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!

Tabletopper
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 29092011 17:45

ghostwalker
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 29092011 18:10
(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!
From this you want to show that is rational.
And there's your contradiction, since is irrational. 
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 29092011 20:56
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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 30092011 14:08
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 rearranging: 2ba=b(root2),
squaring (2ba)^2=2b^2 which implies that (2ba)^2 is even, implies that 2ba is even,
ie. 2ba=2k, some k in Q
implies that (rearranging) a is even  so a has a common factor of 2 (first contradiction)
...call a = 2m, some m in Q, plug back in, rearrange, get b having common factor 2  Bang! contradiction! 
ghostwalker
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 30092011 17:46
(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 rearranging: 2ba=b(root2),
squaring (2ba)^2=2b^2 which implies that (2ba)^2 is even, implies that 2ba is even,
ie. 2ba=2k, some k in Q
implies that (rearranging) a is even  so a has a common factor of 2 (first contradiction)
...call a = 2m, some m in Q, plug back in, rearrange, get b having common factor 2  Bang! contradiction!
You're essentially using the standard method of showing that is irrational.
Subject to above comments that's fine, though more than is required for the OP's question. 
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 30092011 19:36
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? 
ghostwalker
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 30092011 21:28
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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 30092011 23:47
(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 rearranging: 2ba=b(root2),
squaring (2ba)^2=2b^2 which implies that (2ba)^2 is even, implies that 2ba is even,
ie. 2ba=2k, some k in Q
implies that (rearranging) a is even  so a has a common factor of 2 (first contradiction)
...call a = 2m, some m in Q, plug back in, rearrange, get b having common factor 2  Bang! contradiction!
Nonsense:
Spoiler:Show
*****
As for the original statement:
Proof by contradiction:
, where and
, which is a contradiction. 
ghostwalker
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 01102011 07:35
So we had the implicit assumption that
(Original post by Hasufel)
...call a = 2m, some m in Q, plug back in, rearrange, get b having common factor 2  Bang! contradiction!
However my brain doesn't seem to be firing on all cylinders at present, so best if someone else checks. 
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 01102011 10:31
(Original post by ghostwalker)
Good point, though since then a=0, and as such hcf(a,b) is not defined.
Definition
Spoiler:Show
?
However, it is not equal to one anyway, and the two integers are not coprime.
It is not necessary to be . It would be exactly the same if any perfect square that is multiple of two is chosen? 
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 01102011 11:42
I do however still have problems with Hasufel's "call a = 2m, some m in Q, plug back in, rearrange, get b having common factor 2  Bang! contradiction!".
But like I said, my brain's not firing on all cylinders. 
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 02102011 00:14
from the point of...
...a is even, so a has a factor of 2, we therefore have (3rd line of hasufel):
b(2root2)=2m, for some m in Z, (excluding zero) giving:
b=[2m/(2root2)] implies, squaring:
b^2=4m^2/[82root 2] imples:
b^2=2m^2/(4root2)] 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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 02102011 10:03

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 02102011 15:11
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^2/3root2], then b is also irrational  which we stated at the start it wasn`t!

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 02102011 15:51
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 is an irrational, I will probably start like this.
, where and
then, I would rearrange
, then prove that
and turn the statement into:
with the property that .
Hence, show the standard proof. 
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 02102011 18:24
that still only proves that root is irrational, not 2  root2

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 02102011 18:30
Do you learn proof by contradiction at Alevel? If so what module is it in?

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 02102011 20:25
(Original post by Hasufel)
that still only proves that root is irrational, not 2  root2
In fact, I can prove by the same way that all integers, and the most of the rationals, are irrational, when subtracted from . 
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 02102011 20:28
(Original post by TheJ0ker)
Do you learn proof by contradiction at Alevel? If so what module is it in?
However, so far, I have seen only individual people to be introduced to them. Mostly the Oxbridge ones. 
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 02102011 20:31
(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.
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