Question

Prove by contradiction that there is no rational number r such that r^3 + r + 1 = 0.

Assume r^3 + r + 1 = 0.

Let r = a/b. Then

(a/b)^3 + a/b + 1/1 = 0
a^3/b^3 + a/b + 1/1 = 0
(ab^2 + a^3 + b^3)/b^3 = 0
ab^2 + a^3 + b^3 = 0

I'm not seeing the proof... Help?

Answer 1

better to ask, does that equation have a rational x intercept?

Answer 2

use the rational roots theorem on f(r) = r^3 + r + 1

Answer 3

you know the proof by contradiction for proving sqrt(2) is irrational right ?

Answer 4

if f (p/q) = 0 , then p divides 1 and q divides 1.
therefore
the potential candidates for p/q can only be 1, -1

f(1) = 3
f(-1) = -1 , so both of these are not rational roots. youre done

Answer 5

i think that works, ill have to think about it

Answer 6

thats not proof by contradiction perl even though it works

Answer 7

there might be a way to present it as a proof by contradiction.
if you assume there exists p/q such that f(p/q) = 0, for f(r) = r^3 + r + 1

Answer 8

then the rational roots theorem says p/q = {1, -1} . but both of these f(1) , f(-1) does not equal to zero. contradiction

Answer 9

its a contradiction because we assumed there does exist a solution f(p/q) = 0

Answer 10

is zero a rational number ?

Answer 11

0 = 0/1 , so 0 is rational

Answer 12

my point is rational root theorem should not be used in a proof by contradiction

Answer 13

why not?

Answer 14

if u are using rational root theorem anyways, you could present it as direct proof itself... no need to twist

Answer 15

its not what is expected in a contradiction proof. it works though

Answer 16

ok there might be a way to show it using number theoretic properties

Answer 17

looks hard without rational root thereom :O

Answer 18

ohk

r^3+r+1=0
r^3+r=-1
r(r^2+1)=-1

so four cases
case 1:-
r=-1 and r^2+1=1...r^2=0 ignore
case 2:-
r=1 , r^2+1=1 r^2=2 ignore
case 3:-
r=a/b , and r^2+1=-a/b
case 3:-
r=-a/b and r^2+1=a/b

lets work on case 3 and 4

Answer 19

sorry i made a typo xD
case 3:-
r=a/b , and r^2+1=-b/a
case 4:-
r=-a/b and r^2+1=b/a

we might see also other two cases
case 5:-
r=b/a , and r^2+1=-a/b
case 4:-
r=-b/a and r^2+1=a/b

Answer 20

case 3:-
r=a/b , and r^2+1=-b/a

right ?

Answer 21

yep :)

Answer 22

doesn't case3 cover all other cases ?

Answer 23

just let a, b to be integers

Answer 24

ok thats work i just create this up lol
im gonna see where it gonna end :)
now we assume a/b is rational why ?

cuz that what we need to proof , if its not rational like pi/something then we are done :)

Answer 25

now case 3 :- (PS :a and b are integers s.t b>=a)
r=a/b

r^2+1=-b/a
r^2=-b/a-1
=-(b+a)/a
(a/b)^2=-(b+a)/a which is a contradiction

Answer 26

same with other cases xD

Answer 27

why is the last line a contradiction ?

Answer 28

last line u have one side is positive and the other is negative

Answer 29

Oh you're assuming a,b are positive to start with

Answer 30

yeah thats why i gave all cases , sorry for not mention that xD

Answer 31

i see, so we are done if we show all cases lead to contradiction

Answer 32

yep , i still need to write the rest

Answer 33

ok case 4 r=-a/b and r^2+1=b/a
r^2=b/a-1
blah blah
a^3-b^3=-ab^2 which is a contradiction since the only integer solution are 0,0

Answer 34

got it :)

Answer 35

oh , got other method ? :O

Answer 36

nope im saying i got ur method :)

Answer 37

ohkk ...

Answer 38

ok so these are only cases
other cases are like this
b>a

r=b/a

r^2+1=-a/b contradiction r^2+1>0

Answer 39

case 6

b>a
r=-b/a
r^2=(b/a)^2>r
r^2+1=(b/a)^2+1>r+1

r^2+1=a/b contradiction

Answer 40

done :O

Answer 41

This thing finally loaded for me. I sent @ganeshie8 a message thinking I would never be able to get back here. I will post it.

Answer 42

This is what it said: Hey I can't get back to that question. Submit this if you think it is a good idea. I will only put in some things since I'm limited here. r^3+r+1+r^2-r^2=0 where r=a/b is a rational. Let f(r)=r^3+r^2+r+1. Assuming r=a/b we have r^3+r^2+r+1=(a+b)(a^2+b^2)=0. So a=-b. So r=-1. So r+1 is a factor of f(r). So we have r^3+r+1+r^2-r^2 can be written as (r+1)(r^2+1)-r^2=0. This can be written as r=-1/(r^2+1) which is not true for any rational r. Therefore r^3+r+1 has no rational zeros.

Answer 43

nice :)

Answer 44

wow! looks very neat

r=-1/(r^2+1) is not possible

is this because , left side and right side can never have same signs ?

Answer 45

oh no, when r is negative my reason fails.. so i think u have a better reason

Answer 46

r=-1/(r^2+1)
r(r^2+1)=-1
i think this is what we nee to prove lol xD

Answer 47

ok i guess this leads to
a/b=-1/((a^2+b^2)/b^2)=-b^2/a^2+b^2
a/b=-b^2/a^2+b^2
a(a^2+b^2)=-b^3
a^3+b^3+ab^2=0
has no integer solution exept 0,0

Answer 48

i have somehting like this in head:

r^3+r+1+r^2-r^2=0 where r=a/b is a rational.

(a+b)(a^2+b^2)=a^2b

if a, b are odd then left side is even, right side is odd - contradiction
if a, b are of opposite parity, left side is odd and right side is even - contradiction
\(\square \)

Answer 49

nice also hmm

Answer 50

but but second case hmm

Answer 51

ok it works :S

Answer 52

how about this :

a^3 + ab^2 + b^3 = 0
a(a^2+b^2) = -b^3

since a, b are coprime, a|1
similarly we can show that b|1

Answer 53

proof art lol

Answer 54

thats just a copy of rational root theorem proof ;p
http://en.wikipedia.org/wiki/Rational_root_theorem

Answer 55

haha , i liked rational theorem thingy :O

Answer 56

i dont think that i would write such a beautiful or neat proof like that xD

Answer 57

and yeah I guess \[-r=\frac{1}{r^2+1} \text{ is only obvious if } r<0\]
Also I think a lot of that work I did was unnecessary :p.

Answer 58

\[r^3+r+1=0 \\ r(r^2+1)+1=0 \\ r(r^2+1)=-1 \\ r=\frac{-1}{r^2+1} \\ -r=\frac{1}{r^2+1}\]

Answer 59

i mean the equality is only possible if we have r<0* unless I can think of way to show it is not possible