floor function inequality

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chrislav

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prove: [math]\lfloor a\rfloor<\lfloor b\rfloor \implies a\leq b[/math]
We can use contradiction i suppose,so
let [math]b<a\implies b<a<\lfloor a\rfloor+1[/math] and then?
 
You could use cases.

[math]\text {CASE I: } a \text { and } b \text { are integers.}[/math]
[math]\therefore \lfloor a \rfloor = a \text { and } \lfloor b \rfloor = b.[/math]
[math]\text {By hypothesis, } \lfloor a \rfloor < \lfloor b \rfloor.[/math]
[math]\therefore a < b \implies a \le b.[/math]
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.
 
chrislav said:
prove: [math]\lfloor a\rfloor<\lfloor b\rfloor \implies a\leq b[/math]We can use contradiction i suppose,so
let [math]b<a\implies b<a<\lfloor a\rfloor+1[/math] and then?
Click to expand...
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].
Please post a proof for comments.
 
chrislav said:
prove: [math]\lfloor a\rfloor<\lfloor b\rfloor \implies a\leq b[/math]
We can use contradiction i suppose,so
let [math]b<a\implies b<a<\lfloor a\rfloor+1[/math] and then?
Click to expand...
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.
 
JeffM said:
You could use cases.

[math]\text {CASE I: } a \text { and } b \text { are integers.}[/math]
[math]\therefore \lfloor a \rfloor = a \text { and } \lfloor b \rfloor = b.[/math]
[math]\text {By hypothesis, } \lfloor a \rfloor < \lfloor b \rfloor.[/math]
[math]\therefore a < b \implies a \le b.[/math]
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.
Click to expand...
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
 
pka said:
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].
Please post a proof for comments.
Click to expand...
i think i can add one more step: [math]\lfloor b\rfloor\leq b<a<\lfloor a\rfloor+1[/math]
 
chrislav said:
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
Click to expand...
I disagree.

[math]\text {Case IV: Neither } a \text { nor } b \text { is an integer.}[/math]
[math]\therefore \exists \text { integers } p \text { and } q \text { such that } p < a < a + 1 \text { and } q < b < q + 1.[/math]
[math]\therefore p = \lfloor a \rfloor \text { and } q = \lfloor b \rfloor \text { by definition of floor function.}[/math]
[math]\text {By hypothesis, } \lfloor a \rfloor < \lfloor b \rfloor \implies p < q \implies p < p + 1 \le q.[/math]
Now what?
 
Dr.Peterson said:
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.
Click to expand...
how can you prove: [imath]\lfloor a\rfloor<\lfloor b\rfloor \iff \lfloor a\rfloor+1\le\lfloor b\rfloor[/imath],
 
chrislav said:
how can you prove: [imath]\lfloor a\rfloor<\lfloor b\rfloor \iff \lfloor a\rfloor+1\le\lfloor b\rfloor[/imath],
Click to expand...
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.
 
JeffM said:
I disagree.

[math]\text {Case IV: Neither } a \text { nor } b \text { is an integer.}[/math]
[math]\therefore \exists \text { integers } p \text { and } q \text { such that } p < a < a + 1 \text { and } q < b < q + 1.[/math]
[math]\therefore p = \lfloor a \rfloor \text { and } q = \lfloor b \rfloor \text { by definition of floor function.}[/math]
[math]\text {By hypothesis, } \lfloor a \rfloor < \lfloor b \rfloor \implies p < q \implies p < p + 1 \le q.[/math]
Now what?
Click to expand...
I think that:
there exists an integer p for a such that : [math]p\leq a<p+1[/math]and there exists an ineger q for b such that:[math]q\leq b<q+1[/math]then you can have :[math]p=\lfloor a\rfloor[/math]and [math]q=\lfloor b\rfloor[/math]However i cannot continue with your last implication
 
Dr.Peterson said:
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.
Click to expand...
yes you are right but i am looking for the particular theorem in natural Nos that justify that
 
chrislav said:
yes you are right but i am looking for the particular theorem in natural Nos that justify that
Click to expand...
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?
 
chrislav said:
I think that:
there exists an integer p for a such that : [math]p\leq a<p+1[/math]and there exists an ineger q for b such that:[math]q\leq b<q+1[/math]then you can have :[math]p=\lfloor a\rfloor[/math]and [math]q=\lfloor b\rfloor[/math]However i cannot continue with your last implication
Click to expand...
We are dealing with case IV, which means neither a nor b is an integer.

[math]\therefore \exists \ p = \lfloor a \rfloor\implies p < a < p + 1.[/math]
Any questions?

[math]\text {And } \exists \text { integer } q = \lfloor b \rfloor \implies q < b < q + 1.[/math]
[math]\text {By hypothesis, } \lfloor a \rfloor < \lfloor b \rfloor \implies p < q \implies p < p + 1 \le q.[/math]
[math]p < a < p + 1 \le q < b < q + 1 \implies a < q < b \implies a < b \implies a \le b. \text { Q.E.D.}[/math]
 
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.
 
JeffM said:
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.
Click to expand...
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 :[math](x+y)^2 = x^2+2xy+y^2[/math] and you consider cases
 
chrislav said:
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 :[math](x+y)^2 = x^2+2xy+y^2[/math] and you consider cases
Click to expand...
No, my last case does not cover all cases. You can generalize my last case to cover all four cases.
 
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