Assuming that the square root of 2 is rational and is = to an even number divided by an odd number.2 multiplied by the odd number squared is= to the even number squared
It is not clear what your question is, but perhaps you are trying to prove that sqrt2 is irrational. If you assume it is a ratio of two integers reduced to lowest terms (and therefore rational), you get a contradiction.
For a proof, see
http://www.homeschoolmath.net/teaching/proof_square_root_2_irrational.php
To determine whether the statement is true or false, we can apply the properties of rational and irrational numbers.
Let's assume that the square root of 2 is rational and can be expressed as an even number divided by an odd number, such as (2k/m), where k and m are integers.
According to the given statement, (2 × odd number^2) should be equal to (even number^2). Let's analyze this further:
(2 × odd number^2) = (2 × (2k + 1)^2) = 2(4k^2 + 4k + 1) = 8k^2 + 8k + 2
(even number^2) = (2k/m)^2 = (4k^2/m^2)
Now we need to compare these two expressions:
8k^2 + 8k + 2 ≠ 4k^2/m^2
Since the left-hand side is not equal to the right-hand side, we can conclude that our assumption is incorrect. Therefore, the square root of 2 is not a rational number and cannot be expressed as an even number divided by an odd number in order for the given statement to be true.