floor function inequality

chrislav

Junior Member
prove: $\lfloor a\rfloor<\lfloor b\rfloor \implies a\leq b$
We can use contradiction i suppose,so
let $b<a\implies b<a<\lfloor a\rfloor+1$ and then?

JeffM

Elite Member
You could use cases.

$\text {CASE I: } a \text { and } b \text { are integers.}$
$\therefore \lfloor a \rfloor = a \text { and } \lfloor b \rfloor = b.$
$\text {By hypothesis, } \lfloor a \rfloor < \lfloor b \rfloor.$
$\therefore a < b \implies a \le b.$
How many cases do you need to consider? What would be your approach?

I do not say that this is an elegant approach, but it pretty clearly will work.

• Jomo

pka

Elite Member
prove: $\lfloor a\rfloor<\lfloor b\rfloor \implies a\leq b$We can use contradiction i suppose,so
let $b<a\implies b<a<\lfloor a\rfloor+1$ and then?
First lets investigate the floor function. The floor of [imath]\lfloor x\rfloor[/imath] is the greatest integer which does not exceed [imath] x[/imath].
Examples: [imath]\lfloor 3.4\rfloor=3[/imath], [imath]\lfloor 2\rfloor=2[/imath], [imath]\lfloor \pi^2\rfloor=9[/imath], [imath]\lfloor -1.2\rfloor=-2[/imath] and [imath]\lfloor -\pi\rfloor=-4[/imath] You should study each of those examples, particularly the last two.
It needs to be stressed that [imath]\lfloor x\rfloor[/imath] is an integer having the property that [imath]\lfloor x\rfloor\le x<\lfloor x\rfloor+1[/imath].

• Subhotosh Khan

Dr.Peterson

Elite Member
prove: $\lfloor a\rfloor<\lfloor b\rfloor \implies a\leq b$
We can use contradiction i suppose,so
let $b<a\implies b<a<\lfloor a\rfloor+1$ and then?
Your start is valid, if you want to continue that way.

In addition to the very useful facts that pka emphasized, you might observe that [imath]\lfloor a\rfloor<\lfloor b\rfloor \iff \lfloor a\rfloor+1\le\lfloor b\rfloor[/imath], because both sides are integers.

• Jomo and Cubist

chrislav

Junior Member
You could use cases.

$\text {CASE I: } a \text { and } b \text { are integers.}$
$\therefore \lfloor a \rfloor = a \text { and } \lfloor b \rfloor = b.$
$\text {By hypothesis, } \lfloor a \rfloor < \lfloor b \rfloor.$
$\therefore a < b \implies a \le b.$
How many cases do you need to consider? What would be your approach?

I do not say that this is an elegant approach, but it pretty clearly will work.
when we come to the case where a,b are both real we short off come back to the begining of the problem because this is the most difficult solution , so considering cases does not help much

chrislav

Junior Member
First lets investigate the floor function. The floor of [imath]\lfloor x\rfloor[/imath] is the greatest integer which does not exceed [imath] x[/imath].
Examples: [imath]\lfloor 3.4\rfloor=3[/imath], [imath]\lfloor 2\rfloor=2[/imath], [imath]\lfloor \pi^2\rfloor=9[/imath], [imath]\lfloor -1.2\rfloor=-2[/imath] and [imath]\lfloor -\pi\rfloor=-4[/imath] You should study each of those examples, particularly the last two.
It needs to be stressed that [imath]\lfloor x\rfloor[/imath] is an integer having the property that [imath]\lfloor x\rfloor\le x<\lfloor x\rfloor+1[/imath].
i think i can add one more step: $\lfloor b\rfloor\leq b<a<\lfloor a\rfloor+1$

• Dr.Peterson

JeffM

Elite Member
when we come to the case where a,b are both real we short off come back to the begining of the problem because this is the most difficult solution , so considering cases does not help much
I disagree.

$\text {Case IV: Neither } a \text { nor } b \text { is an integer.}$
$\therefore \exists \text { integers } p \text { and } q \text { such that } p < a < a + 1 \text { and } q < b < q + 1.$
$\therefore p = \lfloor a \rfloor \text { and } q = \lfloor b \rfloor \text { by definition of floor function.}$
$\text {By hypothesis, } \lfloor a \rfloor < \lfloor b \rfloor \implies p < q \implies p < p + 1 \le q.$
Now what?

