k, q/k}.

This would mean that:

$\frac { \Sigma y }{ n } =\frac { \frac { p }{ k } +\frac { q }{ k } }{ n } \\ \\ =\frac { \frac { 1 }{ k } \left( p+q \right) }{ n } \\ \\ =\frac { 1 }{ k } \cdot \frac { \left( p+q \right) }{ n } \\ \\ =\frac { 1 }{ k } \cdot \frac { \Sigma x }{ n } \\$

It would also mean that:

$\frac { \Sigma { y }^{ 2 } }{ n } =\frac { { \left( \frac { p }{ k } \right) }^{ 2 }+{ \left( \frac { q }{ k } \right) }^{ 2 } }{ n } \\ \\ =\frac { \frac { { p }^{ 2 } }{ { k }^{ 2 } } +\frac { { q }^{ 2 } }{ { k }^{ 2 } } }{ n } \\ \\ =\frac { \frac { 1 }{ { k }^{ 2 } } \left( { p }^{ 2 }+{ q }^{ 2 } \right) }{ n } \\ \\ =\frac { 1 }{ { k }^{ 2 } } \cdot \frac { \left( { p }^{ 2 }+{ q }^{ 2 } \right) }{ n } \\ \\ =\frac { 1 }{ { k }^{ 2 } } \cdot \frac { \Sigma { x }^{ 2 } }{ n } \\$

And if the above is true:

${ \sigma }_{ y }=\sqrt { \frac { \Sigma { y }^{ 2 } }{ n } -{ \left( \frac { \Sigma y }{ n } \right) }^{ 2 } } \\ \\ =\sqrt { \frac { 1 }{ { k }^{ 2 } } \cdot \frac { \Sigma { x }^{ 2 } }{ n } -{ \left( \frac { 1 }{ k } \cdot \frac { \Sigma x }{ n } \right) }^{ 2 } } \\ \\ =\sqrt { \frac { 1 }{ { k }^{ 2 } } \cdot \frac { \Sigma { x }^{ 2 } }{ n } -\frac { 1 }{ { k }^{ 2 } } \cdot { \left( \frac { \Sigma x }{ n } \right) }^{ 2 } } \\ \\ =\sqrt { \frac { 1 }{ { k }^{ 2 } } \left( \frac { \Sigma { x }^{ 2 } }{ n } -{ \left\{ \frac { \Sigma x }{ n } \right\} }^{ 2 } \right) } \\ \\ =\sqrt { \frac { 1 }{ { k }^{ 2 } } } \cdot \sqrt { \frac { \Sigma { x }^{ 2 } }{ n } -{ \left( \frac { \Sigma x }{ n } \right) }^{ 2 } } \\ \\ =\frac { 1 }{ k } \cdot { \sigma }_{ x }\\$

# Coded Data Proofs, Mean & Standard Deviation: y=kx

Coded Data Proofs (1):

Say y=kx, and also that: x={p, q}, y={kp, kq}.

This would mean that:

$\frac { \Sigma y }{ n } =\frac { kp+kq }{ n } \\ \\ =\frac { k\left( p+q \right) }{ n } \\ \\ =k\cdot \frac { \Sigma x }{ n } \\$

Therefore, when y=kx:

$\frac { \Sigma y }{ n } =k\cdot \frac { \Sigma x }{ n } \\$

What we’d also be able to conclude is that:

$\frac { \Sigma { y }^{ 2 } }{ n } =\frac { { \left( kp \right) }^{ 2 }+{ \left( kq \right) }^{ 2 } }{ n } \\ \\ =\frac { { k }^{ 2 }{ p }^{ 2 }+{ k }^{ 2 }{ q }^{ 2 } }{ n } \\ \\ =\frac { { k }^{ 2 }\left( { p }^{ 2 }+{ q }^{ 2 } \right) }{ n } \\ \\ ={ k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } \\$

When considering the above, we can deduce that:

