proofs not based on pomi1 non pomi based proofs 2 some contradiction proofs 3 triangle inequalities...

MATH 4530: Analysis One

Proofs Not Based On POMI

James K. Peterson

Department of Biological Sciences and Department of Mathematical SciencesClemson University

February 12, 2018


Outline

1 Non POMI Based Proofs

2 Some Contradiction Proofs

3 Triangle Inequalities

4 Homework


Some Contradiction Proofs

√2 is not a rational number

Proof

We will prove this technque using a technique calledcontradiction. Let’s assume we can find positive integers p and qso that 2 = (p/q)2 with p and q having no common factors. Whenthis happens we say p and q are relatively prime. This tells usp2 = 2q2 which also tells us p2 is divisible by 2. Thus, p2 is even.Does this mean p itself is even? Well, if p was odd, we could writep = 2`+ 1 for some integer `. Then, we would know

p2 = (2`+ 1)2 = 4`2 + 4`+ 1.

The first two terms, 4`2 and 4` are even, so this implies p2 wouldbe odd. So we see p odd implies p2 is odd. Thus, we see p mustbe even when p2 is even. So we now know p = 2k for some integerk as it is even. But since p2 = 2q2, we must have 4k2 = 2q2. Butthis says q2 must be even.



Proof

The same reasoning we just used to show p odd implies p2 is odd, thentells us q odd implies q2 is odd. Thus q is even too.Now here is the contradiction. We assumed p and q were relativelyprime; i.e. they had no common factors. But if they are both even, theyshare the factor 2. This is the contradiction we seek. Hence, our originalassumption must be incorrect and we can not find positive integers p andq so that 2 = (p/q)2.



√3 is not a rational number

Proof

Let’s assume we can find positive integers p and q so that3 = (p/q)2 with p and q being relatively prime. This tells usp2 = 3q2 which also tells us p2 is divisible by 3. Does this mean pitself is divisible by 3? Well, if p was not divisible by 3, we couldwrite p = 3`+ 1 or 3`+ 2 for some integer `. Then, we would know

p2 = (3`+ 1)2 = 9`2 + 6`+ 1.

or

p2 = (3`+ 2)2 = 9`2 + 12`+ 4.

The first two terms in both choices are divisible by 3 and the lastterms are not. So we see p2 is not divisible by 3 in both cases.Thus, we see p must be divisible by 3 when p2 is divisible by 3.



Proof

So we now know p = 3k for some integer k as it is divisible by 3. Butsince p2 = 3q2, we must have 9k2 = 3q2. But this says q2 must bedivisible by 3. The same reasoning we just used to show p2 divisible by 3implies p is divisible by 3, then tells us q2 divisible by 3 implies q isdivisible by 3.Now here is the contradiction. We assumed p and q were relativelyprime; i.e. they had no common factors. But if they are both divisible by3, they share the factor 3. This is the contradiction we seek. Hence, ouroriginal assumption must be incorrect and we can not find positiveintegers p and q so that 3 = (p/q)2.



Let’s introduce some notation:1 if p and q are relatively prime integers, we say (p, q) = 1.2 if the integer k divides p, we say k |p.

Now let’s modify the two proofs we have seen to attack themore general problem of showing

√n is not rational if n is

prime.



√n is not a rational number when n is a prime number

Proof

Let’s assume there are integers u and v with (u, v) = 1 so then = (u/v)2 which implies nv2 = u2. This tells us n|u2. Now wewill invoke a theorem from number theory or abstract albebrawhich tells us every integer u has a prime factor decomposition:

u = (p1)r1 (p2)r2 · · · (ps)rs

for some prime numbers p1 to ps and positive integers r1 to rs . Forexample, here are two such prime decompositions.

66 = 2 · 3 · 11

80 = 24 · 5

It is easy to see what the integers p1 to ps and r1 to rs are in eachof these two examples and we leave that to you!



Proof

Thus, we can say

u2 = (p1)2r1 (p2)2r2 · · · (ps)2rs

Since n|u2, n must divide one of the terms in the prime factordecomposition: i.e. we can say n divides the term prii . Now the term priiis a prime number to a positive integer power ri . The only number thatcan divide into that evenly are appropriate powers of pi . But, we know nis a prime number too, so n must divide pi itself. Hence, we can concluden = pi . But this tells us immediately that n divides u too.



