An integer (from the Latin integer meaning "whole") is colloquially defined as a number that can be written without a fractional component. For example, 21, 4, 0, and −2048 are integers, while 9.75, 5+1/2, and √2 are not.
The set of integers consists of zero (0), the positive natural numbers (1, 2, 3, ...), also called whole numbers or counting numbers, and their additive inverses (the negative integers, i.e., −1, −2, −3, ...). The set of integers is often denoted by the boldface (Z) or blackboard bold
(
Z
)
{\displaystyle (\mathbb {Z} )}
letter "Z"—standing originally for the German word Zahlen ("numbers").ℤ is a subset of the set of all rational numbers ℚ, which in turn is a subset of the real numbers ℝ. Like the natural numbers, ℤ is countably infinite.
The integers form the smallest group and the smallest ring containing the natural numbers. In algebraic number theory, the integers are sometimes qualified as rational integers to distinguish them from the more general algebraic integers. In fact, (rational) integers are algebraic integers that are also rational numbers.
This is not homework but just a mental exercise. I suspect there are no integer square roots in this set but can't prove it. I mean all digits are used once in each number. I am interested in arguments from principal and not algorithms. Note the number ##3## is missing. Thanks.
I am really struggling in how to begin this problem. So far I have considered using the Euclidean Algorithm and trying to find the gcd of each number like gcd(9,10) but each time they give me 1 so that doesn't work. My next idea is to do a proof by contradiction where I start with assuming that...
Hello,
Am re-studying math & calculus aiming to start pure math studying later.
However, I got this problem in Stewart calculus.
Typically, this is a straightforward IVT application.
x = x^3 + 1, call f(x)= x^3 - x + 1 & apply IVT.
However I have two things to discuss. First thing is simple...
Consider a certain integer between ## 1 ## and ## 1200 ##.
Then ## x\equiv 1\pmod {9}, x\equiv 10\pmod {11} ## and ## x\equiv 0\pmod {13} ##.
Applying the Chinese Remainder Theorem produces:
## n=9\cdot 11\cdot 13=1287 ##.
This means ## N_{1}=\frac{1287}{9}=143, N_{2}=\frac{1287}{11}=117 ## and...
Proof:
Suppose that ## 6 ## divides ## N ##.
Let ## N=a_{m}10^{m}+\dotsb +a_{2}10^{2}+a_{1}10+a_{0} ##, where ## 0\leq a_{k}\leq 9 ##, be the
decimal expansion of a positive integer ## N ##.
Note that ## 6=2\dotsb 3 ##.
This means ## 2\mid 6 ## and ## 3\mid 6 ##.
Then ## 2\mid...
Proof:
Let ## P(x)= \Sigma^{m}_{k=0} a_{k} x^{k} ## be a polynomial function.
Then ## N=a_{m}10^{m}+a_{m-1}10^{m-1}+\dotsb +a_{1}10+a_{0} ## for ## 0\leq a_{k}\leq 9 ##.
Since ## 10\equiv 1\pmod {3} ##, it follows that ## P(10)\equiv P(1)\pmod {3} ##.
Note that ## N\equiv (a_{m}+a_{m-1}+\dotsb...
Proof:
Suppose ## N ## is the integer and ## x ## is the units digit of ## N ##.
Then ## N=10k+x ## for some ## k\in\mathbb{Z} ## where ## x={0, 1, 2, 3, 4, 5, 6, 7, 8, 9} ##.
Note that ## 10k\equiv 0\pmod {2}\implies N\equiv x\pmod {2} ##.
Thus ## 2\mid N\implies N\equiv 0\pmod {2}\implies...
Proof:
Suppose ## a\equiv b\pmod {n_{1}} ## and ## a\equiv c\pmod {n_{2}} ## where the integer ## n=gcd(n_{1}, n_{2}) ##.
Then ## a-b=n_{1}k_{1} ## and ## a-c=n_{2}k_{2} ## for some ## k_{1}, k_{2}\in\mathbb{Z} ##.
This means ## b-c=n_{2}k_{2}-n_{1}k_{1} ##.
Since ## n=gcd(n_{1}, n_{2}) ##, it...
## y-x \gt 1 \implies y \gt 1+x##
Consider the set ##S## which is bounded by an integer ##m##, ## S= \{x+n : n\in N and x+n \lt m\}##.
Let's say ##Max {S} = x+n_0##, then we have
$$
x+n_0 \leq m \leq x+(n_0 +1)$$
We have,
$$
x +n_0 \leq m \leq (x+1) +n_0 \lt y+ n_0 $$
Thus,
##x+n_0 \leq m \lt...
Proof:
Suppose ## a ## is an odd integer.
Then ## a=2k+1 ## for some ## k\in\mathbb{Z} ##.
