prove or desprove by example

**transgalactic** · September 2nd 2011, 04:40 AM

1.prove or desprove by example:
a)
$sup\{\frac{[x]}{[x]+1}|x\geq0\}=1$
b)prove or desprove that $x^{3}$ 'uniformly' continuous in R
c)if f(x) monotonicly decreasing in $[0,\infty)$ and $lim_{x->\infty}=0$ then f(x)>0 for $x\in[0,\infty)$
d)f,g are continues in [a,b] if $x\in[a,b] \int_{a}^{x}f(t)dt=\int_{a}^{x}g(t)dt$ then f=g in [a,b]

regarding the first one i couldnt find a value for which the supremum is 1
for part a) i cant say if its correct or not

?

**HallsofIvy** · September 2nd 2011, 06:48 AM

Originally Posted by transgalactic

1.prove or desprove by example:
a)
$sup\{\frac{[x]}{[x]+1}|x\geq0\}=1$
b)prove that $x^{3}$ 'uniformly' continuous in R
c)if f(x) monotonicly decreasing in $[0,\infty)$ and $lim_{x->\infty}=0$ then f(x)>0 for $x\in[0,\infty)$
d)f,g are continues in [a,b] if $x\in[a,b] \int_{a}^{x}f(t)dt=\int_{a}^{x}g(t)dt$ then f=g in [a,b]

regarding the first one i couldnt find a value for which the supremum is 1

That makes no sense. The supremum does not depend upon a "value". It should be obvious that 1 is an upper bound for this. Is there a smaller upper bound? That is, is there a number $\alpha< 1$ such that $\frac{x}{x+1}< \alpha$ for all x> 1? What do you get if you solve that inequality for x?

for part a) i cant say if its correct or not

?

**transgalactic** · September 2nd 2011, 07:25 AM

ok so 1 is the upper bound and in order to prove that its supremum ,we need to show
that its the smallest upper bound.

suupose 1 is not the smallest and $sup\{\frac{[x]}{[x]+1}|x\geq0\}=t<1$ .
now i think we should use $lim_{x->\infty}\frac{[x]}{[x]+1}$
but i cant use lhopital law here like normal variable because we have to get rid of [x].
fo x>0
x-1<[x]<x+1
so $lim_{x->\infty}\frac{[x]}{[x]+1}$ > $lim_{x->\infty}\frac{x-1}{x+1+1}$
so i have the result of a smaller limit.
how i know the result of the bigger one
?

**emakarov** · September 2nd 2011, 08:11 AM

Instead of using the L'Hopital's rule, I would follow the suggestion above and solve [x]/([x] + 1) < t for x. If [x] means the integer part of x, I would solve it when x is an integer.

**transgalactic** · September 2nd 2011, 08:20 AM

so you want me to replace [x] with x,and prove that t is its supremum
?

**emakarov** · September 2nd 2011, 09:49 AM

Originally Posted by transgalactic

so you want me to replace [x] with x,and prove that t is its supremum
?

No,

Originally Posted by emakarov

I would follow the suggestion above and solve [x]/([x] + 1) < t for x

I have not said anything about proving that t is a supremum. The suggestion is to solve the inequality on integers, so you can remove the square brackets.

**transgalactic** · September 2nd 2011, 09:53 AM

ahhhh instead of R i solve it on N
but how to make the transition back to R
?

**emakarov** · September 2nd 2011, 11:39 AM

but how to make the transition back to R
?

Each integer is a real number.

We know that 1 is an upper bound of $\{[x]/([x]+1)\mid x\ge0\}$ . The idea is to prove that anything less than 1 is not an upper bound. Suppose t < 1. Then t is not an upper bound iff the inequality $[x]/([x]+1) > t$ is true for some $x\ge0$ . Well, the suspicion is that it is true for infinitely many $x\ge 0$ , and even true for infinitely many positive integers. So, try to solve this inequality when x is an integer and see if this is true.

