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Archangel Macsika

Use a proof by contradiction to show that there is no rational number r for which r+r+1=0. (Assume that r=a/h is a root, where a and b are integers and a/b is in lowest terms.obtain an equation involving integers by multiplying by then look at whether a and bare each odd or even.

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Use a proof by contradiction to show that there is no rational number r for which r^3 + r + 1 = 0.

Proof

Assume that r = a/b is a root, where a and b are integers and a/b is in lowest terms.

We see that

if a is odd, then b is even,

if a is even, then b is odd.

Substitute


\dfrac{a^3}{b^3}+\dfrac{a}{b}+1=0

Multiply by b^3\not=0


a^3+ab^2+b^3=0

If a is odd, b is even, then

a^3is odd, ab^2 is even, b^3 is even. Hence the sum a^3+ab^2+b^3 is odd. But zero is an even number.

We have a contradiction in this case.


If a is even, b is odd, then

a^3is even, ab^2 is even, b^3 is odd. Hence the sum a^3+ab^2+b^3 is odd. But zero is an even number.

We have a contradiction in this case.


Therefore our assumption leads to the contradiction.

Therefore there is no rational number r for which r^3 + r + 1 = 0.


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Question ID: mtid-5-stid-8-sqid-2627-qpid-1097