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Help with Proof by Contradiction

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Reply 20
Original post by TheJ0ker
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


1. Have a look here.

2. Have a read here.

3. [URL="http://docs.google.com/viewer?a=v&q=cache:hRzGM03P8fQJ:cheng.staff.shef.ac.uk/proofguide/proofguide.pdf+proofs+filetype[excludedFace]tongue[/excludedFace]df&hl=en&gl=uk&pid=bl&srcid=ADGEESgxgLVlM61hDjwss6m95TZexOA4dVzngw1rI8fly2wBa3-hjwznEbqlaWI7nPb3KguvVHHWLUxCtqlsjwxFpXNc1_H9g7A6on19xAUKSehlFhDPz4Izo5tqOPF1Fr7V8SsDmnS2&sig=AHIEtbRe6AbMD5-kR9wjaedVu9v0fcfvdA"]More to read here.


These are good ones
*. Yet more to read here.
**. Yet more to read here 2.

... and these are all from Google ..
(edited 12 years ago)
Reply 21
Original post by gff
1. Have a look here.

2. Have a read here.

3. [URL="http://docs.google.com/viewer?a=v&q=cache:hRzGM03P8fQJ:cheng.staff.shef.ac.uk/proofguide/proofguide.pdf+proofs+filetype[excludedFace]tongue[/excludedFace]df&hl=en&gl=uk&pid=bl&srcid=ADGEESgxgLVlM61hDjwss6m95TZexOA4dVzngw1rI8fly2wBa3-hjwznEbqlaWI7nPb3KguvVHHWLUxCtqlsjwxFpXNc1_H9g7A6on19xAUKSehlFhDPz4Izo5tqOPF1Fr7V8SsDmnS2&sig=AHIEtbRe6AbMD5-kR9wjaedVu9v0fcfvdA"]More to read here
.


These are good ones
*. Yet more to read here.
**. Yet more to read here 2.

... and these are all from Google ..

Cheers :smile:

Those are really really good!
(edited 12 years ago)
Reply 22
Original post by gff

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

......


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


Could you please clarify what you did there? When I substitute 2b2k2b-2k for aa, I end up with b=kb=k.

How does that lead to hcf(a,b)1\therefore hcf(a, b) \not= 1?
Reply 23
Original post by und
Could you please clarify what you did there? When I substitute 2b2k2b-2k for aa, I end up with b=kb=k.

How does that lead to hcf(a,b)1\therefore hcf(a, b) \not= 1?


This is NOT a proof of anything, so don't even try to follow it. I should have made it clearer.
Reply 24

Original post by gff
This is NOT a proof of anything, so don't even try to follow it. I should have made it clearer.


Well, it's not a proof because there's no contradiction. However, Hasufel's proof has a clear contradiction: a and b are both even, therefore they have a common factor. What's wrong with that proof?
Reply 25
Original post by und
Well, it's not a proof because there's no contradiction. However, Hasufel's proof has a clear contradiction: a and b are both even, therefore they have a common factor. What's wrong with that proof?


Perhaps you did notice that I reduced my statement to the statement of Hasufel?

What you are saying has nothing to do with my proposition.
Reply 26

Original post by gff
Perhaps you did notice that I reduced my statement to the statement of Hasufel?

What you are saying has nothing to do with my proposition.


And what exactly is your proposition? I'm now quite confused as you can probably tell.

Hence, by your argument 24=22=02 - \sqrt{4} = 2 - 2 = 0 is an irrational number.


Should I ignore this too? Whose argument are you referring to?
Reply 27
Original post by und
And what exactly is your proposition? I'm now quite confused as you can probably tell.



Should I ignore this too? Whose argument are you referring to?


I don't see why we should waste more time with this?

216=ab,hcf(a,b)=12 - \sqrt{16} = \frac{a}{b}, hcf(a, b) = 1

2ba=b16\Rightarrow 2b - a = b\sqrt{16}

(2ba)2=16b2\Rightarrow (2b - a)^2 = 16b^2

2ba=2k,kZ\Rightarrow 2b - a = 2k, k \in \mathbb{Z}

.. from now on you can continue with the proof given above.


What I'm asking is, what makes it different? Probably that you know in advance that 2\sqrt{2} is irrational? (what you pretend you aren't assuming)

Reduce it to contradiction and let me know. :tongue:
Reply 28

Original post by gff
I don't see why we should waste more time with this?

216=ab,hcf(a,b)=12 - \sqrt{16} = \frac{a}{b}, hcf(a, b) = 1

2ba=b16\Rightarrow 2b - a = b\sqrt{16}

(2ba)2=16b2\Rightarrow (2b - a)^2 = 16b^2

2ba=2k,kZ\Rightarrow 2b - a = 2k, k \in \mathbb{Z}

.. from now on you can continue with the proof given above.


What I'm asking is, what makes it different? Probably that you know in advance that 2\sqrt{2} is irrational? (what you pretend you aren't assuming)

Reduce it to contradiction and let me know. :tongue:


It's quite different, isn't it? If you continue with your 'proof', there's no contradiction, whereas when you're trying to prove that 222-\sqrt2 is rational you encounter a clear contradiction. Therefore, I don't see what you're getting at. In fact, one should assume that that 222-\sqrt2 is rational until finding the contradiction.

I'm sorry if I've misunderstood you.

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