• Cubist

chrislav

Junior Member
Your start is valid, if you want to continue that way.

In addition to the very useful facts that pka emphasized, you might observe that [imath]\lfloor a\rfloor<\lfloor b\rfloor \iff \lfloor a\rfloor+1\le\lfloor b\rfloor[/imath], because both sides are integers.
how can you prove: [imath]\lfloor a\rfloor<\lfloor b\rfloor \iff \lfloor a\rfloor+1\le\lfloor b\rfloor[/imath],

JeffM

Elite Member
how can you prove: [imath]\lfloor a\rfloor<\lfloor b\rfloor \iff \lfloor a\rfloor+1\le\lfloor b\rfloor[/imath],
Because they are integers.

Dr.Peterson

Elite Member
how can you prove: [imath]\lfloor a\rfloor<\lfloor b\rfloor \iff \lfloor a\rfloor+1\le\lfloor b\rfloor[/imath],
The next integer after [imath]\lfloor a\rfloor[/imath] is [imath]\lfloor a\rfloor+1[/imath]; so [imath]\lfloor b\rfloor[/imath] must be at least that.

chrislav

Junior Member
I disagree.

$\text {Case IV: Neither } a \text { nor } b \text { is an integer.}$
$\therefore \exists \text { integers } p \text { and } q \text { such that } p < a < a + 1 \text { and } q < b < q + 1.$
$\therefore p = \lfloor a \rfloor \text { and } q = \lfloor b \rfloor \text { by definition of floor function.}$
$\text {By hypothesis, } \lfloor a \rfloor < \lfloor b \rfloor \implies p < q \implies p < p + 1 \le q.$
Now what?
I think that:
there exists an integer p for a such that : $p\leq a<p+1$and there exists an ineger q for b such that:$q\leq b<q+1$then you can have :$p=\lfloor a\rfloor$and $q=\lfloor b\rfloor$However i cannot continue with your last implication

chrislav

Junior Member
The next integer after [imath]\lfloor a\rfloor[/imath] is [imath]\lfloor a\rfloor+1[/imath]; so [imath]\lfloor b\rfloor[/imath] must be at least that.
yes you are right but i am looking for the particular theorem in natural Nos that justify that

Dr.Peterson

Elite Member
yes you are right but i am looking for the particular theorem in natural Nos that justify that
If you're looking in some particular list of theorems, you'll have to tell us what that list is! Perhaps it is incomplete. Perhaps it is not immediately obvious that a theorem in the list applies here.

But this is a theorem; it is true. The successor of any given natural number n is n + 1. So any natural number greater than n must be at least n + 1. What do you need in order to believe this?

JeffM

Elite Member
I think that:
there exists an integer p for a such that : $p\leq a<p+1$and there exists an ineger q for b such that:$q\leq b<q+1$then you can have :$p=\lfloor a\rfloor$and $q=\lfloor b\rfloor$However i cannot continue with your last implication
We are dealing with case IV, which means neither a nor b is an integer.

$\therefore \exists \ p = \lfloor a \rfloor\implies p < a < p + 1.$
Any questions?

$\text {And } \exists \text { integer } q = \lfloor b \rfloor \implies q < b < q + 1.$
$\text {By hypothesis, } \lfloor a \rfloor < \lfloor b \rfloor \implies p < q \implies p < p + 1 \le q.$
$p < a < p + 1 \le q < b < q + 1 \implies a < q < b \implies a < b \implies a \le b. \text { Q.E.D.}$

• chrislav

JeffM

Elite Member
You still may be correct that a proof by contradiction is simpler. But a proof by cases is quite feasible. Try it for cases II and III.

I went for a proof by cases because the relationship between a number and its floor is straightforward by cases.

chrislav

Junior Member
You still may be correct that a proof by contradiction is simpler. But a proof by cases is quite feasible. Try it for cases II and III.

I went for a proof by cases because the relationship between a number and its floor is straightforward by cases.
your last proof covers all the other cases since a,b are reals and either of them can be an integer
IT is like when you are asked to prove :$(x+y)^2 = x^2+2xy+y^2$ and you consider cases

JeffM

Elite Member
your last proof covers all the other cases since a,b are reals and either of them can be an integer
IT is like when you are asked to prove :$(x+y)^2 = x^2+2xy+y^2$ and you consider cases
No, my last case does not cover all cases. You can generalize my last case to cover all four cases.