Maths Survivor
  • Member for 5 years, 3 months
  • Last seen more than a month ago
  • Pensilvania, USA
Prove that if $f$ and $g$ are continuous functions then $f/g$ is also continuous
Accepted answer
3 votes

Let $f(x):(a,b)\to R$ and $g(x):(a,b)\to R$ be continuous at the point $x_o$ $ϵ$ $(a,b)$. Then $f(x)/g(x)$ is continuous at the point $x_o$ $ϵ$ $(a,b)$ , for $g(x_0)$ different from zero. Proof: ...

View answer
How to show that the composition of two surjective functions is injective?
2 votes

That's not true in general.We can take a counterexample to show one case when it is not true: Let $f:A\to B$ and $g:B\to C$ be two surjective functions and let $h:A\to C$ be their composition such ...

View answer
An alternative proof of Cauchy's Mean Value Theorem
1 votes

Or we could just consider the function: $F(x)=f(x)-f(a)-\frac{f(b)-f(a)}{g(b)-g(a)}(g(x)-g(a))$ Clearly $F(x)$ is defined iff $g(b) $ is different from $g(a)$ , otherwise if $g(a)=g(b)$ then $F(x)$ ...

View answer
Prove that $||a|-|b|| \leq |a-b|$?
1 votes

You know that $|x+y|\leq |x|+|y|$ holds for every real number , then taking $x=a-b$ and $y=b$ we get: $|a-b|\geq|a|-|b|$ ... (1) if we exchange $a$ and $b$ we'll have: $|a-b|\geq -(|a|-|b|)$, or $...

View answer
Basic Field Properties: multiplication
Accepted answer
1 votes

a) In order to prove it using the axioms of multiplication first we use the axiom that states that there exists $1\in \mathbb R $ such that for every $x\in \mathbb R $ it holds that $x*1=x$ ,so we ...

View answer
When does the cyclist get overtaken
1 votes

The speed of the cyclist is v1=16km/h and the speed of motorcyclist is v2=48km/h, when they two will meet the distance passed by each of them will be equal so d1=d2 but the time that took the ...

View answer
Why the set of rational numbers is not an order complete field as it is the subset of real numbers which is an order complete field?
Accepted answer
1 votes

Let x be an irrational number , then take two rational sequences an and bn (rational sequences means that every term of that sequence is a rational number) such that both of them converge to x and an ...

View answer
Prove that if $ \sup\{r \in \mathbb{R} | r>0, r^2<2\}=A$ then $A^2\geq 2$
Accepted answer
0 votes

From the definition of supremum we know that: 1.$A\geq a$ for every $a$ $\epsilon$ $\{r \in \mathbb{R} | r>0, r^2<2\}$ and 2.for every $\epsilon>0$ there exists an $a$ $\epsilon$ $\{r \in ...

View answer
Property of continuous functions
Accepted answer
0 votes

In mathematics, the qualifier pointwise is used to indicate that a certain property is defined by considering each value ${\displaystyle f(x)}$ of some function ${\displaystyle f.}$ An important ...

View answer
Find with proof the infimum and supremum of a sequence
0 votes

Starting from $2-1/n<2$ for every natural number $n$ we see that $2$ is an upper bound of the set $A={\{a_n|n∈N\}}$. We also see that for every $\epsilon>0$ there exists an $n ∈ N$ such that : ...

View answer
Bound a natural by two consecutive powers
0 votes

You can easily prove the first one using Archimedean Property for real numbers $\log m$ and $\log n$ And for the second one again use the Archimedean Property for real number $b*\log_{m} n$ Knowing ...

View answer
Given $a_n, b_n$ are both convergent sequences such that $a_n \leq b_n$ for all $n$, show $\lim_{n\to\infty} a_n \leq \lim_{n\to\infty} b_n$.
0 votes

Let $\lim_{n\to\infty} (a_n)=a$ and $\lim_{n\to\infty} (b_n)$=b ,then using epsilon definition of limit and this property of real numbers: $|a-b|\leq|a|+|b|$ you can prove easily that $\lim_{n\to\...

View answer
Proof that $1 > 0$ using the field and order axioms
0 votes

Based on an order axiom we know that if $ a$ and $b $ have the same sign then $a*b>0$ let $a=b=1$ , clearly a and be have the same sign therefore $a*b=1*1=1>0$

View answer
$T(n)=2T(n-1)+n,T(1)=1,n\ge2$
0 votes

You can see that if $T(n)=2\cdot T(n-1)+n$ and $T(1)=1$ then you'll have a sequence of $T(1),T(2),T(3),T(4),T(5),\dots$ that looks like this: $1,4,11,26,57,120,\dots$ and so on. From this you can see ...

View answer
How to quickly calculate this expression?
0 votes

We can write 1+2-3+4+5-6+7+8-9+10+⋯+95+95-96+97+98-99+100 like this : 0+3+6+9+⋯+93+96+100 now using the formula of finding the arithmetic sum for 0+3+6+9+⋯+93+96 Sn=n/2(a1+an) we'll have S33=33/...

View answer