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#1
A question is my textbook goes:
Prove by contradiction that the cube root of 2 is irrational.

Halfway through the proof i get to a^3=2b^3, can i just state that if a^3 is even then a is even. Or would i also have to write the proof for that, it's a simple proof but would take up more time.
0
1 month ago
#2
(Original post by Skiwi)
A question is my textbook goes:
Prove by contradiction that the cube root of 2 is irrational.

Halfway through the proof i get to a^3=2b^3, can i just state that if a^3 is even then a is even. Or would i also have to write the proof for that, it's a simple proof but would take up more time.
You are fairly safe to state that a must have a factor of 2 or its even- prime factor theorem if necessary.
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#3
(Original post by mqb2766)
You are fairly safe to state that a must have a factor of 2 or its even- prime factor theorem if necessary.
Ok i thought it would be fine, but just wanted to check as it would be a fairly stupid mark to lose if otherwise.

Another question i'm not too sure on at the moment is:
Prove by contradiction that there are no integer solutions to x^2-y^2=2

So far i've got-
Assume there is at least one integer value for x and y that satisfies x^2+y^2=2
x^2+y^2=2 can be written in the form (x-y)(x+y)=2
let m=x-y and n=x+y, both m and n are integer as x and y are.
so mn=2
the only integer values that m and n can be are
m=1,n=2
m=-1,n=-2
m=2, n=1
m=-2,n=-1

from here i thought about subbing all these value back and getting two simultaneous equation, which could be solved to show that x and y are never integers. But this is starting to look a lot like proof by exhaustion, is it still viable or is there another way i'm missing? I'm still showing a contradiction but it just seems very unlike every other one i've had to do.
Last edited by Skiwi; 1 month ago
0
1 month ago
#4
(Original post by Skiwi)
Ok i thought it would be fine, but just wanted to check as it would be a fairly stupid mark to lose if otherwise.

Another question i'm not too sure on at the moment is:
Prove by contradiction that there are no integer solutions to x^2+y^2=2

So far i've got-
Assume there is at least one integer value for x and y that satisfies x^2+y^2=2
x^2+y^2=2 can be written in the form (x-y)(x+y)=2
let m=x-y and n=x+y, both m and n are integer as x and y are.
so mn=2
the only integer values that m and n can be are
m=1,n=2
m=-1,n=-2
m=2, n=1
m=-2,n=-1

from here i thought about subbing all these value back and getting two simultaneous equation, which could be solved to show that x and y are never integers. But this is starting to look a lot like proof by exhaustion, is it still viable or is there another way i'm missing? I'm still showing a contradiction but it just seems very unlike every other one i've had to do.
Is the bold right? Youve gone from x^2+y^2 to x^2-y^2?
Also is the question right x=y=1 seems to work? Or is the original question x^2-y^2=2?
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#5
(Original post by mqb2766)
Is the bold right? Youve gone from x^2+y^2 to x^2-y^2?
Also is the question right x=y=1 seems to work? Or is the original question x^2-y^2=2?
Sorry, mistyped the question. It should be a minus
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1 month ago
#6
(Original post by Skiwi)
Sorry, mistyped the question. It should be a minus
Thought so.
There are a few ways you could prove this result. An elegant one (not contradiction so for interest only) is to note that every square number is 0,1 or 4 mod 8. So there are no two square numbers which differ by 2.
So back to your problem, there are two ways you could go
x^2 = y^2 + 2
and justify why this cant occur. Or dots on the left as you did
(x+y)(x-y) = 2
Then reason about the two factors. Taking a bit of inspiration from the previous proof, you could think about different combinations of parity (odd/even) of x and y.
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#7
(Original post by mqb2766)
Thought so.
There are a few ways you could prove this result. An elegant one (not contradiction so for interest only) is to note that every square number is 0,1 or 4 mod 8. So there are no two square numbers which differ by 2.
So back to your problem, there are two ways you could go
x^2 = y^2 + 2
and justify why this cant occur. Or dots on the left as you did
(x+y)(x-y) = 2
Then reason about the two factors. Taking a bit of inspiration from the previous proof, you could think about different combinations of parity (odd/even) of x and y.
looking at x^2=y^2+2 :
As both x and y are squared we can look at only positive values,
is it enough to say that the smallest difference between two positive square numbers is 3, as the difference form an arithmetic sequence of 2n+1, there will never be a difference that is smaller than 3 thus no integer values of of x and y can exist.

