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- Thread starter chrislav
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Please show what you did -- you should know that by now!is it provable the following

[math]\lfloor x^n\rfloor\geq(\lfloor x\rfloor)^n[/math] for all naturals n and reals x

I can prove it for n=2,3,4

But i cannot generilised it to n

It may be that we can help you generalize it. But we're not just going to do it for you.

Hint: it is provable. Can you post you proof for n=3 and 4?

for n=3 we have:

Let x=n+a where n is a natural No and [math]0\leq a<1[/math]then:

[math]\lfloor(n^3+3n^2a+3na^2+a^3\rfloor\geq n^3\Leftrightarrow\lfloor(3n^2a+3na^2+a^3)\rfloor[/math][math]\Leftrightarrow(3n^2a+3na^2+a^3)\geq 0[/math]Now for a=0 we have [math]0\geq 0[/math] which is correct

For [math]a\neq 0[/math] we have:

[math](3n^2+3na+a^2)\geq 0[/math] which is correct

Hence :

[math]\lfloor x^3\rfloor\geq(\lfloor x\rfloor)^3[/math]

I would definitely attack this one by cases. For one thing, I am not sure it is even true for negative x. (Not saying it is false; just saying it is not intuitively obvious to me that it is true for negative x.)is it provable the following

[math]\lfloor x^n\rfloor\geq(\lfloor x\rfloor)^n[/math] for all naturals n and reals x

I can prove it for n=2,3,4

But i cannot generilised it to n

My first case would be the trivial one that n = 1.

My next case would be the trivial one that x is an integer.

[math]x \in \mathbb Z \implies (\lfloor x \rfloor = x \text { and } x^n \in \mathbb Z \implies[/math]

[math]\lfloor x^n \rfloor = x^n = (\lfloor x \rfloor)^n.[/math]

Then I would go for proofs by induction. Based on my first case, we can certainly say

[math]\exists \text { integer } k \ge 1 \text { such that } \lfloor x^k \rfloor \ge (\lfloor x \rfloor )^k \text { for non-integer } x > 0.[/math]

Now see what that entails for k + 1 given that

[math]\exists \text { integer } p \text { such that } 0 \le p < x < p + 1 \implies p^k < x^k < (p + 1)^k.[/math]

Sorry but there is a mistake here it should be;for n=3 we have:

Let x=n+a where n is a natural No and [math]0\leq a<1[/math]then:

[math]\lfloor(n^3+3n^2a+3na^2+a^3\rfloor\geq n^3\Leftrightarrow\lfloor(3n^2a+3na^2+a^3)\rfloor[/math][math]\Leftrightarrow(3n^2a+3na^2+a^3)\geq 0[/math]Now for a=0 we have [math]0\geq 0[/math] which is correct

For [math]a\neq 0[/math] we have:

[math](3n^2+3na+a^2)\geq 0[/math] which is correct

Hence :

[math]\lfloor x^3\rfloor\geq(\lfloor x\rfloor)^3[/math]

for k=3 where k is natural No

and n should be an integer

how do you know that if x belongs to integers then [math] x^n [/math] belongs to integers as wellI would definitely attack this one by cases. For one thing, I am not sure it is even true for negative x. (Not saying it is false; just saying it is not intuitively obvious to me that it is true for negative x.)

My first case would be the trivial one that n = 1.

My next case would be the trivial one that x is an integer.

[math]x \in \mathbb Z \implies (\lfloor x \rfloor = x \text { and } x^n \in \mathbb Z \implies[/math]

[math]\lfloor x^n \rfloor = x^n = (\lfloor x \rfloor)^n.[/math]

Then I would go for proofs by induction. Based on my first case, we can certainly say

[math]\exists \text { integer } k \ge 1 \text { such that } \lfloor x^k \rfloor \ge (\lfloor x \rfloor )^k \text { for non-integer } x > 0.[/math]

Now see what that entails for k + 1 given that

[math]\exists \text { integer } p \text { such that } 0 \le p < x < p + 1 \implies p^k < x^k < (p + 1)^k.[/math]

Because the integers are closed with respect to multiplication: the product of integers is an integer.how do you know that if x belongs to integers then [math] x^n [/math] belongs to integers as well

yes but how do know that holds for all nBecause the integers are closed with respect to multiplication: the product of integers is an integer.

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That depends on what your starting point is in your class.yes but how do know that holds for all n

Presumably you are starting from some definitions and axioms, or perhaps they are just called properties. Closure may just be given, or might have been proved; but you wouldn't be assigned this proof without knowing that it is true.

Please tell us about your class, as well as why you aren't sure that the integers are closed with respect to multiplication.

This is the problem with questions about proofs. We do not know what axioms you start with and what theorems have already been proved prior to the specific theorem you are asking about.yes but how do know that holds for all n

In most courses, you are not expected to prove that the sum of two integers is itself an integer or that the product of two integers is itself an integer. Such propositions are treated as axioms along with commutivity and the distribution law. Your course probably starts by assuming that the integers have the properties of a commutative group with respect to addition and the properties of a commutative monoid with respect to multiplication.

In a foundations course, you might start with just the Peano postulates. In an abstract algebra course, you might start with the axioms and definitions of a semi-group.

