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Welcome to the ninth video lecture in this
discrete math. So this will be the fourth
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video lecture of this week and we will be
concluding our study on the proof techniques
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using, and particularly the proof technique
study of constructive proof. In next lecture
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we will continue our study of proof techniques
into proof by contradiction and other techniques.
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So to recap what we have done so far.
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So we saw that to proof a statement like A
implies B there can be various techniques
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that are there and we have seen that for some
problems some of the techniques can be useful.
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One of the most important thing is to understand
is which technique is to apply and we have
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discussed this thing earlier also that which
technique to apply depends fully on the problem.
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Some problems can be split up into small smaller
parts or some problem can be viewed in a different
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way which might help in solving the problem,
but at the end of the day whether to split
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a problem or how to split a problem or how
to look at a problem differently, everything
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will depend upon your creative mind. So in
fact it will be you will be giving lots of
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thumb rule for which one to apply when and
so on.
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But in the end your skill that you have to
develop using lots of lots practice will be
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the only useful thing.
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Now we started with understanding how to split
a problem into two smaller parts if the deduction
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or namely to prove A implies B that part B
can be split up into two separate things or
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in other words if B can be written as C AND
D then we saw that A implies B is same as
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proving A implies C AND A implies D. Now we
looked at this example and we saw that if
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the problem is something like if b is an odd
prime then something and something.
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Then that can be split up into two parts namely,
we can split up you can since there is ‘and’
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in the beginning in the deduction so we can
split up into two of the parts. Part a and
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part one and part two or first part and second
part. So this was the first technique we learnt
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first time to solve a problem. Second one
was the fact that to understand what properties
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of the assumption that are required.
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Namely there can be a lot of redundant assumptions
and in there are redundant assumptions we
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would like to throw them away. Just to simplify
the whole problem. The idea is simple simply
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that if A implies B then adding some more
assumptions for example A AND C will also
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imply B. So in such cases one would like to
just throw out the C.
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So in fact if we have to prove A and C implies
B, and we can prove A implies B then we would
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call this C as a redundant assumption and
throwing it away would be useful for making
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this problem simple. Now, again which assumption
to be throw or which assumption can be thrown
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will fully depend upon your intelligence.
There are times when you will not be able
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to understand which one to throw and which
one not to throw. In that case you have to
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continue with various attempts.
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So for example in this problem that we saw
after we split the main problem into two parts,
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in both of these parts part a and part first
part and second part, the assumption was b
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is an odd prime and we saw that what properties
of b is required in either of the cases. So
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what we saw is the first in the first part
of what we need is that just the fact that
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b is an odd integer and the second one is
that b is a real number greater than or equal
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to three.
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So these are the two parts that are there.
Now after i split this one the next technique
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that we apply was the proof, where the direct
proof techniques so we use the direct proof
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to solve both the first and second part.
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So namely these are the constructive proofs.
So a construction constructive proof has two
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parts. The direct proof namely A implies B,
you work with A and you want to prove B or
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second part case study is when we split the
problem into smaller problems.
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So the first part that was there or in other
words if n is an odd integer then n square
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is congruent to 1 mod 4. This we did using
direct proof technique. I am not going through
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the proof technique fully. I am not going
through the full proof right now.
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We saw that in second example namely where
we have to prove that if b is a real number
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greater than or equal to 3 then 2 b square
is greater than b plus 1 whole square. And
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we saw that we can come up with a direct proof
also but the direct proof did look a bit magical.
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So the question was that sometimes this direct
proof, do look a bit magical and it is hard
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to understand how to obtain that. So in such
cases we discussed another approach namely
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a backward proof. The main idea is that if
we have to proof A implies B, we start with
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the deduction that we have to get B and we
message it we message it or change it. And
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if we proof the C is identical to B so C is
IFF B then proving A implies B is same as
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proving A implies C.
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And since we have simplified B to C we would
possibly be able to prove A implies C in a
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much simpler way.
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We saw the application of this particular
technique when we solve this, the second proof
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of this example. And we get the backward proof
so if we worked out, what started our proof
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from the b in this case the 2 b square is
greater than b plus 1 whole square and we
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worked our way through till what was referred
to proof was simple enough that we could prove
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it directly.
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Okay so in other words for proving, for the
direct proof or proving A implies B we can
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start with A and go step by step and prove
B or we can also go do a backward proof in
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which case we kind of play with B simplified
to C and then A implies B with same as proving
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A implies C. This is a very crucial technique
that it applied along.