For (x+y)(x-y)=2
if x is even then y cannot be odd as even+odd=odd and a even-odd=odd and odd*odd=odd , 2 is not an odd number so both x and y must be even
then x is in the form 2n and y is in the form 2m
(2n+2m)(2n-2m)=4n^2-4m^2
n^2-m^2=0.5
however n and m are integers, and thus the equation n^2-m^2 can never produce 0.5 which is not an integer but a rational.
if x is odd then y cannot be even, follows same justification as before. So x and y must both be odd
x is in the form 2n-1 and y=2m-1
(2n+2m-2)(2n-2m)=4n^2-4m^2-4n+4m
so n^2-n-m^2+m=0.5, follows same justification as before as to why we can reject this.
therefore no integer solutions exist.

I think the second one gives a more convincing argument, i could probably be more thorough but i think it's enough.
0
1 month ago
#8
(Original post by Skiwi)
looking at x^2=y^2+2 :
As both x and y are squared we can look at only positive values,
is it enough to say that the smallest difference between two positive square numbers is 3, as the difference form an arithmetic sequence of 2n+1, there will never be a difference that is smaller than 3 thus no integer values of of x and y can exist.

For (x+y)(x-y)=2
if x is even then y cannot be odd as even+odd=odd and a even-odd=odd and odd*odd=odd , 2 is not an odd number so both x and y must be even
then x is in the form 2n and y is in the form 2m
(2n+2m)(2n-2m)=4n^2-4m^2
n^2-m^2=0.5
however n and m are integers, and thus the equation n^2-m^2 can never produce 0.5 which is not an integer but a rational.
if x is odd then y cannot be even, follows same justification as before. So x and y must both be odd
x is in the form 2n-1 and y=2m-1
(2n+2m-2)(2n-2m)=4n^2-4m^2-4n+4m
so n^2-n-m^2+m=0.5, follows same justification as before as to why we can reject this.
therefore no integer solutions exist.

I think the second one gives a more convincing argument, i could probably be more thorough but i think it's enough.
For the first one, think about a binomial-type expansion as the right hand side is like that but with no linear term. So
x = y+d
for an integer d, then
(y+d)^2 = y^2 + 2
so what can you say about d? Youd end up similar to the dots factorization.

For the second one, it reads roughly right, but Id be a bit more concise. If x and y have the same parity (both odd or both even) then x+y and x-y are both even. So the lhs is divisiblel by 4 and the rhs isnt. If x and y have different parity, then x+y and x-y are both odd so ...

For an exam, Id probably be a bit longer than my terse proof, but try to shorten yours as it may be 4-5 marks so not too long. For the first one, be careful about making unnecessary assumptions as it makes the proof longer / harder to make sure youve covered all cases.
Last edited by mqb2766; 4 weeks ago
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#9
(Original post by mqb2766)
For the first one, think about a binomial-type expansion as the right hand side is like that but with no linear term. So
x = y+d
for an integer d, then
(y+d)^2 = y^2 + 2
so what can you say about d? Youd end up similar to the dots factorization.

For the second one, it reads roughly right, but Id be a bit more concise. If x and y have the same parity (both odd or both even) then x+y and x-y are both even. So the lhs is divisiblel by 4 and the rhs isnt. If x and y have different parity, then x+y and x-y are both odd so ...

For an exam, Id probably be a bit longer than my terse proof, but try to shorten yours as it may be 4-5 marks so not too long. For the first one, be careful about making unnecessary assumptions as it makes the proof longer / harder to make sure youve covered all cases.
Ok, going for the second one first. I see how you can cut it down by a half by considering the both even/odd at the same time. I've re done it and i've managed to write what i think is enough in my exam in about 5-6 lines. Happy with this now.

