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1. 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?
2. (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.
3. (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...
4. (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.
5. (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
6. (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.
7. (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:
Show

Let n be an irrational number for which there exist two distinct equations x^3 + ax - b = 0 and x^3 + cx - d = 0, where n is a root of each of them.

We then have:

n^3 + an - b = 0..............(1)
n^3 + cn - d = 0..............(2)

(I avoid f(n) since this is one specific function. If you want to use that then you'd have f(n) and g(n))

Subtracting we get:

(a-c)n -(b-d)=0

If a=c, then (0)n- (b-d) = 0 and it follows that b-d = 0 and b=d, and they are the same equation.

So, a not = c and we then have n=(b-d)/(a-c)

Since all the terms on the right are rational, it follows that n is rational.

This is a contradiction, since n is irrational.

Therefore no such n can exist and any irrational number can the root of at most one equation of the form x^3 + cx - d = 0

QED.

8. (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:
Show

Let n be an irrational number for which there exist two distinct equations x^3 + ax - b = 0 and x^3 + cx - d = 0, where n is a root of each of them.

We then have:

n^3 + an - b = 0..............(1)
n^3 + cn - d = 0..............(2)

(I avoid f(n) since this is one specific function. If you want to use that then you'd have f(n) and g(n))

Subtracting we get:

(a-c)n -(b-d)=0

If a=c, then (0)n- (b-d) = 0 and it follows that b-d = 0 and b=d, and they are the same equation.

So, a not = c and we then have n=(b-d)/(a-c)

Since all the terms on the right are rational, it follows that n is rational.

This is a contradiction, since n is irrational.

Therefore no such n can exist and any irrational number can the root of at most one equation of the form x^3 + cx - d = 0

QED.

Thank you :-)
9. (Original post by rdj201)
Thank you :-)

Can't recall seeing a proof by contradiction being required at A-level before - could just be my memory though.
10. (Original post by ghostwalker)

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
11. (Original post by ghostwalker)

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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