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can anyone tell me why the transcendental numbers are uncountable?

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can anyone tell me why the transcendental numbers are uncountable?

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Hurkyl

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How many real numbers are there? How many algebraic numbers are there?

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HallsofIvy

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i didn't understand this. could you elaborate a little?HallsofIvy said:Since a polynomial of degree n has exactly n coefficients, there are a countable number of such polynomials.

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Consider this definition of an algebraic number:

Let A denote the set of algebraic numbers. Then z is in A IFF z is a root of a ploynomial with integer coefficients (sometimes stated with rational coefficients, but this is equivalent).

That is z is in A IFF there exists integers a0,a1,a2,...,an not all zero such that the polynomial P(z):=sum(ak*z^k, k=0..n)=0.

You can prove A is countably infinite by:first defining the height of a polynomial as the sum of the absolute values of its coefficients and its degree, e.g., |P(z)|=|a0|+|a1|+...+|an|+n for the above polynomial.

Then prove that there are finitely many polynomials of a given height, and that each such polynomial has finitely many roots (use the fundamental theorem of algebra for the second part).

Let the (finite) union of the sets of the roots of polynomials of height m be denoted by Hm. Then A is the (countable) union of Hm over all positive integers m (why?) Therefore A is countable (why?)

The set of transdental numbers must then be uncountable (why?)

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how??? can you explain further? i understood the rest.benorin said:You can prove A is countably infinite by:first defining the height of a polynomial as the sum of the absolute values of its coefficients and its degree, e.g., |P(z)|=|a0|+|a1|+...+|an|+n for the above polynomial.

Then prove that there are finitely many polynomials of a given height, and that each such polynomial has finitely many roots (use the fundamental theorem of algebra for the second part).

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HallsofIvy

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Do you know this theorem: If A and B are countable sets then AxB (Cartesian product: ordered pairs where first member is in A and second in B) is countable. To prove that, imagine that you make a table bymurshid_islam said:i didn't understand this. could you elaborate a little?

The set of all pairs (a,b) where a and b are both integers is countable because it is NxN and N is countable. Of course, we can identify the linear polynomial a+ bx with (a,b).

It is then easy to show that the Cartesian product of

But we can identify the polynomial a+ bx+ cx

In general, the Cartesian product of any finite number of countable sets is countable and we can identify a+ bx+ cx

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