floor function inequality

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?
 

JeffM

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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.
 

pka

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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?
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.
 

Dr.Peterson

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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?
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.
 

chrislav

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Jun 22, 2017
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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.
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

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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.
i think i can add one more step: [math]\lfloor b\rfloor\leq b<a<\lfloor a\rfloor+1[/math]
 

JeffM

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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.

[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?
 

chrislav

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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],
 

Dr.Peterson

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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

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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?
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
 

chrislav

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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

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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

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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
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]
 

JeffM

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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

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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 :[math](x+y)^2 = x^2+2xy+y^2[/math] and you consider cases
 

JeffM

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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
No, my last case does not cover all cases. You can generalize my last case to cover all four cases.
 
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