Last edited:

my question was clear how do we know(prove) that [math] x^n[/math] belongs to integers for all nThat depends on what your starting point is in your class.

Presumably you are starting from some definitions and axioms, or perhaps they are just called properties. Closure may just be given, or might have been proved; but you wouldn't be assigned this proof without knowing that it is true.

Please tell us about your class, as well as why you aren't sure that the integers are closed with respect to multiplication.

given that x belongs to integers

I know about axioms theorems definitions and all that stuff.

In 1st order theories in any axiomatic development inclusion is not mentioned at all

But in other axiomatic developments some books use the inclusion property as an axiom

I do not belong to any class.

For example in any 1st order axiomatic development the following theorem substitutes the inclusion property

[math]\forall x\forall y\exists z![x+y=z][/math] for the operation of addition

or [math]\forall x\forall y\exists z![x.y=z][/math] for the operation of multiplication

In some other axiomatic formulations [math]\forall x\forall y\exists z![x+y=z][/math] is taken as an axiom.There is no need for it because informality says again that addition is a binary operation and formality says it can be proved.

But in our case we have not only two Nos but n Nos that cover an infinite amount of Nos and i wonder how can we prove that

By induction ,how?

This is the problem with questions about proofs. We do not know what axioms you start with and what theorems have already been proved prior to the specific theorem you are asking about.

In most courses, you are not expected to prove that the sum of two integers is itself an integer or that the product of two integers is itself an integer. Such propositions are treated as axioms along with commutivity and the distribution law. Your course probably starts by assuming that the integers have the properties of a commutative group with respect to addition and the properties of a commutative monoid with respect to multiplication.

In a foundations course, you might start with just the Peano postulates. In an abstract algebra course, you might start with the axioms and definitions of a semi-group.

[math]\text {Given } n \text { is an integer } \ge 0, \text { define }\\ x^n = 1 \text { if } n = 0;\\ x^n = x * x^{(n-1)} \text { if } n > 0.[/math]

[math]\text {Prove, given } x, \ n \in \mathbb Z \text { and } n \ge 0, \text { that } x^n \in \mathbb Z.[/math]

[math]n = 0 \implies x^n = 1 \in \mathbb Z.[/math]

[math]\therefore \exists \text { integer } k \ge 0 \text { such that } x^k \in \mathbb Z.[/math]

[math]\therefore k + 1 \text { is an integer} > 0 \implies x^{(k+1)} = x * x^{\{(k+1)-1\}} = x * x^k.[/math]

[math]x \text { and } x^k \text { are integers} \implies x * x^k \text { is an integer} \implies[/math]

[math]x^{(k+1)} \text { is an integer.}[/math]

yes but behind every definition we must have a theorem of existence

[math]\text {Given } n \text { is an integer } \ge 0, \text { define }\\ x^n = 1 \text { if } n = 0;\\ x^n = x * x^{(n-1)} \text { if } n > 0.[/math]

[math]\text {Prove, given } x, \ n \in \mathbb Z \text { and } n \ge 0, \text { that } x^n \in \mathbb Z.[/math]

[math]n = 0 \implies x^n = 1 \in \mathbb Z.[/math]

[math]\therefore \exists \text { integer } k \ge 0 \text { such that } x^k \in \mathbb Z.[/math]

[math]\therefore k + 1 \text { is an integer} > 0 \implies x^{(k+1)} = x * x^{\{(k+1)-1\}} = x * x^k.[/math]

[math]x \text { and } x^k \text { are integers} \implies x * x^k \text { is an integer} \implies[/math]

[math]x^{(k+1)} \text { is an integer.}[/math]

For example behind the definition of sqrt we have a theorem of existence

behind the definition of logarith we have a theorem of existence e.t.c e.t.c

In our case what would be the theorem of existence for the definition you produced for the definition: [math]x^n=x.x^{n-1}[/math]?

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Using the structure of the integers and the fact that if [imath]a\in\Re[/imath] then [imath]\exists! k\in \mathbb{N}[n\le a<k+1[/imath].yes but behind every definition we must have a theorem of existence

For example behind the definition of sqrt we have a theorem of existence

behind the definition of logarith we have a theorem of existence e.t.c e.t.c

In our case what would be the theorem of existence for the definition you produced for the definition: [math]x^n=x.x^{n-1}[/math]?

Your question about [imath]x=x\cdot x^{n-1}[/imath] does not need proof beyond the nature of operation on exponents.

But it is known and proven that there is no integer between [imath]k~\&~k+1[/imath] forall [imath]k\in\mathbb{N}[/imath].

Moreover, we know that [imath]\lfloor a \rfloor[/imath] is the largest integer that does not exceed [imath]a[/imath].

So that gives us [imath]\lfloor a \rfloor \le a<\lfloor a \rfloor +1[/imath] How know that [imath]\lfloor a^k \rfloor\le a^k<\lfloor a^k \rfloor+1~?[/imath] (do we even know that?)

[imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath]

That is plain ridiculous. How can you possibly prove the existence of something that is not defined?yes but behind every definition we must have a theorem of existence

For example behind the definition of sqrt we have a theorem of existence

behind the definition of logarith we have a theorem of existence e.t.c e.t.c

In our case what would be the theorem of existence for the definition you produced for the definition: [math]x^n=x.x^{n-1}[/math]?