Proof

Hence, we know now u = nw for some integer w . This tells us(nw)2 = nv2 or nw2 = v2. Thus, n|v2. The previous argument appliedto v then tells us n|v too. Hence u and v share the factor n which is acontradiction. Thus we conclude

√n is irrational.


Triangle Inequalities

Definition

Absolute ValuesLet x be any real number. We define the absolute value of x ,denoted by |x |, by

|x | =

{x , if x ≥ 0−x , if x < 0.

For example, | − 3| = 3 and |4| = 4.



Theorem

Triangle InequalityLet x and y be any two real numbers. Then

|x + y | ≤ |x |+ |y ||x − y | ≤ |x |+ |y |

and for any number z .

|x − y | ≤ |x − z |+ |z − y |



Proof

We know (|x + y |)2 = (x + y)2 which implies

(|x + y |)2 = x2 + 2xy + y2.

But 2xy ≤ 2|x ||y | impyling

(|x + y |)2 ≤ x2 + 2|x | |y |+ y2 = |x |2 + 2|x | |y |+ |y |2

= (|x |+ |y |)2.

Taking square roots, we find |x + y | ≤ |x |+ |y |. Of course, the argumentfor x − y is similar as x − y = x + (−y). To do the next part, we know|a + b| ≤ |a|+ |b| for any a and b. Let a = x − z and b = z − y . Then

|(x − z) + (z − y)| ≤ |x − z |+ |z − y |.



Comment

The technique where we do x − y = (x − z) + (z − y) is called the Addand Subtract Trick and we will use it a lot!

Comment

Also note |x | ≤ c is the same as −c ≤ x ≤ c and we use this other wayof saying it a lot too.

Theorem

Backwards Triangle InequalityLet x and y be any two real numbers. Then

|x | − |y | ≤ |x − y ||y | − |x | ≤ |x − y |

| |x | − |y | | ≤ |x − y |



Proof

Let x and y be any real numbers. Then by the Triangle Inequality

|x | = |(x − y) + y | ≤ |x − y |+ |y | ⇒ |x | − |y | ≤ |x − y |

Similary,

|y | = |(y − x) + x | ≤ |y − x |+ |x | ⇒ |y | − |x | ≤ |x − y |

Combining these two cases we see

−|x − y | ≤ |x | − |y | ≤ |x − y |

But this is the same as saying | |x | − |y | | ≤ |x − y |.



Lemma

Proving a Number x is Zero via EstimatesLet x be a real number that satisfies |x | < ε, ∀ε > 0. Then, x = 0.

Proof

We will prove this by contradiction. Let’s assume x is not zero. Then|x | > 0 and |x |/2 is a valid choice for ε. The assumption then tells usthat |x | < |x |/2 or |x |/2 < 0 which is not possible. So our assumption iswrong and x = 0.



Theorem

Extended Triangle InequalityLet x1 to xn be a finite collection of real numbers with n ≥ 1. Then

|x1 + · · · xn| ≤ |x1| + · · · + |xn|

or using summation notation∣∣∑n

i=1 xi∣∣ ≤∑n

i=1 |xi |.

Proof

BASIS : P(1) is the statement |x1| ≤ |x1|; so the basis step is verified.

INDUCTIVE : We assume P(k) is true for an arbitrary k > 1. Hence,we know ∣∣∣∣∣

k∑i=1

xi

∣∣∣∣∣ ≤k∑

i=1

|xi |.



Proof

Now look at P(k + 1). We note by the triangle inequality applied to

a =∑k

i=1 xi and b = xk+1, we have |a + b| ≤ |a|+ |b| or∣∣∣∣∣k+1∑i=1

xi

∣∣∣∣∣ ≤( ∣∣∣∣ k∑

i=1

xi

∣∣∣∣)

+ |xk+1|

Now apply the induction hypothesis to see∣∣∣∣∣k+1∑i=1

xi

∣∣∣∣∣ ≤k∑

i=1

|xi | + |xk+1| =k+1∑i=1

|xi |

This shows P(k + 1) is true and by the POMI, P(n) is true for all n ≥ 1.