**transgalactic** · September 2nd 2011, 11:56 AM

ahhh if i will prove for integer that $\frac{x}{x+1}>t$ then t cannot be an upper bound and specially supremum.
ok the integer group is not bounded from top so $N> \frac{x}{x+1}$
but i dont know how to continue and i dont know where to put "t"
i only remmember the part when N is not bounded and we put it to be larger then the fracture.

another way i remmber is
$lim_{x->\infty}\frac{x}{x+1}=1$
so for every $\epsilon >0$ there is $M >0$ for which every $x>M$
$|\frac{x}{x+1}-1|<\epsilon$
i can open this
to
$1-\epsilon<\frac{x}{x+1}<\epsilon+1$
so i can choose $1-\epsilon=t$
and
$t<\frac{x}{x+1}$
correct?

**emakarov** · September 2nd 2011, 05:20 PM

Originally Posted by transgalactic

ahhh if i will prove for integer that $\frac{x}{x+1}>t$ then t cannot be an upper bound and specially supremum.
ok the integer group is not bounded from top so $N> \frac{x}{x+1}$
but i dont know how to continue and i dont know where to put "t"

You should review how to solve linear inequalities such as x / (x + 1) > t. Without it, frankly, talking about limits, continuity, or any other calculus topic looses meaning. (I am assuming this is the issue because this thread has four suggestions to solve this inequality, and my last post explains how this would help solve the problem.)

Originally Posted by transgalactic

another way i remmber is
$lim_{x->\infty}\frac{x}{x+1}=1$
so for every $\epsilon >0$ there is $M >0$ for which every $x>M$
$|\frac{x}{x+1}-1|<\epsilon$
i can open this
to
$1-\epsilon<\frac{x}{x+1}<\epsilon+1$
so i can choose $1-\epsilon=t$
and
$t<\frac{x}{x+1}$
correct?

There is sense in this, but, again, I think it's meaningless doing limits without a solid grasp of algebra.

**transgalactic** · September 2nd 2011, 10:00 PM

i know how to solve innequalitties
you didnt say which one.i have presented two ways of continuing this solution
could you say which one to follow
?
be more spesific

**emakarov** · September 3rd 2011, 03:17 AM

Originally Posted by emakarov

We know that 1 is an upper bound of $\{[x]/([x]+1)\mid x\ge0\}$ . The idea is to prove that anything less than 1 is not an upper bound. Suppose t < 1. Then t is not an upper bound iff the inequality $[x]/([x]+1) > t$ is true for some $x\ge0$ . Well, the suspicion is that it is true for infinitely many $x\ge 0$ , and even true for infinitely many positive integers. So, try to solve this inequality when x is an integer and see if this is true.

.

**transgalactic** · September 3rd 2011, 03:26 AM

ahhhhhhhhhhhhh ok
x/(x+1) >t
x>t(x+1)
x-tx>1
x>1/(1-t)
0<t<1 from the data in the question
but how it proves that t is not upper bound
?

**emakarov** · September 3rd 2011, 03:32 AM

Since 0<t<1, 1/(1-t) is well-defined positive number. You can take x to be any integer greater than 1/(1-t), and then [x] / ([x] + 1) > t, disproving that t is the upper bound of {[x] / ([x] + 1) | x >= 0}.

**transgalactic** · September 3rd 2011, 04:02 AM

ahh i understand now part a

regarding part b:
i know a guide line that any curve above f(x)=x is not uniformly continues

so we must prove that there is $\epsilon>0$ so for all $\delta>0$
there is x,y in R which $|x-y|<\delta$ and $|f(x)-f(y)|=>\epsilon$

$|x^3-y^3|=|(x-y)(x^2+xy+y^2)|<|\delta (x^2+xy+y^2)|$
so i choose $\epsilon=2$
$x=\frac{2}{\sqrt{\delta}}$
$y=\frac{3}{\sqrt{\delta}}$
but i know only that $|x-y|<\delta$
i need to find some delta expression so $|x-y|$ will be greater that it.

so the result of the x,y value input will give use greater then 2.

but here i have smaller then delta only expression
what do?

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