${ \sigma }_{ y }=\sqrt { \frac { \Sigma { y }^{ 2 } }{ n } -{ \left( \frac { \Sigma y }{ n } \right) }^{ 2 } } \\ \\ =\sqrt { { k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } -{ \left( k\cdot \frac { \Sigma x }{ n } \right) }^{ 2 } } \\ \\ =\sqrt { { k }^{ 2 }\cdot \frac { \Sigma { x }^{ 2 } }{ n } -{ k }^{ 2 }\cdot { \left( \frac { \Sigma x }{ n } \right) }^{ 2 } } \\ \\ =\sqrt { { k }^{ 2 }\left( \frac { \Sigma { x }^{ 2 } }{ n } -{ \left\{ \frac { \Sigma x }{ n } \right\} }^{ 2 } \right) } \\ \\ =\sqrt { { k }^{ 2 } } \cdot \sqrt { \frac { \Sigma { x }^{ 2 } }{ n } -{ \left( \frac { \Sigma x }{ n } \right) }^{ 2 } } \\ \\ =k\cdot { \sigma }_{ x }\\$

Therefore, when y=kx:

${ \sigma }_{ y }=k\cdot { \sigma }_{ x }\\$

# Further Pure Maths: Complex Number Proof (1)

In this post, I’ll be proving that: $\left| { z }_{ 1 }\cdot { z }_{ 2 } \right| =\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right|$

First of all, let’s say that:

${ z }_{ 1 }=x+iy$

Whereby, $\left\{ x\in R,\quad y\in R \right\}$.

And also that:

${ z }_{ 2 }=p+iq$

Whereby, $\left\{ p\in R,\quad q\in R \right\}$.

If this is the case, this means that:

${ z }_{ 1 }\cdot { z }_{ 2 }=\left( x+iy \right) \left( p+iq \right) \\ \\ =px+iqx+ipy+{ i }^{ 2 }qy\\ \\ =px-qy+i\left( qx+py \right)$

Therefore:

$LHS\\ \\ =\left| { z }_{ 1 }\cdot { z }_{ 2 } \right| \\ \\ =\sqrt { { \left( px-qy \right) }^{ 2 }+{ \left( qx+py \right) }^{ 2 } } \\ \\ =\sqrt { \left( px-qy \right) \left( px-qy \right) +\left( qx+py \right) \left( qx+py \right) } \\ \\ =\sqrt { { p }^{ 2 }{ x }^{ 2 }-2pqxy+{ q }^{ 2 }{ y }^{ 2 }+\left\{ { q }^{ 2 }{ x }^{ 2 }+2pqxy+{ p }^{ 2 }{ y }^{ 2 } \right\} } \\ \\ =\sqrt { { p }^{ 2 }{ x }^{ 2 }+{ q }^{ 2 }{ y }^{ 2 }+{ q }^{ 2 }{ x }^{ 2 }+{ p }^{ 2 }{ y }^{ 2 } } \\ \\ =\sqrt { \left( { x }^{ 2 }+{ y }^{ 2 } \right) \left( { p }^{ 2 }+{ q }^{ 2 } \right) } \\ \\ =\sqrt { { x }^{ 2 }+{ y }^{ 2 } } \cdot \sqrt { { p }^{ 2 }+{ q }^{ 2 } } \\ \\ =\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right| \\ \\ =RHS$

Hence we’ve proven that:

$\left| { z }_{ 1 }\cdot { z }_{ 2 } \right| =\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right|$

# How to derive the formula for the area of an equilateral triangle

In this post I’ll be showing you how to derive the formula for the area of an equilateral triangle – in easy steps. In order to understand this derivation properly, you need to be familiar with Pythagoras’ theorem and also a few algebraic rules. What you’ll also need is a ruler, pair of compasses, a pencil and a sheet of paper.

Step 1: Put a point on a blank sheet of paper and name it A.

Step 2: Put the needle of your compass on the point A and draw a circle around it.

Step 3: Add a point B to this circle, on its edge.

Step 4: Put the needle of your compass on the point B and your pencil on the point A.

Step 5: Draw another circle with a radius the length AB.

Step 6: Now add a few extra points to your drawing. Call these points C and D.

Step 7: Connect the points A, B and C forming a triangle.

Step 8: Draw a line going through the points C and D.

Step 9: Where the line going through C and D intersects the triangle, place the point E.

Step 10: Now look at your latest work very carefully… What you will notice is that the lengths AB, AC and BC are all equal to one another. This is because both the circles you drew – are exactly the same size. They each have radiuses equal in proportion. In simple terms, AB=AC=BC.

What you have to do now is name these lengths (r) for radius. Here’s the thing though, because the line going through C and D splits the triangle (equilateral, as each of its sides has the same length) down its middle, the length AE is equal to 1/2 x r, and similarly the length BE is equal to 1/2 x r. Together, the length AE + BE = AB = r.

