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    I am trying to figure this out... My lecturer explained the answer in class, but I didnt follow....


    Show that 2 −√2 is irrational..... In the Question he assumed that the Square root of 2 was irrational.... Any help would help greatly... Im trying to write up my Notes!
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    (Original post by Tabletopper)
    I am trying to figure this out... My lecturer explained the answer in class, but I didnt follow....


    Show that 2 −√2 is irrational..... In the Question he assumed that the Square root of 2 was irrational.... Any help would help greatly... Im trying to write up my Notes!
    For proof by contradiction, we assume that 2-\sqrt{2} is rational, so it can be written in the form p/q where p,q are in Z.

    From this you want to show that \sqrt{2} is rational.

    And there's your contradiction, since \sqrt{2} is irrational.
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    I think the question is to prove that 2 −√2 is also irrational. You are allowed to assume without argument that √2 is irrational.

    Suppose for a contradiction that 2 −√2 = p/q, where p and q are integers. By rearranging you can conclude that √2 is rational. But √2 is not rational. This is a contradiction.

    There are two uses of the word "assumption". There is where you assume that something is an absolutely true fact and go ahead to use it in your everyday business. Basically, the common language use of the word assumption.

    The other is where you say things like "assume for a contradiction..." or "assume that the statement is true for n...". The former is found in proofs by contradiction, the latter is found in proofs by induction. These are "argumentative assumptions", "assumptions for the moment".
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    Spoiler alert!

    Someone tell me if you think this answer is`nt kosher:

    assume that 2-(root2) is rational - which implies that it is in Q, then:

    2-(root2)=a/b, some a,b in Q, with no common factors, then:

    b[2-(root2)]=a, so multiplying out, then re-arranging: 2b-a=b(root2),

    squaring (2b-a)^2=2b^2 which implies that (2b-a)^2 is even, implies that 2b-a is even,

    ie. 2b-a=2k, some k in Q

    implies that (re-arranging) a is even - so a has a common factor of 2 (first contradiction)

    ...call a = 2m, some m in Q, plug back in, re-arrange, get b having common factor 2 - Bang! contradiction!
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    (Original post by Hasufel)
    Spoiler alert!

    Someone tell me if you think this answer is`nt kosher:

    assume that 2-(root2) is rational - which implies that it is in Q, then:

    2-(root2)=a/b, some a,b in Q, with no common factors, then:

    b[2-(root2)]=a, so multiplying out, then re-arranging: 2b-a=b(root2),

    squaring (2b-a)^2=2b^2 which implies that (2b-a)^2 is even, implies that 2b-a is even,

    ie. 2b-a=2k, some k in Q

    implies that (re-arranging) a is even - so a has a common factor of 2 (first contradiction)
    At this point, "a" has a factor of 2, but I don't see what's common about it or why that's a contradiction.

    ...call a = 2m, some m in Q, plug back in, re-arrange, get b having common factor 2 - Bang! contradiction!
    So b has a factor of 2, thus a and b have a common factor contradicting their definition. Hence....


    You're essentially using the standard method of showing that \sqrt{2} is irrational.

    Subject to above comments that's fine, though more than is required for the OP's question.
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    Of course! i meant factor of 2 - at that stage we don`t know it`s common!

    d`you think the proof is rigorous enough, though?:confused:
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    (Original post by Hasufel)

    d`you think the proof is rigorous enough, though?:confused:
    You would need to expand "...call a = 2m, some m in Q, plug back in, re-arrange, get b having common factor 2".

    Your proof is rigorous enough otherwise; however it is overkill as far as the OP's requirements go, since you've not relied on sqrt(2) being irrational, but taken it back a stage further.
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    (Original post by Hasufel)
    Spoiler alert!

    Someone tell me if you think this answer is`nt kosher:

    assume that 2-(root2) is rational - which implies that it is in Q, then:

    2-(root2)=a/b, some a,b in Q, with no common factors, then:

    b[2-(root2)]=a, so multiplying out, then re-arranging: 2b-a=b(root2),

    squaring (2b-a)^2=2b^2 which implies that (2b-a)^2 is even, implies that 2b-a is even,

    ie. 2b-a=2k, some k in Q

    implies that (re-arranging) a is even - so a has a common factor of 2 (first contradiction)

    ...call a = 2m, some m in Q, plug back in, re-arrange, get b having common factor 2 - Bang! contradiction!

    Non-sense:
    Spoiler:
    Show


    Proposition:

    2 - \sqrt{4} is an irrational number.

    "Proof":

    Let us assume that 2 - \sqrt{4} \in \mathbb{Q}. Therefore,

    \displaystyle 2 - \sqrt{4} = \frac{a}{b} where a, b \in \mathbb{Z} and hcf(a, b) = 1.

    \therefore \ \ \  b[2 - \sqrt{4}] = a \ \ \ \Rightarrow \ \ \ 2b - a = b\sqrt{4}

    By squaring both sides, one can obtain:

    (2b - a)^2 = 4b^2

    \Rightarrow 2b - a = 2k for some k \in \mathbb{Z}

    ...


    \therefore hcf(a, b) \not= 1

    Hence, by your argument 2 - \sqrt{4} = 2 - 2 = 0 is an irrational number.



