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_{K}, to be 0.000±0.005, consistent with a flat universe. Many observational cosmological probes revealed that the universe is flat obeying the classical Euclidean geometry. But till this day, there is no mathematical formulation/proof for the geometry of our universe. In this work, the attempts to establish that the shape of our universe is flat.

The density parameter Ω, the curvature parameter k and the Hubble parameter H are related as [1-5].

(1 – Ω ) = - kc^{2} / H^{2}R^{2} (1)

If omega less than 1, k is less than 1

If omega is equal to 1, k is zero

If omega greater than 1, k is +1.

If k is -1, the geometry of the universe is open,

If it is greater than one, the shape of the universe is closed

And the universe obeys Euclidean geometry if k is equal to zero.

I.e if Ω = 1 , the universe is Euclidean,

If Ω = less than 1 , the geometry of the universe is open,

And if Ω = greater than 1, the universe is closed,

Let –n = - kc^{2} / H^{2}R^{2} in (1) (1a)

Assuming (1a) in (2), 1 – Ω + n = 0

Squaring, 1 + Ω^{2} + n^{2} - 2 Ω – 2n Ω – 2n = 0

i.e (Ω – n ) ^{2} = 2 Ω +2n - 1 = 0

From (1a) we have, Ω – n = 1. Putting this in the first factor of the above equation,

= 2 Ω +2n - 1

Simplifying, 1 = Ω +n

But from eqn. (1a) ,1 = Ω - n

Adding the above two relations, 1 = Ω (2)

As we have previously seen in Ω is equal to 1, the curvature of our universe is zero and the geometry of our universe is flat [6].

The density parameter Ω, the curvature parameter k and the Hubble parameter H are related as [7,8]

(1 – Ω ) = - kc^{2} / H^{2}R^{2} (1)

If omega less than 1, k is less than 1

If omega is equal to 1, k is zero

If omega greater than 1, k is +1.

If k is -1, the geometry of the universe is open,

If it is greater than one, the shape of the universe is closed

And the universe obeys Euclidean geometry if k is equal to zero.

I.e if Ω = 1 , the universe is Euclidean,

If Ω = less than 1 , the geometry of the universe is open,

And if Ω = greater than 1, the universe is closed.

For our convenience, let us assume in (1), – n = - kc^{2} / H^{2}R^{2}

So, (1 – Ω ) = - n (1a)

Applying (1a) and cubing (1) we get that,

1 – Ω^{3} -3 Ω ( 1 – Ω) = - n^{3}

i.e( n^{3} – Ω^{3}) + 1 – 3 Ω (1 – Ω ) = 0

By applying the famous algebraic cubic formula a3 – b3 =

(a-b)3 + 3ab ( a b ) in the first factor of the above relation we obtain that,

( n – Ω ) ^{3} + 3n Ω (n – Ω ) = -1 +3 Ω (1 – Ω )

From (1a) we have , n – Ω = - 1

Putting this relation in the above eqn. we have, n (n – Ω ) = Ω (1 – Ω ) (1b)

Again applying (1a) in RHS, n (n – Ω ) = - n Ω (3)

From (1a) we also have, n – Ω = -1

By assuming the above relation in the LHS of (2) we get - n = -n Ω

By simplifying we get that Ω = 1 (4)

From equations (2)and (4) we can conclude that the geometry of our Universe is FLAT.

The shape of the entire relies on the following properties:

1. Finite or infinite

2. The geometry is flat, hyperbolic or elliptic

Simply connected space or multiply connected space

The exact shape is a burning problem in physical cosmology. Several experimental and observational data WMAP, PLANCK, BOOMERanG confirm that the universe is flat with only a 0.4% margin of error. Theorists believe that the universe is flat and infinite. Our finding adds more and more favorable arguments for the shape and fate of our universe.

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© 2020 Kalimuthu S. This is an open-ampcess article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.

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