Step 11: Remember that I said that the line going through C and D splits the triangle down its middle. Also, notice that this exact line is perpendicular to the length AB. Now, because of this, at the point E, you’ve got two right angles. Name these two right angles big R.

[Knowing that these two angles are equal to 90 degrees is vital – because you’ll be able to use Pythagoras’ theorem to find the length CE.]

Step 12: Find the length CE using Pythagoras’ theorem, Adjacent² + Opposite² = Hypotenuse². You will need this length to find the area of the equilateral triangle you’ve produced.

*Algebraic skills will be required from this point…

${ AE }^{ 2 }+{ CE }^{ 2 }={ AC }^{ 2 }\\ \\ \Rightarrow \quad { \left( \frac { 1 }{ 2 } r \right) }^{ 2 }+{ CE }^{ 2 }={ r }^{ 2 }\\ \\ \Rightarrow \quad { CE }^{ 2 }={ r }^{ 2 }-{ \left( \frac { 1 }{ 2 } r \right) }^{ 2 }\\ \\ \Rightarrow \quad { CE }^{ 2 }=\frac { 4r^{ 2 } }{ 4 } -\frac { { r }^{ 2 } }{ 4 } \\ \\ \Rightarrow \quad { CE }^{ 2 }=\frac { 3{ r }^{ 2 } }{ 4 } \\ \\ \Rightarrow \quad CE=\sqrt { \frac { 3{ r }^{ 2 } }{ 4 } } \\ \\ \therefore \quad CE=\frac { r\sqrt { 3 } }{ 2 }$

Step 13: Derive the formula for the area (A) of the equilateral triangle. Remember that the area of a right angled triangle is L x W x 1/2.

$A=\frac { 1 }{ 2 } r\cdot \frac { r\sqrt { 3 } }{ 2 } \cdot \frac { 1 }{ 2 } +\frac { 1 }{ 2 } r\cdot \frac { r\sqrt { 3 } }{ 2 } \cdot \frac { 1 }{ 2 } \\ \\ =\frac { 1 }{ 8 } { r }^{ 2 }\sqrt { 3 } +\frac { 1 }{ 8 } { r }^{ 2 }\sqrt { 3 } \\ \\ =2\cdot \frac { 1 }{ 8 } { r }^{ 2 }\sqrt { 3 } \\ \\ =\frac { 1 }{ 4 } { r }^{ 2 }\sqrt { 3 }$

Presto!!! Keep in mind that you can transform the variable (r) into any variable you wish. This variable (r) is the length of each side of the equilateral triangle you were working with. The formula you’ve derived can be used to find the area of any equilateral triangle.

# Proof: Thales’ Theorem

In this post I’ll be demonstrating how you can prove that Thales’ Theorem is true. To follow the steps in this post (11 in total), what you will require is a ruler, pair of compasses and a pencil.

##### Step 11: Prove that the angle at point D is equal to 90 degrees.

Thales’ Theorem is as follows:

Because AC is the diameter of the circle you drew, the angle at the point D (α+β) must be equal to 90 degrees. In more specific and general terms, if you have the points A, C and D lying on a circle – and the line AC is in fact the diameter of this circle – then the angle at point D (α+β) must be a right angle.

Proof (which must be derived using the diagram you’ve created):

All angles within a triangle (in 2 space) must add up to 180 degrees.

Mathematically, this means that:

$\alpha +\alpha +\beta +\beta =180\\ \\ \Rightarrow \quad 2\alpha +2\beta =180\\ \\ \Rightarrow \quad 2\left( \alpha +\beta \right) =180\\ \\ \Rightarrow \quad \frac { 2\left( \alpha +\beta \right) }{ 2 } =\frac { 180 }{ 2 } \\ \\ \therefore \quad \alpha +\beta =90$

And as a result, Thales’ theorem must be true. The angle α+β is the angle at point D.

# Proof: Opposite angles formed when two lines intersect, are equal to one another

How can we prove that opposite angles (when two lines intersect) are in fact equal to one another?

Well, first of all – let’s draw a circle…

We know that in a full circle, there are 360 degrees. This is an indisputable fact. Now, what happens if we split this circle in two with a straight line (going through its centre)?

