Mathematical Awareness #2
Statement S : All real transcendental numbers are irrational numbers.
Statement A : Some irrational numbers are transcendental numbers.
Statement N : ( − 1 ) − 1 is not a transcendental number.
Statement D : ( π + e ) and ( π × e ) : Both are not transcendental numbers.
Statement E : 2 and 3 are both transcendental as well as irrational numbers.
Statement E : Sum of two transcendental numbers is always a transcendental numbers.
Statement P : Some irrational numbers are not transcendental numbers.
Which of the above statements are correct?
Note : There are two statements above named as "E". It is provided that either both the "E" statements are correct or both are incorrect. In the former case "EE" will be included in the answer and in the later case, no "E" statement will be counted in the answer.
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1 solution
For those who have problem with Statement D :
For any two transcendental numbers p and q , at least one of ( p + q ) and p q must be transcendental.
Proof :
Lets make a polynomial equation with roots p and q . i.e. ( x − p ) ( x − q ) = 0
Now we can write : ( x − p ) ( x − q ) = x 2 − ( p + q ) x + p q .
If both ( p + q ) and p q are algebraic , then x 2 − ( p + q ) x + p q will be a polynomial with algebraic coefficients, which implies the roots of the polynomial equation x 2 − ( p + q ) x + p q = 0 must be algebraic. But it is a contradiction to our initial supposition that both p and q are transcendental numbers.
So we can say that at least one of ( p + q ) and p q must be transcendental.
Same is with the case of π and e , both being the transcendental numbers.
Great solution. However, you're missing the second "E" statement.
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It was provided in the problem that either both statements were true or both were false. However i updated the solution.
Then I will answer that part to a certain degree.
This is not a definite proof, but a way of how I thought that the sum of two transcendental numbers do not necessarily add up to a trancendental number. It is because there are infinitely many transcendental numbers, also meaning its addition can also be anything. This gains more credibility with Cantor's work that establishes that there are as many transcendental numbers as there are algebraic numbers. If the summation can be anything, why can't it be an algebraic number?
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Nope. That reasoning is fallacious. For example, replace "transcendental" with "integer" and "algebraic" with "rational numbers with ∣ q ∣ = 1 ."
Your reasoing would then imply that the sum of two integers can be a strictly rational number.
Just consider π and 1 − π .
S − All transcendental numbers are irrational since all rational numbers are algebraic (all numbers of the form b a are a solution to the equation b x − a = 0 ).
A − An example can be π .
N − − 1 − 1 = i i = e − 2 π which is transcendental by the Gelfond–Schneider theorem .
D − Quote from this page
E 1 − Neither 2 nor 3 are transcendental as they're solutions to the equations x 2 − 2 = 0 and x 2 − 3 = 0 respectively.
E 2 − Since π is transcendental and any non-constant algebraic function of a single variable yields a transcendental value when applied to a transcendental argument then 1 − π is also transcendental. We can clearly see that π + ( 1 − π ) = 1 is algebraic.
P − One example can be 2 as it's known to be irrational but not transcendental.
The only correct option is P A S .