Prove that if p(x) is polynomial with integer coefficients and p(√2)=0 then p( -√2 )=0.

Prove that if p(x) is polynomial with integer coefficients and p(√2)=0 then p( -√2 )=0.

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TopNewestdavid mcmillan see this

Same thing Sandeep Bhardwaj said – Megh Choksi · 2 years, 5 months ago

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– Brian Charlesworth · 2 years, 5 months ago

But \(p(2) = 0\) doesn't imply that \(p(-2) = 0\). How else should I be reading "\(p(2)\) then \(( -2 ) = 0\)" ?Log in to reply

– Megh Choksi · 2 years, 5 months ago

Sir he meant - \(p(\sqrt{2})\)Log in to reply

– David Mcmillan · 2 years, 5 months ago

Sorry for the typo. I meant \( p(-\sqrt{2})\)Log in to reply

To save some time, I'm just going to give you a link to a proof of the general theorem, (which was surprisingly hard to find). Scroll down to theorem 16 on page 14 for the desired proof. You can adapt this proof for the special case \(a = 0, b = 1\) to simplify it a bit. Good question; I don't think I've ever come across a proof of this theorem before. :) – Brian Charlesworth · 2 years, 5 months ago

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Do you mean that if \(p(2) = 0\) then \(p(-2) = 0\)? If so, then this is not the case. If \(p(x) = x^{2} - x - 2\) then \(p(2) = 0\) but \(p(-2) = 4\). Perhaps you meant something else with the terminology "( -2 ) = 0"?

What can be said is that if a complex number \(z\) is a root of a polynomial then so is its conjugate \(\bar {z}\), i.e., \(p(z) = 0 \Leftrightarrow p(\bar {z}) = 0\). – Brian Charlesworth · 2 years, 5 months ago

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@Calvin Lin @brian charlesworth @Sreejato Bhattacharya Please help. – David Mcmillan · 2 years, 5 months ago

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@Sudeep Salgia @Sandeep Bhardwaj Also try and help – David Mcmillan · 2 years, 5 months ago

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– Sandeep Bhardwaj · 2 years, 5 months ago

This is the basic property of a polynomial equation. In fact, this is true for all rational coefficients. You try to prove it by taking a quadratic equation and then try to generalize that.Log in to reply