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Now there is one more technique that was there,
namely sometimes we see that proving something
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harder can be actually be easier or in other
word if we have to proof A implies B and if
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C implies B then if i can prove A implies
C, then i would also be able to prove A implies
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B. So in other words we can prove something
harder. So A implies C is clearly a harder
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thing to prove than A implies B but sometimes
proving the harder thing can be easier.
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So we solve the problem namely if b is a real
number and b is greater than equal 2 then
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2 b cube is strictly greater than 3 b plus
2.
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If this is the problem, we basically realize
that there is a way of making it harder or
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in other words we saw that
we kept on making this whole thing harder
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and harder we started. So for example here
like the b square is greater than 2 b plus
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2 is a harder statement to prove than b square
plus not harder than 2 but it is a stronger
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statement than this the earlier one which
says that the 2 b square is bigger than 3
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b plus 2.
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But somehow by making this problem harder
we are able to simplify the whole statement
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and that helps us to prove the main statement.
This was one more technique that we have seen
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that can be applied for proving a problem.
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The last of the technique that we have learned
which we saw in the last video was the proof
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by case study, namely we talked about splitting
the assumption into cases.
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So sometimes the assumptions can be split
with the assumptions in particular OR, AND
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like A is equals to C OR D, so the assumptions
are ‘OR’ so A is C OR D then A implies
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B is same as proving C implies B AND D implies
B. And this is the main thinking to do. The
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problem is that how to split the assumptions
in two parts or meaning how to write A equals
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to C or D. Now this is not easy, for some
problems there is a kind of standard split
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for some it's not.
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So in the last class for example, we saw the
example that if a and b are two positive integers
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then can we prove a square minus 4 b cannot
be equal to 2 or in other words a square is
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not congruent to 2 mod 4.
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And in this particular case, the main technique
that we applied was that we split the problem
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up based on whether the integer a has remainder
0, 1, 2 or 3 when divisible when divided by
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4. So it basically solved it in the in a case
by case basis depending on the remainders.
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Okay so to let us take up a new problem in
this video lecture and we will again solve
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this problem particular problem.
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So to start with, we start with prime numbers.
We have defined prime numbers in your introductory
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week but just to just for people just to recap,
refresh your minds here it is. So we say a
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prime number, p is the prime number if no
other number less than p divides p okay. And
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this is an important thing if I have two numbers
a and b if p divides a and p doesn't divide
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b then p does not divide a plus b. This is
something that we did in the first week and
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we will be needing particular number theoretic
observation in this in this proof. So what
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is the problem?
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The problem is that proof that a square of
a prime number is always 1 mod 6, when the
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prime is greater than or equal to 5. So in
other words if a prime, if p is greater than
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equal to 5 then p square minus 1 is divisible
by 6. Now here to start with let us understand
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what is the A? So if you looking A implies
B right, so the A is p is a prime number greater
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than or equal to 5. That is the prime number;
let us say what that the assumptions that
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are there.
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And what is the thing that we have to prove?
To prove p square minus 1 is divisible by
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6, so how can we split up the assumptions
namely a p is a prime number into different
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cases.
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Okay, here we again using case by case basis.
We split it by looking at the remainders when
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divided by 6. Now when a number is divided
by 6, what are the remainders that can remain?
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The remainders that can remain are of course
0, 1, 2, 3, 4 and 5. So I have this, 6 cases
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depending on what remainders to use that will
be left when you divide by 6 and these are
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the cases for which we will be solving the
problem. So let us start with the first case.
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Guess what? The remainder is divisible by
6, sorry remainder is zero. That means that
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the prime is divisible by 6 but can a prime
be divisible by 6? No. Of course not, if a
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prime is divisible by 6 that means the prime
p is equals to 6 k in which case it is of
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course divisible by 2 or divisible by 3 right.
So it cannot be a prime, by the definition
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of a prime it cannot be divisible by any smaller
number.
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But here we can see that a number which is
6 times some k, integer k is of course divisible
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by 2 or 3. So thus a prime cannot be of the
order of cannot be 6 times k and hence the
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case 1 cannot exist okay. So case 1 does not
exist, it is not a valid case. So done.
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Similarly, let us go for this case 3. So case
3 again, the remainder is 2. In other words,
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p is of the form 6 k plus 2 and as you can
see again since 6 k plus 2 is divisible by
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2, it was 2 divides 6 k plus 2. It leaves
3 k plus 1 as the quotient and hence again
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this case cannot be a valid case. So if p
is a prime number then p cannot be cannot
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have a remainder 2 when divided by 6.
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Now, here I have left something for you to
check. Note that just like we did the other
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two cases, I claim that the only two remainders
that can be there when you divide by 6 are
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1 and 5. So 0, 2, 3, 4 are not possible remainders
when you divide by 6. When you divide by 6
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only two remainders are 1 and 5. I have proved
that two of the cases cannot happen. You have
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to go home and prove that the other two cases
does not happen.
