basic question

prover

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Oct 16, 2021
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so in my first calaules class the professor presented a problem i found hard to solve. will very appreciate the help.

x is irrational
x^3 is rational

prove or disprove

x^2 + x + 1 is rational
 

mmm4444bot

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Hello. Please share your thoughts and any work that you tried on paper, to give us a starting point for tutoring. Thank you.

 

Jomo

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You asked for help. Excellent! What do you need help with?
 

prover

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Oct 16, 2021
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i think that is it irrational, so i tried to assume by cintridiction that it is rational.
i comperd the expression to n/m (n and m are integers, m is not zero) and tried to do differnt algebric manipulations to the equation.
i also tried to divide the entire expresion by x^3 and then to assume it is rational but encuterd the same problems.
i tried to find a counter example but culdent find one i can prove is irrational without a calcultor.
i am a little lost at the moment and not sure which direction is practical and would very appreciate sume guidance.
thank you
 

Dr.Peterson

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Nov 12, 2017
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12,662
i think that is it irrational, so i tried to assume by cintridiction that it is rational.
i comperd the expression to n/m (n and m are integers, m is not zero) and tried to do differnt algebric manipulations to the equation.
i also tried to divide the entire expresion by x^3 and then to assume it is rational but encuterd the same problems.
i tried to find a counter example but culdent find one i can prove is irrational without a calcultor.
i am a little lost at the moment and not sure which direction is practical and would very appreciate sume guidance.
thank you
Have you observed that x^2 + x + 1 is a factor of an expression involving x^3?
 

Subhotosh Khan

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so in my first calaules class the professor presented a problem i found hard to solve. will very appreciate the help.

x is irrational
x^3 is rational

prove or disprove

x^2 + x + 1 is rational
Use the fact that:

a3 - b3 = (a - b) * (a2 + ab + b2)
 

Otis

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Apr 22, 2015
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i tried to find a counter example but [couldn't] find one
Hi prover. That was my first thought because it's the easiest way to disprove that the given quadratic polynomial must represent a Rational number.

Here are some things to consider. Is the cube root of a prime number Rational? Are you familiar with the standard proof that [imath]\sqrt{2}[/imath] is Irrational? We can do something similar.

Let p = prime number
Let a/b = reduced Rational number

[imath]\sqrt[3]{p} = \frac{a}{b}[/imath]

Therefore [imath]a^3 = b^3 \cdot p[/imath]

Each side is divisible by a, but b itself is not. What can you conclude? Does that help you pick [imath]x^3[/imath] and [imath]x[/imath] values for a counter example?

😎
 
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