If $a$ is an algebraic number and $b$ is a rational number, then show that $ab$ is algebraic number. My attempt is to prove it by contradiction, but it failed. Can someone please help me?
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Try proving all rational number are algebraic and the product of two numbers is algebraic. If you are having trouble, refer to https://math.stackexchange.com/questions/527508/showing-rational-numbers-are-algebraic and https://math.stackexchange.com/questions/155122/how-to-prove-that-the-sum-and-product-of-two-algebraic-numbers-is-algebraic – JC12 Apr 21 '20 at 23:50
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1Does this answer your question? How to prove that the sum and product of two algebraic numbers is algebraic? – Feng Apr 21 '20 at 23:55
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@JC12 The proof for the fact that product of two algebraic numbers is algebraic isn't elementary, however this question can be answer by elementary methods as in the answer below – Anshuman Agrawal May 17 '22 at 18:55
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Let $$f(x)=x^n + r_{n-1}x^{n-1}+\cdots +r_1x + r_0$$ be a polynomial with rational coefficients such that $f(a)=0$.
Then $$a^n + r_{n-1}a^{n-1} + \cdots + r_1a + r_0 = 0.$$ Multiply through by $b^n$. We get $$b^na^n + r_{n-1}bb^{n-1}a^{n-1} + \cdots + r_1b^{n-1}ba + r_0b^n = 0.$$ Can you finish it off?
Arturo Magidin
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