The first one i'm still not sure about,

expanding and comparing coefficients i get that d is root2 so that the constant terms match but d must be 0 for there to not be a linear term. Therefore integer x exists where x is in the form y+d. The only case missing is where x=y, which obviously doesn't work because y^2<y^2+2.

Not entirely sure if this works?
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4 weeks ago
#10
(Original post by Skiwi)
Ok, going for the second one first. I see how you can cut it down by a half by considering the both even/odd at the same time. I've re done it and i've managed to write what i think is enough in my exam in about 5-6 lines. Happy with this now.

The first one i'm still not sure about,

expanding and comparing coefficients i get that d is root2 so that the constant terms match but d must be 0 for there to not be a linear term. Therefore integer x exists where x is in the form y+d. The only case missing is where x=y, which obviously doesn't work because y^2<y^2+2.

Not entirely sure if this works?
The first one is similar to the second, I was just trying to get you to think about different ways of representing the algebraic expression and how to analyse it. For the second one, its "efficient" if you analyse the parity of x and y in the "right" way as you now seem to be doing. For the first
(y+d)^2 = y^2 + 2
or expanding
2yd + d^2 = 2
Using a similar argument about parity of d. Even d means the lhs is divislbe by 4. Odd d means odd lhs. You could factorize it as
d(2y+d)
which is the same as
(x-y)(x+y)
So little different to the second approach, but a different way of getting there. It could be considered a bit more efficient as you only have to consider the parity of a single variable, but equally you could have got there from dots by just considering dots in terms of the difference.
Last edited by mqb2766; 4 weeks ago
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#11
(Original post by mqb2766)
The first one is similar to the second, I was just trying to get you to think about different ways of representing the algebraic expression and how to analyse it. For the second one, its "efficient" if you analyse the parity of x and y in the "right" way as you now seem to be doing. For the first
(y+d)^2 = y^2 + 2
or expanding
2yd + d^2 = 2
Using a similar argument about parity of d. Even d means the lhs is divislbe by 4. Odd d means odd lhs. You could factorize it as
d(2y+d)
which is the same as
(x-y)(x+y)
So little different to the second approach, but a different way of getting there. It could be considered a bit more efficient as you only have to consider the parity of a single variable, but equally you could have got there from dots by just considering dots in terms of the difference.
Ok i understand what you wanted with the first one now. I assume the problem with how i approached the first one is that i just stated that no value of d can satisfy it, whereas by taking both odd/even cases you actually show why?
0
4 weeks ago
#12
(Original post by Skiwi)
A question is my textbook goes:
Prove by contradiction that the cube root of 2 is irrational.

Halfway through the proof i get to a^3=2b^3, can i just state that if a^3 is even then a is even. Or would i also have to write the proof for that, it's a simple proof but would take up more time.
which textbook is this please? and which exam board? thank you
0
#13
which textbook is this please? and which exam board? thank you
The standard Pearson edexcel pure book( the blue one), page 5 of the year 2 book.
Last edited by Skiwi; 4 weeks ago
0
4 weeks ago
#14
(Original post by Skiwi)
Ok i understand what you wanted with the first one now. I assume the problem with how i approached the first one is that i just stated that no value of d can satisfy it, whereas by taking both odd/even cases you actually show why?
Thats the key thing with proof by contradictions. If youre argueing about a value like here, you need to be careful to make sure you justify why all values are excluded. By considering parity, "all" reduces to just 2 cases (for d). The expansion for x^2
y^2 + 2yd + d^2
but then youd have to argue about sign of y and how the second term subtracts from d^2 and why that could never be 2. Its certainly possible, but possibly messy in an exam.
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#15
(Original post by mqb2766)
Thats the key thing with proof by contradictions. If youre argueing about a value like here, you need to be careful to make sure you justify why all values are excluded. By considering parity, "all" reduces to just 2 cases (for d). The expansion for x^2
y^2 + 2yd + d^2
but then youd have to argue about sign of y and how the second term subtracts from d^2 and why that could never be 2. Its certainly possible, but possibly messy in an exam.
Thanks for the help 0
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