least upper bound
A set A of reals can have at most one least upper bound.
if a is LUB, then a≤b
if b is LUB, then b≤a
if a, b are both two LUBs, then b=a,
such as {42} has the only element 42 as both an upper bound and a lower bound.
lower bound
if a set A of reals has a lower bound, it has infinitely many lower bounds.
if a is a lower bound of A, then so is any b<a.
Set of reals is infinite
if a set A of reals has both a lower bound and an upper bound, then it is infinite.
i.e. let A=[0,1], it is an infinite set of real numbers.
Least upper bound of integers set
-1 is the least upper bound of the set of negative integers, considered as a subset of the reals.
the set of negative integers {…… n, -3, -2, -1}
Sandwich Theorem(Squeeze Theorem)
the Sandwich Theorem say that
if $\left{{a_n}\right}{\left(n=1\right)}^∞$, $\left{{b_n}\right}{\left(n=1\right)}^∞$, $\left{{c_n}\right}_{\left(n=1\right)}^∞$ are sequences such that,
from some point $n_0$ onwards, $a_n$ ≤ $b_n$ ≤ $c_n$,
and if $\lim _{n→∞}a_n$ = $L$ , $\lim _{n→∞}c_n$ = $L$
then $\left{{b_n}\right}_{\left(n=1\right)}^∞$ is convergent, and $\lim _{n→∞}b_n$ = $L$
how to prove by the Sandwich Theorem ?
Theorem $\lim _{n→∞}$$\frac{\left(\sin n\right)^{2\ }}{3^n}$ = 0
proof: for any n, 0≤$\frac{\left(\sin n\right)^{2\ }}{3^n}$ ≤ $\frac{1\ }{3^n}$
clearly, $\lim _{n→∞}$$\frac{1\ }{3^n}$ = 0
0≤$\frac{\left(\sin n\right)^{2\ }}{3^n}$ ≤ 0
hence by the Sandwich Theorem,
$\lim _{n→∞}$$\frac{\left(\sin n\right)^{2\ }}{3^n}$ = 0
as required.
Peer graded Assignment
- $(∃m∈ N)(∃n∈ N)(3m+5n=12)$
Answer: this statement is false.
Prove: set n=2, the closer possible solution is m=1, then 3m+5n=13;
if n>2, then 3m+5n>13;
set m=2, the closer possible solution is n =, then 3m+5n=11;
if m>2, then 3m+5n>11, but never could be 12 by natural number;
There doe not exist nature number n, m that satisfy the equation.
$(¬∃m ∈ N) (¬∃n ∈ N)$(3m+5n=12)
So, it is proved.
- The sum of any five consecutive integers is divisible by 5 (without remainder).
Answer: the statement is True.
Prove: set (n ∈ Z), n could be anyone of the integer set,
the sequence of any five consecutive integers are: $n, n+1, n+2, n+3, n+4$
The sum of above numbers sequence: $Σ(n,n+1,n+2,n+3,n+4)= 5n+10 = 5(n+2)$
obviously, 5(n+2) is divisible by 5.
The statement is proved.
- For any integer n, the number $n^2+ n +1$ is odd.
Answer: the statement is True.
Prove: by induction
set n =1, then $n^2+ n +1$ = 3, it is odd.
Assume for any integer n, it is true.
$∀n[A(n)⇒A(n+1)]$
set n = n+1, then $(n+1)^2+(n+1)+1=n^2+3n+3=(n^2+n+1)+2(n+1)$
clearly that 2(n+1) is even or zero, and $n^2+ n +1$ is odd,
when number odd add even(or zero), it could only produce odd possibly,
so it is proved.
- every odd natural number is of one of the forms 4n+1 or 4n+3, where n is an integer.
Answer: the statement is true.
Prove: by induction set (n ∈ Z), n could be anyone of the integers. when n=0, 4n+1=1, 4n+3=3; when n=1, 4n+1=5, 4n+3=7;
{1, 3, 5, 7} are odd natural numbers.
Assume for any integer n, it is true.
$∀n[A(n)⇒A(n+1)]$
set n = n+1, then the forms are 4(n+1)+1,4(n+1)+3
4(n+1)+1=(4n+1)+4,4(n+1)+3=(4n+3)+4
clearly that any number odd add an even number 4, it could only produce odd possibly,
so, it is proved.
- for any integer n, at least one of the integers n, n+2, n+4 is divisible by 3.
Answer:
the statement is true.(the prompt does not require to answer true or false)
Prove: by the division theorem. by induction(the main method is not induction).
set n=1, n+2= 1x3 ; n=2, n+4 = 2x3, so on, etc.
assume for any integer n, the statement is true.
set n=3k, then n is divisible by 3;
set n=3k+1, then n+2=3(k+1) is divisible by 3;
set n=3k+2, then n+4=3(k+2) is divisible by 3;
for n = n+1,
the integers n+1, n+3, n+5
set n=3k, then n+3= 3(k+1) is divisible by 3;
set n=3k+1, then n+5=3(k+2) is divisible by 3;
set n=3k+2, then n+1=3(k+1) is divisible by 3;
by reasoning they are all divisible by 3
It is proved.