Note that ## a^{2}=(2k+1)^{2}=4k^{2}+4k+1=4k(k+1)+1 ##.
Since ## k(k+1) ## is the product of two consecutive integers,
it follows that ## k(k+1) ## must be even.
This means ## k(k+1)=2m ## for some ##...
The last three digits of ##x^3## must be solely dependent on the last 3 digits of ##x##. So let ##x=a+10b+100c## for integers ##a,b,c##. Then ##x^3 = a^3 + 30 a^2 b + 300 a b^2 + 300 a^2 c +O(1000)## where of course ##O(1000)## don't affect the last 3 digits. Evidently ##a^3## is the only...
Proof:
First, we will prove that the integer ## 53^{103}+103^{53} ## is divisible by ## 39 ##.
Note that ## 53\equiv 14 \pmod {39}\implies 53^{2}\equiv 14^{2}\pmod {39}\equiv 196\pmod {39}\equiv 1\pmod {39} ##.
Now observe that ## 103\equiv 25\pmod {39}\equiv -14\pmod {39}\implies 103^{2}\equiv...
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}=\frac{1}{2021}$$
$$\frac{y+x+1}{xy}=\frac{1}{2021}$$
$$xy = 2021y + 2021 x + 2021$$
Then I am stuck. How to continue?
Thanks
Distance between point (-4, 5) and point on circle:
$$d=\sqrt{(x+4)^2+(y-5)^2}$$
$$=\sqrt{x^2+8x+16+y^2-10y+25}$$
Then substitute ##y^2## from equation of circle:
$$d=\sqrt{x^2+8x+16-x^2+4x-6y+12-10y+25}$$
$$=\sqrt{12x-16y+53}$$
After this, I need to try the points one by one to check whether...
Closer to odd number implies ##|y/x - (2n+1)| < 1/2## for ##n = 0,1,2...##. Then
$$-\frac 1 2 < \frac y x - (2n+1) < \frac 1 2 \implies\\
y < (2n + 1.5)x,\\
y > (2n + 0.5)x$$
for each ##n##. We note ##x \in (0,1)## implies ##y## can be larger than 1 since the slope is greater than 1 (but we know...
Proof:
Let ## a>5 ## be an integer.
Now we consider two cases.
Case #1: Suppose ## a ## is even.
Then ## a=2n ## for ## n\geq 3 ##.
Note that ## a-2=2n-2=2(n-1) ##,
so ## a-2 ## is even.
Applying Goldbach's conjecture produces:
## 2n-2=p_{1}+p_{2} ## as a sum of two primes ## p_{1} ## and ##...
Proof:
Let ## n ## be an integer.
Then ## 2n=p_{1}+p_{2} ## for ## n\geq 2 ## where ## p_{1} ## and ## p_{2} ## are primes.
Suppose ## n=k-1 ## for ## k\geq 3 ##.
Then ## 2(k-1)=p_{1}+p_{2} ##
## 2k-2=p_{1}+p_{2} ##
## 2k=p_{1}+p_{2}+2 ##.
Thus ## 2k+1=p_{1}+p_{2}+3 ##...
Proof:
Suppose for the sake of contradiction that there exists a prime divisor of ## n!+1 ##,
which is an odd integer that is not greater than ## n ##.
Let ## n>1 ## be an integer.
Since ## n! ## is even,
it follows that ## n!+1 ## is odd.
Thus ## 2\nmid (n!+1) ##.
This means every prime factor...
Proof:
Suppose ## a ## is a positive integer and ## \sqrt[n]{a} ## is rational.
Then we have ## \sqrt[n]{a}=\frac{b}{c} ## for some ## b,c\in\mathbb{Z} ##
such that ## gcd(b, c)=1 ## where ## c\neq 0 ##.
Thus ## \sqrt[n]{a}=\frac{b}{c} ##
## (\sqrt[n]{a})^{n}=(\frac{b}{c})^n ##...
Hello there, I am having trouble understanding what parts b-d of the question are asking. By solving the Schrodinger equation I got the following for the Landau Level energies:
$$E_{n,k} = \hbar \omega_H(n+\frac 12)+\frac {\hbar^2k^2}{2m}\frac{\omega^2}{\omega_H^2}$$
Where ##\omega_H =...
Proof:
Suppose ## n>1 ## is a positive integer.
Let ## n=p_{1}^{k_{1}} p_{2}^{k_{2}}\dotsb p_{r}^{k_{r}} ## be the prime factorization of ## n ##
such that each ## k_{i} ## is a positive integer and ## p_{i}'s ## are prime for ## i=1,2,3,...,r ## with
## p_{1}<p_{2}<p_{3}<\dotsb <p_{r} ##...
Proof:
Suppose an integer ## n>1 ## is square-free.
