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Proof

Prove that any irrational number can be a root of at most one cubic equation of the form x^3+ax =b where a and b are rational.

How do I get started?
Original post by rdj201
Prove that any irrational number can be a root of at most one cubic equation of the form x^3+ax =b where a and b are rational.

How do I get started?


Proof by contradiction. Assume there is an irrational number that is the root of two equations of that form, and show this leads to a contradiction.
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rdj201
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Original post by ghostwalker
Proof by contradiction. Assume there is an irrational number that is the root of two equations of that form, and show this leads to a contradiction.


Thank you...still struggling at the moment, can I have a bit more help...
(edited 7 years ago)
Original post by rdj201
Thank you...still struggling at the moment, can I have a bit more help...


Based on the suggested assumption, form and solve two simultaneous eqns.
Original post by ghostwalker
Proof by contradiction. Assume there is an irrational number that is the root of two equations of that form, and show this leads to a contradiction.


PRSOM
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rdj201
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Original post by atsruser
Based on the suggested assumption, form and solve two simultaneous eqns.
.

Is this sufficient...
If n is an irrational number.
If assume it works for two equations x^3 + ax - b = 0 and x^3 + cx - d = 0.
Then
f(n) = n^3 + an - b = 0
and
f(n) = n^3 + cn - d = 0

So n^3 + an - b = n^3 + cn - d
an - b = cn - d
an - cn = b - d
n = (b-d)/(a-c)
If a, b, c and d are rational numbers, then n must be rational.
Original post by rdj201
.

Is this sufficient...
If n is an irrational number.
If assume it works for two equations x^3 + ax - b = 0 and x^3 + cx - d = 0.
Then
f(n) = n^3 + an - b = 0
and
f(n) = n^3 + cn - d = 0

So n^3 + an - b = n^3 + cn - d
an - b = cn - d
an - cn = b - d
n = (b-d)/(a-c)
If a, b, c and d are rational numbers, then n must be rational.


In essence, yes. I would tighten up the phraseology a bit. Also you need to deal with the case a=c, since you've dividing by a-c.

Here's my take:

Spoiler

(edited 7 years ago)
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rdj201
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Original post by ghostwalker
In essence, yes. I would tighten up the phraseology a bit. Also you need to deal with the case a=c, since you've dividing by a-c.

Here's my take:

Spoiler




Thank you :-)
Original post by rdj201
Thank you :-)


Your welcome. Your answer was spot on - I was only dotting i's and crossing t's.

Can't recall seeing a proof by contradiction being required at A-level before - could just be my memory though.
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rdj201
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Original post by ghostwalker
Your welcome. Your answer was spot on - I was only dotting i's and crossing t's.

Can't recall seeing a proof by contradiction being required at A-level before - could just be my memory though.


It is a suggestion for the new A Level
Original post by ghostwalker
Your welcome. Your answer was spot on - I was only dotting i's and crossing t's.

Can't recall seeing a proof by contradiction being required at A-level before - could just be my memory though.


The combination of the words "proof" and "A level" switched my brain off - I couldn't see what they were getting at, using standard A level methods.

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