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- Thread starter chrislav
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[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.

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First lets investigate the floor function. The floor of [imath]\lfloor x\rfloor[/imath] is theprove: [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?

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.

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

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.

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

[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.

i think i can add one more step: [math]\lfloor b\rfloor\leq b<a<\lfloor a\rfloor+1[/math]First lets investigate the floor function. The floor of [imath]\lfloor x\rfloor[/imath] is thegreatest 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 disagree.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

[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?

how can you prove: [imath]\lfloor a\rfloor<\lfloor b\rfloor \iff \lfloor a\rfloor+1\le\lfloor b\rfloor[/imath],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.

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

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The nexthow can you prove: [imath]\lfloor a\rfloor<\lfloor b\rfloor \iff \lfloor a\rfloor+1\le\lfloor b\rfloor[/imath],

I think that: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?

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

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

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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.yes you arerightbut i am looking for the particulartheoremin natural Nos that justify that

But this

We are dealing with case IV, which means neither a nor b is an integer.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

[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]

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

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

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.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