Then we have ## a^2\nmid n, \forall a\in\mathbb{Z} ##.
Let ## n=p_{1}^{a_{1}} p_{2}^{a_{2}}\dotsb p_{r}^{a_{r}} ## be the
prime factorization of ## n ## such that each ## a_{i} ## is a positive integer
and ## p_{i}'s ## are prime for ##...
Proof:
Suppose a positive integer ## a>1 ## is a square.
Then we have ## a=b^2 ## for some ## b\in\mathbb{Z} ##,
where ## b=p_{1}^{n_{1}} p_{2}^{n_{2}} \dotsb p_{r}^{n_{r}} ##
such that each ## n_{i} ## is a positive integer and ## p_{i}'s ##
are prime for ## i=1,2,3,...,r ## with ##...
Proof:
Suppose ## n>1 ## is an integer not of the form ## 6k+3 ##.
Then we have ## n=6k ## for some ## k\in\mathbb{Z} ##.
Thus ## n^{2}+2^{n}=(6k)^{2}+2^{6k} ##
## =36k^{2}+2^{6k} ##
## =2(18k^{2}+2^{6k-1}) ##...
Proof:
Suppose for the sake of contradiction that ## p=2^{k}-1 ## is prime
but ## k ## is not an odd integer.
That is, ## k ## is an even integer.
Then we have ## k=2a ## for some ## a\in\mathbb{Z} ##.
Thus ## p=2^{k}-1
=2^{2a}-1
=4^{a}-1. ##
Note that ## 3\mid...
Proof:
Suppose n is an integer such that ## n>11 ##.
Then n is either even or odd.
Now we consider these two cases separately.
Case #1: Let n be an even integer.
Then we have ## n=2k ## for some ## k\in\mathbb{Z} ##.
Consider the integer ## n-6 ##.
Note...
Proof:
Suppose ##a=8^n+1 ## for some ##a \in\mathbb{Z}## such that n##\geq##1.
Then we have ##a=8^n+1 ##
=## (2^3)^n+1 ##
=## (2^n+1)(2^{2n} -2^n+1) ##.
This means ## 2^n+1\mid 2^{3n} +1 ##.
Since ##2^n+1>1## and ##2^{2n} -2^n+1>1## for all...
Proof: Suppose a=n^4+4 for some a##\in\mathbb{Z}## such that n>1.
Then we have a=n^4+4=(n^2-2n+2)(n^2+2n+2).
Note that n^2-2n+2>1 and n^2+2n+2>1 for n>1.
Therefore, every integer of the form n^4+4, with n>1, is composite.
Proof: First, we will show that gcd(a, 0)=abs(a).
Suppose a is a nonzero integer such that a##\neq##0.
Note that gcd(a, 0)##\le##abs(a) by definition of the greatest common divisor.
Since abs(a) divides both a and 0,
we have that...
Find the smallest positive integer N that satisfies all of the following conditions:
• N is a square.
• N is a cube.
• N is an odd number.
• N is divisible by twelve prime numbers.
How many digits does this number N have?Please Explain your steps in detail.
Look at this example,
>>> a = 97
>>> type(a)
<class 'int'>
>>> bin(a)
'0b1100001'
>>> b = ord('a')
>>> b
97
>>> type(b)
<class 'int'>
>>> bin(b)
'0b1100001'
Is this means that the string 'a' and the integer 97 stored as the same binary in the memory ? If so then how can python tell the...
Compute the number of positive integer divisors of 10!. By the fundamental theorem of arithmetic and the factorial expansion:
10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
= 2 x 5 x 3^2 x 7 x 2 x 3 x 5 x 2^2 x 3 x 2 x 1
= 2^8 x 3^4 x 5^2 x 7
Then there are 9 possibilities for 2, 5 for 3, 3 for...
i do not seem to understand part ##ii## of this problem...mathematical induction proofs is one area in maths that has always boggled me :oldlaugh:
let ##n=3, p=7, ⇒m=4##
therefore,
##7=(4-3)(4+3)##
##7=1⋅7##
##1, 7## are integers...##p## is prime.
i am attempting part ##iii## in a moment...
Hello,
I was wondering how to prove that the Binomial Series is not infinite when k is a non-negative integer. I really don't understand how we can prove this. Do you have any examples that can show that there is a finite number when k is a non-negative integer?
Thank you!
Given the definition of floor value :
For all A,B we define the floor value of A denoted by [A] to be an iteger B such that : $[A]=B\Leftrightarrow B\leq A<B+1$
And in symbols $\forall A\forall B ( [A]=B\Leftrightarrow B\leq A<B+1\wedge B\in Z)$,then prove:
For all A $ [A]\leq A<[A]+1$