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 August 21st, 2017, 08:07 PM #1 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2 Algebraic Solutions I didn't understand the meaning of the below sentences, if it means the coefficients/ solutions of an Algebraic equations only lies in Real numbers then it will not be true since we can have a polynomial of the below type with Complex coefficients and Complex solutions as well. If it does mean something else please let me know "The solution of an Algebraic equation of the form $\displaystyle a_0 x^n+a_1 x^{n-1}+....+a_n = 0$ where $\displaystyle a_0, a_1,a_2,....a_n$ are integers, are called Algebraic numbers. These Algebraic numbers are either Rational or Irrational numbers i.e., Reals. The numbers which are not solution of an Algebraic equations are known as transcendental numbers." eg : $\displaystyle \pi,e$
 August 21st, 2017, 08:17 PM #2 Senior Member   Joined: Aug 2012 Posts: 1,956 Thanks: 547 Yes you're right, that's slightly ambiguous. $x^2 + 1 = 0$ is a polynomial with integer coefficients with its roots in the complex numbers. In general, we'd call those roots algebraic numbers. If we intersect the complex algebraic numbers with the reals we get the real algebraics. Typically when we say "algebraic number" we often mean real algebraic number. But in general we mean the complex algebraic numbers too. I'd call that paragraph a bit imprecise but it's fairly common. The default is to consider the algebraic numbers to be a subset of the reals unless the problem states otherwise. This is a common imprecision since the context is usually clear. Wiki takes the general point of view. "An algebraic number is any complex number [my emphasis] that is a root of a non-zero polynomial in one variable with rational coefficients (or equivalently – by clearing denominators – with integer coefficients)." https://en.wikipedia.org/wiki/Algebraic_number If memory serves, a complex number is algebraic if and only if its real and imaginary parts are real algebraic. Might be worth trying to prove. Last edited by Maschke; August 21st, 2017 at 08:20 PM.
August 21st, 2017, 08:21 PM   #3
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 Originally Posted by Maschke Yes you're right, that's slightly ambiguous. $x^2 + 1 = 0$ is a polynomial with integer coefficients with its roots in the complex numbers. In general, we'd call those roots algebraic numbers. If we intersect the complex algebraic numbers with the reals we get the real algebraics. Typically when we say "algebraic number" we often mean real algebraic number. But in general we mean the complex algebraic numbers too. I'd call that paragraph a bit imprecise but it's fairly common. The default is to consider the algebraic numbers to be a subset of the reals unless the problem states otherwise. This is a common imprecision since the context is usually clear. If memory serves, a complex number is algebraic if and only if its real and imaginary parts are real algebraic. Might be worth trying to prove.
Does it mean, since the context only talking about the Reals, we are considering the Algebraic numbers as Reals and when we abstract the topic it will also include the complex numbers ??

 August 21st, 2017, 08:30 PM #4 Senior Member   Joined: May 2016 From: USA Posts: 1,047 Thanks: 430 The definition you gave makes clear that the coefficients are integers. That same definition implies, without being explicitly clear, that the solutions being discussed are real solutions. Real algebraic numbers are real zeros of polynomial functions with integer coefficients. Transcendental numbers are real numbers that are not algebraic numbers. Notice that all rational numbers are algebraic. $px - q = 0 \implies x = \dfrac{q}{p}, \text { where } p, q \in \mathbb Z \text { and } p \ne 0.$ All integer roots of positive rational numbers are algebraic numbers. $x^n - r = 0 \implies x = \sqrt[n]{r}, \text { where } r \in \mathbb Q,\ r > 0, \text { and } n \in \mathbb Z.$ Note that an algebraic number may be irrational. Example. $x^2 - 2 = 0 \implies x = \sqrt{2} \text { or } x = \sqrt{2}.$ So all transcendental numbers are irrational, but not all irrational numbers are algebraic. Last edited by skipjack; August 21st, 2017 at 09:02 PM.
August 21st, 2017, 08:31 PM   #5
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 Originally Posted by Lalitha183 Does it mean, since the context only talking about the Reals, we are considering the Algebraic numbers as Reals and when we abstract the topic it will also include the complex numbers ??
Depends on the context. When in doubt, ask. If you're studying real analysis they probably mean the real algebraics if they don't specify.

August 21st, 2017, 08:33 PM   #6
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 Originally Posted by JeffM1 The definition you gave makes clear that the coefficients are integers. That same definition implies, without being explicitly clear, that the solutions being discussed are real solutions.
Then what are $i$ and $-i$? They're algebraic. But not real. They satisfy the Wiki definition but not the OP's definition. That's why the definition of algebraic numbers is often ambiguous in this way.

August 21st, 2017, 08:48 PM   #7
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 Originally Posted by Maschke Then what are $i$ and $-i$? They're algebraic. But not real. They satisfy the Wiki definition but not the OP's definition. That's why the definition of algebraic numbers is often ambiguous in this way.
It's why I said the definition implied that only real solutions were being considered. "These algebraic numbers are ... reals."

It was a sloppy definition. I really do not think we have a big disagreement. I was trying to clarify what was intended by a poor definition. You were saying that the term is frequently used ambiguously and suggesting that "real algebraic number" is a better term. I agree. I could perhaps have phrased my clarification a bit better than I did.

 August 21st, 2017, 10:59 PM #8 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2 I just want to make clear that, I got these notes from a lecture on Dedekind's theory of Irrationals. So they might have considered only Reals in this context.

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