- prove the only prime triple (i.e. three primes, each 2 from the next) is 3, 5, 7.
by contradiction
Assume there are more prime triples.
set the prime triples are P, P+2, P+4
and set P is the set of Prime.
when P=3, the prime triple is 3, 5, 7.
according to question 5: for any integer n, at least one of the integers n, n+2, n+4 is divisible by 3.
except P=3, the other all results will get sets that could be divisible by 3,
so they are not prime.
it proves the only prime triple is 3, 5, 7.
- Prove that for any natural number $2^2+2^2+2^3+…+2^n=2^(n+1)−2$
Prove: by induction.
set n = 1 , left side is 2, left side is 2, they are equal to each other.
Assume for any integer n, it is true.
$∀n[A(n)⇒A(n+1)]$
set n = n+1, for left side of euqation:
$2^2+2^2+2^3+…+2^n+2^(n+1)$ =
$2^(n+1)−2 +2^(n+1)$ = $2[2^(n+1)−1]$
for the right side of equation: $2^((n+1)+1)−2=2[2^(n+1)]−2=2[2^(n+1)−1]$
both sides are equal to each other.
So it is proved
- the sequence $\left{Ma_n\right}_{n=1}^{\ ∞}$ tends to the limit $ML.$
Prove:
give 𝜀>0,
$lim _{n→∞}a_n = L$ , $M>0$
set |A| = | $\left{a_n\right}_{n=1}^{\ ∞}$ $−L$ | < 𝜀
such that |A| < 𝜀/𝑀
$M |A|$ < 𝜀, $|MA|$ < 𝜀,
| $\left{Ma_n\right}_{n=1}^{\ ∞}$ $−ML$ | < 𝜀
by the definition of limts.(very important for the conclusion statement)
hence, it proves the sequence $\left{Ma_n\right}_{n=1}^{\ ∞}$ tends to the limit $ML.$
- Prove that your example has the stated property.
${∩}(n=1) ^∞$$∩(n=1) ^∞$ An= { ${x∣(∀n)(x∈An)}$ }
Prove :
∩𝑛=1∞𝐴𝑛=∩^∞_{𝑛=1}𝐴_𝑛 =
∩n=1∞An= { x∣(∀n)(x∈An)x|(∀n) (x ∈ An)x∣(∀n)(x∈An) }
set x=1/nx= 1/n x=1/n
∩𝑛=1∞𝐴𝑛=∩^∞_{𝑛=1}𝐴_𝑛 =
∩n=1∞An= { x∣(∀n)(x∈An)x |(∀n) (x ∈ An)x∣(∀n)(x∈An) }
so AnAnAn = ( 0,10, 10,1],
set x=1/(n+1)x= 1/(n+1) x=1/(n+1), ∩𝑛=1∞𝐴(𝑛+1)=∩^∞{𝑛=1}𝐴(𝑛+1) =
∩n=1∞A(n+1)= { x∣(∀(n+1)[x∈A(n+1)]x |(∀(n+1) [x ∈ A(n+1)]x∣(∀(n+1)[x∈A(n+1)] } ,
A(n+1)A(n+1)A(n+1) = ( 0,1/20, 1/20,1/2],
1/(n+1)<1/n1/(n+1) < 1/n 1/(n+1)<1/n, 1/2<1 1/2 < 11/2<1,
so A(n+1)⊂AnA_(n+1) ⊂ A_nA(n+1)⊂An
When take n as a big number,
∩𝑛=1∞𝐴𝑛∩^∞_{𝑛=1}𝐴_𝑛
∩n=1∞An = ∩𝑛=1∞A(n+1)∩^∞_{𝑛=1}A(n+1)
∩n=1∞A(n+1) = ( 0, 0) =∅= ∅=∅ so A(n+1)=AnA_(n+1) = A_nA(n+1)=An
it proves that for all nnn and ∩𝑛=1∞𝐴𝑛=∅∩^∞_{𝑛=1}𝐴_𝑛=∅
∩n=1∞An=∅
such that A(n+1)⊂AnA_(n+1) ⊂ A_nA(n+1)⊂An
- Prove that your example has the stated property.
Prove:
n∈Zn ∈ Zn∈Z, A(n+1)⊂AnA_(n+1) ⊂ A_nA(n+1)⊂An
set x=0nx=\frac{0}{n}x=n0 ,
for ∩𝑛=1∞𝐴𝑛=∩^∞_{𝑛=1}𝐴_𝑛 =
∩n=1∞An= {(x∣(∀n)(x∈An) (x |(∀n) (x ∈ An)(x∣(∀n)(x∈An) }
for any n, and n+1, x=0,
such that ∩𝑛=1∞𝐴𝑛∩^∞_{𝑛=1}𝐴_𝑛
∩n=1∞An ={0}
it is the single real number in the set
so it is proved.
Peer review:
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