Well, each half of the circle (top and bottom) – will now contain 180 degrees. We know this because:

$\frac { 360 }{ 2 } =180$

Ok, so far so good… Now, let’s draw another line through the circle (going through its centre) which intersects the first line we have drawn…

As we can see, because we have done this, we now have 4 different angles. Let’s name the two angles which are situated in the top half of the circle α and β

Earlier in this demonstration, we remarked that the top half of the circle (when it was split in two) contained 180 degrees. Mathematically and logically speaking, as this is the case, we must say that:

$\alpha +\beta =180$

Great, now let’s name the angles in the bottom half of the circle x and y

It follows, because the angles in the top half of the circle add up to 180 degrees, we must deduce that:

$x+y=180$

So, it turns out we now have two useful equations:

$\alpha +\beta =180$

$x+y=180$

Do we have enough to form our proof though? Unfortunately, not quite… We have to look at our most recent figure again, but this time from a different perspective…

You see, there are different top halves and bottom halves…

• There exists top halves α+β and also… β+y
• There exists bottom halves x+y but also α+x

You may ask, why is this important? Well, here’s what’s crucial:

$\beta +y=180$

$\alpha +x=180$

And now we have 4 different equations, 3 of which – will help us finally complete our proof.

$\alpha +\beta =180$

$x+y=180$

$\beta +y=180$

$\alpha +x=180$

Here’s why we need these equations…

α+β and α+x are equivalent (180 degrees), so we can deduce that:

$\alpha +\beta =\alpha +x\\ \\ \therefore \quad \beta =x$

*Subtract α from both sides of the equation.

α+β and β+y are equivalent (180 degrees), so we can deduce that:

$\alpha +\beta =\beta +y\\ \\ \therefore \quad \alpha =y$

*Subtract β from both sides of the equation.

Hence, we’ve proven that: Opposite angles (when two lines intersect) are equal to one another.

β=x and α=y:

# How to prove that sin(A-B)=sin(A)cos(B)-cos(A)sin(B) geometrically

In this post I’ll be demonstrating how one can prove that sin(A-B)=sin(A)cos(B)-cos(A)sin(B) geometrically…

First of all, let me show you this diagram…

sin(A-B)=sin(A)cos(B)-cos(A)sin(B) proof

*If you click on the diagram, you will be able to see its full size version.

Now, to begin with, I will have to write about some of the properties related to the diagram…

Property 1:

Angle B + (A – B) = B + A – B = A

Therefore, angle POR = A.

Property 2:

Angle OPS = 90 degrees

Property 3:

Length OS = 1

Also note:

All angles within a triangle on a flat plane should add up to 180 degrees. If you understand this rule, you will be able to discover why the angles shown on the diagram are correct. Angles which are 90 degrees are shown on the diagram too.

PROVING THAT SIN(A-B)=SIN(A)COS(B)-COS(A)SIN(B)

Since I’ve noted down some of the important properties related to the diagram, I can now focus on demonstrating why the formula above is true. I will demonstrate why the formula above is true using mathematics and the SOH CAH TOA rule…

$\sin { \left( A-B \right) } =\frac { O }{ H } =\frac { ST }{ 1 } =ST$

But it turns out that…

$ST=PR-PQ$

Because:

$QR=ST$

Now, what is PR and what is PQ?

$\sin { \left( B \right) } =\frac { O }{ H } =\frac { PS }{ 1 } =PS\\ \\ \cos { \left( B \right) } =\frac { A }{ H } =\frac { OP }{ 1 } =OP\\ \\ \sin { \left( A \right) } =\frac { O }{ H } =\frac { PR }{ \cos { \left( B \right) } } \quad \\ \\ \therefore \quad \sin { \left( A \right) } \cos { \left( B \right) } =PR\\ \\ \cos { \left( A \right) } =\frac { A }{ H } =\frac { PQ }{ \sin { \left( B \right) } } \\ \\ \therefore \quad \cos { \left( A \right) } \sin { \left( B \right) } =PQ$

And finally, to sum it all up:

$ST=PR-PQ\\ \\ \therefore \quad \sin { \left( A-B \right) =\sin { \left( A \right) } \cos { \left( B \right) } -\cos { \left( A \right) } \sin { \left( B \right) } }$

Need a better explanation? Watch this video…

Related Videos:

Related posts:

Simple But Elegant Way To Prove That sin(A+B)=sinAcosB+cosAsinB (Edexcel Proof Simplified)

# 2 ways to derive Pythagoras’ equation from scratch

The other day I discovered one more way to derive Pythagoras’ equation from scratch, completely by accident. I was deriving Pythagoras’ equation using the usual method, whilst navigating  a diagram similar to the one below, but without (B-A) measurements…