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So in other words we are left with two cases,
namely case two which is that the remainder
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when p has a remainder 1 when divided by 6
and when it has remainder 4 when divided by
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6, 5, sorry remainder 5 when divided by 6.
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So now to prove these two cases one by one,
so case 1 so remainder 1 when divided by 6
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and p is of the form 6 k plus 1, so p square
if just square it up 6 k plus 1 whole square
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which is comes down to some 36 k square plus
12 k plus 1 which is 6 times 6 k square plus
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2 k plus 1. Now you might ask why do I split
up like this, like divisible by 6. Note that
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what is the problem we have to do?
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The problem is that we have to prove that
p square minus 1 is divisible by 6. We have
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a p square here, now we have a 1 here and
we have something 6 times this. That is the
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main idea right. So p square minus 1 is 6
times 6 k square plus 2 k and 6 k square plus
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2 k is a integer, so p square minus 1 is divisible
by six and hence done. So this takes care
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of the case 2.
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Now for the case 6 the last case also has
a similar thing, this time the remainder is
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5. So p is of the order of 6 k plus 5, which
means p square if you do the calculation everything
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its come to 6 times 6 k square plus 10 k plus
4 plus 1. So here 5 square left 25 which I
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can write it as 24 plus 1 and thus 6 times
4 plus 1 and hence again p square minus 1
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is divisible by 6.
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Thus to prove the fact that if p is a prime
number, p square minus 1 is divisible by 6,
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we divided into case by case various cases
depending on what is the remainder when it
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is divisible by 6. Some of the cases cannot
happen because p is a prime and we threw them
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away. We were left with two cases and for
those two cases we did a case by case analysis.
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This was the, this is one more example of
a case study proof.
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In the exercises that will be handed out on
Tuesday or posted on Tuesday, there will be
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a number of, more number of exercises for
direct and case study proofs. We will be solving
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some more of these problems in the video lecture
where we will dedicate to problem solving.
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Thus so far what we have, so for proving A
implies B we have seen a number of various
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techniques, here are some of the big ones.
We split the problem if B is a written as
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C AND D. We remove redundant assumptions.
Sometimes it is easier to prove a stronger
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statement, we have the direct proof, we have
the backward proof and if A is C OR D then
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we split it up into cases. Now okay let us
move on. Let us start with a let us look at
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the following problem.
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Now what are the primes that are there? The
primes are say 2, 3, 5, 7, 11, 13 and so on
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right. Can you prove that the primes are infinite?
That means there are infinite many number
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of primes. So there are many ways of asking
the same question. Can we say that given any
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big enough number there is a number which
is a prime which is bigger than that number
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or is it that the number of prime is less
than some fixed number, say is there only
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thousand numbers of prime or what?
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Okay of course we have been asked to prove
that primes are infinite. So how do you come
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up with a mathematical proof of taking that
there are not their finite number of primes
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are not there or in fact there are infinite
number of primes. Now for this proving this
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thing we will come up with a completely new
proof technique.
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This proof technique is what is called as
proof by contradiction. So basic idea is that
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if we have been asked to prove A implies B,
sometimes it is easier to prove NOT B AND
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A is false. Now I would like you to go and
check it for yourself this statements are,
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these two statements are indeed equivalent.
That means A implies B is equivalent to statement
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NOT B AND A is false. So sometimes we will
be using this thing and this is what is known
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as proof by contradiction.
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Now what is the basic idea behind it?
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The basic idea is say consider this example
that if I have to prove that the earth is
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flat. Now one way of proving it is that okay,
you say that okay there is a ship coming from
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the horizon. When a ship coming from the horizon,
I first see the top of the ship and slowly
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the complete ship comes out. So you argue
that the earth must be round and not flat.
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Right, There is one more attempt of doing
it, namely let us assume that the earth is
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flat, if the earth is flat then when the ship
comes from the horizon the whole ship would
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appear at same time but that does not happen.
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First the mast is seen or the top of the ship
is seen and then the whole ship is seen. Thus
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see the contradiction. What you see or what
happens is not same as what you thought would
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happen and the reason is that we have assumed
something which is, let us assume earth is
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flat and this assumption is wrong and hence
we prove that earth is not flat. So this is
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the main way of attacking a proof by contradiction.
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We will of course come back next week and
do a lot more problems on proof by contradiction
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then. So next week, we will be doing proof
by contradiction, proof using contrapositive
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which is also similar to contradiction and
counter example. Thank you.