    *****

    As for the original statement:

    Proof by contradiction:

    \frac{m}{n} - \not q = \frac{a}{b}, where \not q \not\in \mathbb{Q} and a, b, m, n \in \mathbb{Z}

    \Rightarrow \not q = \frac{m}{n} - \frac{a}{b} = \frac{mb - an}{nb}, which is a contradiction.
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    (Original post by gff)
    "Proof":

    Let us assume that 2 - \sqrt{4} \in \mathbb{Q}. Therefore,

    \displaystyle 2 - \sqrt{4} = \frac{a}{b} where a, b \in \mathbb{Z} and hcf(a, b) = 1..
    Good point, though since 2=\sqrt{4} then a=0, and as such hcf(a,b) is not defined.

    So we had the implicit assumption that 2-\sqrt{2}\not=0

    (Original post by Hasufel)
    ...call a = 2m, some m in Q, plug back in, re-arrange, get b having common factor 2 - Bang! contradiction!
    I assumed this bit worked, though, unless I made a slip, this doesn't show b has a common factor.

    However my brain doesn't seem to be firing on all cylinders at present, so best if someone else checks.
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    (Original post by ghostwalker)
    Good point, though since 2=\sqrt{4} then a=0, and as such hcf(a,b) is not defined.
    Why is it undefined when a = 0?

    Definition
    Spoiler:
    Show

    Let a, b \in \mathbb{Z}. A common factor of a and b is an integer that devides both a and b.
    The highest common factor of a and b, written hcf(a, b) is the largest positive integer that divides both a and b.


    hcf(0, b) = |b|?

    However, it is not equal to one anyway, and the two integers are not coprime.

    It is not necessary to be \sqrt{4}. It would be exactly the same if any perfect square that is multiple of two is chosen?
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    (Original post by gff)
    Why is it undefined when a = 0?

    Definition
    Spoiler:
    Show

    Let a, b \in \mathbb{Z}. A common factor of a and b is an integer that devides both a and b.
    The highest common factor of a and b, written hcf(a, b) is the largest positive integer that divides both a and b.


    hcf(0, b) = |b|?
    I was using the wiki definition that requires the integers to be non-zero, however reading further they contradict themselves.


    It is not necessary to be \sqrt{4}. It would be exactly the same if any perfect square that is multiple of two is chosen?
    Agreed.


    I do however still have problems with Hasufel's "call a = 2m, some m in Q, plug back in, re-arrange, get b having common factor 2 - Bang! contradiction!".

    But like I said, my brain's not firing on all cylinders.
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    from the point of...
    ...a is even, so a has a factor of 2, we therefore have (3rd line of hasufel):

    b(2-root2)=2m, for some m in Z, (excluding zero) giving:

    b=[2m/(2-root2)] implies, squaring:

    b^2=4m^2/[8-2root 2] imples:

    b^2=2m^2/(4-root2)] which implies that b^2 is even, which in turn implies that b is even,

    therefore both a and b both have common factors of 2 - contradicting our original assumption - QED?
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    (Original post by smartb**d)

    b=[2m/(2-root2)] implies, squaring:

    b^2=4m^2/[8-2root 2] imples:
    6-4root2, but I catch the drift of it.
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    notice also that sice b^2 is even implies b is even - contradiction being that both a and b have a factor of 2 - you have a 2nd contradiction - that if b^2 = 2[k^2/3-root2], then b is also irrational - which we stated at the start it wasn`t!
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    I still can't understand why you would even consider doing all this?

    The proof I proposed is a two liner, general proof, for the sum/difference of rational with an irrational.


    However, if I shouldn't assume that \sqrt{2} is an irrational, I will probably start like this.

    \displaystyle 2 - \sqrt{2} = \frac{a}{b}, where a, b \in \mathbb{Z} and hcf(a, b) = 1

    then, I would rearrange

    \sqrt{2} = 2 - \frac{a}{b} = \frac{2b - a}{b}, then prove that hcf(2b - a, b) = 1

    and turn the statement into:

    \sqrt{2} = \frac{2b - a}{b} = \frac{m}{n} with the property that hcf(m, n) = 1.

    Hence, show the standard proof.
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    that still only proves that root is irrational, not 2 - root2
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    Do you learn proof by contradiction at A-level? If so what module is it in?
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    (Original post by Hasufel)
    that still only proves that root is irrational, not 2 - root2
    Why would you think so? Look carefully what's being set up before the standard proof.

    In fact, I can prove by the same way that all integers, and the most of the rationals, are irrational, when subtracted from \sqrt{2}.
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    (Original post by TheJ0ker)
    Do you learn proof by contradiction at A-level? If so what module is it in?
    I believe that in OCR MEI, there is a whole part of the course dedicated on proofs.

    However, so far, I have seen only individual people to be introduced to them. Mostly the Oxbridge ones.
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    (Original post by gff)
    I believe that in OCR MEI, there is a whole part of the course dedicated on proofs.

    However, so far, I have seen only individual people to be introduced to them. Mostly the Oxbridge ones.
    Yeah I'm asking because I'm doing the STEP papers and the only formal proof I learn on Edexcel if proof by induction and they ask you about other methods of proving stuff on the STEP exams
 
 
 
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