*Note (regarding diagram above): x+y = 90 degrees

The usual method goes like this…

The area of the largest square is:

${ \left( A+B \right) }^{ 2 }$

It is also:

$4\cdot \frac { 1 }{ 2 } AB+{ C }^{ 2 }$

Which means that:

${ \left( A+B \right) }^{ 2 }=4\cdot \frac { 1 }{ 2 } AB+{ C }^{ 2 }\\ \\ { A }^{ 2 }+2AB+{ B }^{ 2 }=2AB+{ C }^{ 2 }\\ \\ \therefore \quad { A }^{ 2 }+{ B }^{ 2 }={ C }^{ 2 }$

Now, when I added the lengths (B-A) to my diagram, which are included in the diagram above, I discovered a new way to derive Pythagoras’ equation…

I did this by focusing on the area C^2. It turns out that:

$4\cdot \frac { 1 }{ 2 } AB+{ \left( B-A \right) }^{ 2 }={ C }^{ 2 }$

And since:

${ \left( B-A \right) }^{ 2 }\\ \\ ={ \left( A+B \right) }^{ 2 }-4AB\\ \\ ={ A }^{ 2 }+2AB+{ B }^{ 2 }-4AB\\ \\ ={ B }^{ 2 }-2AB+{ A }^{ 2 }$

I was able to say that:

$4\cdot \frac { 1 }{ 2 } AB+\left\{ { B }^{ 2 }-2AB+{ A }^{ 2 } \right\} ={ C }^{ 2 }\\ \\ 2AB+{ B }^{ 2 }-2AB+{ A }^{ 2 }={ C }^{ 2 }\\ \\ \therefore \quad { A }^{ 2 }+{ B }^{ 2 }={ C }^{ 2 }$

Obviously, I was quite pleased. Have you discovered other ways in which to derive Pythagoras’ equation??

Related:

Video on how to come up with Pythagoras’s equation…

How To Come Up With Pythagoras’s Equation

# How to find the square root of 2 using simple algebraic rules

Ever wondered how to discover the value of the square root of 2 using simple algebraic rules?

I’ve been through plenty of maths books and 99.9% of them haven’t demonstrated how this feat can be accomplished.

Today I’m going to share with you the methods which you can use to find the value of irrational numbers as continued fractions, and in the process you will learn how to write the value of the square root of 2 as a continued fraction.

See the mathematics below…

${ x }^{ 2 }=2\\ \\ { x }^{ 2 }+x=2+x\\ \\ x\left( x+1 \right) =\left( x+1 \right) +1\\ \\ \frac { x\left( x+1 \right) }{ \left( x+1 \right) } =\frac { \left( x+1 \right) }{ \left( x+1 \right) } +\frac { 1 }{ \left( x+1 \right) } \\ \\ x=1+\frac { 1 }{ x+1 } \\ \\ x=1+\frac { 1 }{ \left( 1+\frac { 1 }{ x+1 } \right) +1 }$

$\\ \\ x=1+\frac { 1 }{ 2+\frac { 1 }{ x+1 } } \\ \\ x=1+\frac { 1 }{ 2+\frac { 1 }{ \left( 1+\frac { 1 }{ x+1 } \right) +1 } } \\ \\ x=1+\frac { 1 }{ 2+\frac { 1 }{ 2+\frac { 1 }{ x+1 } } } \\ \\ x=1+\frac { 1 }{ 2+\frac { 1 }{ 2+\frac { 1 }{ \left( 1+\frac { 1 }{ x+1 } \right) +1 } } } \\ \\ x=1+\frac { 1 }{ 2+\frac { 1 }{ 2+\frac { 1 }{ 2+\frac { 1 }{ x+1 } } } } \\ \\ x=1+\frac { 1 }{ 2+\frac { 1 }{ 2+\frac { 1 }{ 2+\frac { 1 }{ \left( 1+\frac { 1 }{ x+1 } \right) +1 } } } }$

$\\ \\ x=1+\frac { 1 }{ 2+\frac { 1 }{ 2+\frac { 1 }{ 2+\frac { 1 }{ 2+... } } } } \\ \\ x=\sqrt { 2 } \\ \\ \therefore \quad \sqrt { 2 } =1+\frac { 1 }{ 2+\frac { 1 }{ 2+\frac { 1 }{ 2+\frac { 1 }{ 2+... } } } }$