Theorem

`2 Norm InequalityLet {a1, . . . , an} and {b1, . . . , bn} be finite collections of real numberswith n ≥ 1. Then∣∣∣∣∣

n∑i=1

aibi

∣∣∣∣∣2

≤

(n∑

i=1

a2i

) (n∑

i=1

b2i

)

Proof

BASIS : P(1) is the statement |a1b1|2 ≤ a21b21; the basis step is true.

INDUCTIVE : We assume P(k) is true for k > 1. Hence, we know∣∣∣∣∣k∑

i=1

aibi

∣∣∣∣∣2

≤

(k∑

i=1

a2i

) (k∑

i=1

b2i

)



Proof

Now look at P(k + 1).∣∣∣∣∣k+1∑i=1

aibi

∣∣∣∣∣2

=

∣∣∣∣∣k∑

i=1

aibi + ak+1bk+1

∣∣∣∣∣2

Let A denote the first piece; i.e. A =∑k

i=1 aibi . Then expanding theterm |A + ak+1bk+1|2, we have∣∣∣∣∣

k+1∑i=1

aibi

∣∣∣∣∣2

=

(k+1∑i=1

aibi

)2

= A2 + 2Aak+1bk+1 + a2k+1 b2k+1

=

∣∣∣∣∣k∑

i=1

aibi

∣∣∣∣∣2

+ 2

(k∑

i=1

aibi

)ak+1bk+1 + a2k+1 b

2k+1



Proofor ∣∣∣∣∣

k+1∑i=1

aibi

∣∣∣∣∣2

≤

∣∣∣∣∣k∑

i=1

aibi

∣∣∣∣∣2

+ 2

∣∣∣∣∣k∑

i=1

ai bi

∣∣∣∣∣ ak+1bk+1 + a2k+1 b2k+1

Now use the induction hypothesis to see∣∣∣∣∣k+1∑i=1

aibi

∣∣∣∣∣2

≤k∑

i=1

a2i

k∑i=1

b2i + 2

√√√√ k∑i=1

a2i

√√√√ k∑i=1

b2i ak+1bk+1 (1)

+a2k+1 b2k+1



Proof

Now let α =√∑k

i=1 a2i bk+1 and β =√∑k

i=1 b2i ak+1. We know for

any real numbers α and β that (α− β)2 ≥ 0. Thus, α2 + β2 ≥ 2α β. Wecan use this in our complicated sum above. We have

2α β = 2

√√√√ k∑i=1

a2i

√√√√ k∑i=1

b2i ak+1 bk+1

α2 + β2 =

(k∑

i=1

a2i

)b2k+1 +

(k∑

i=1

b2i

)a2k+1



Proof

Hence, the middle part of Equation 1 can be replaced by the2αβ ≤ α2 + β2 inequality above to get∣∣∣∣∣

k+1∑i=1

aibi

∣∣∣∣∣2

≤k∑

i=1

a2i

k∑i=1

b2i +

(k∑

i=1

a2i

)b2k+1 (2)

+

(k∑

i=1

b2i

)a2k+1 + a2k+1 b

2k+1

=

(k∑

i=1

a2i + a2k+1

) (k∑

i=1

b2i + b2k+1

)



Proof

But this says ∣∣∣∣∣k+1∑i=1

aibi

∣∣∣∣∣2

≤

(k+1∑i=1

a2i

) (k+1∑i=1

b2i

)

This shows P(k + 1) is true and by the POMI, P(n) is true for all n ≥ 1.



Comment

This is a famous type of theorem. For two vectors

V =

[a1a2

]and W =

[b1b2

]the inner product of V and W , < V ,W > is a1b1 + a2b2 and the normor length of the vectors is

||V || =√a21 + a22 and ||W || =

√b21 + b22

The Theorem above then says | < V ,W > | ≤ ||V || ||W ||. This is calledthe Cauchy - Schwartz Inequality also.

Comment

This works for three and n dimensional vectors too.


Homework

Homework 3

Prove the following propositions.

3.1√

5 is not a rational number using the same sort of argumentwe used in the proof of

√3 is not rational.

3.2√

7 is not a rational number using the same sort of argumentwe used in the proof of

√3 is not rational.

3.3 On the interval [1, 10], use factoring and the triangleinequality to prove |x − y | ≤ 2

√10 |√x −√y |.

proofs not based on pomi1 non pomi based proofs 2 some contradiction proofs 3 triangle inequalities...

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