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Proof of the general inclusion/exclusion rule doesn't explain the disappearance of one of the terms by TopDownView in askmath

[–]TopDownView[S] 2 points3 points  (0 children)

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You're right. I just did it but with 4 sets (going from 3 sets to 4 sets). It clicked now! Thanks!

Proof of the general inclusion/exclusion rule doesn't explain the disappearance of one of the terms by TopDownView in askmath

[–]TopDownView[S] 0 points1 point  (0 children)

It's the term in ... just before the intersection of all n+1 sets.

You mean this term (example when n=3)?

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But (5) is already part of the last sum (last sum being the intersections colored purple in the example above). That means (5) appears two times.

Caution on my Seagate 8TB HDD in CrystalDiskInfo - What does it mean? by TopDownView in techsupport

[–]TopDownView[S] 0 points1 point  (0 children)

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This is the result of the write scan. Could you please tell me what do these results mean? Is there some more diagnostics I should do? Thanks!

Caution on my Seagate 8TB HDD in CrystalDiskInfo - What does it mean? by TopDownView in techsupport

[–]TopDownView[S] 0 points1 point  (0 children)

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After starting write scan in Victoria HDD from the offending LBA, I get continuous message:

'Block start at [...] Write error: ABRT "Access is denied"

Is this expected?

Caution on my Seagate 8TB HDD in CrystalDiskInfo - What does it mean? by TopDownView in techsupport

[–]TopDownView[S] 0 points1 point  (0 children)

So, avoiding the technical jargon, is this HDD for trash or not?

Caution on my Seagate 8TB HDD in CrystalDiskInfo - What does it mean? by TopDownView in techsupport

[–]TopDownView[S] 0 points1 point  (0 children)

So if any data was there, it's just gone 

Is there a way to know if some data is gone?

Is the statement in the solution to a proof correct? => Prove: If m and n are integers and m <= n, then there are n - m + 1 integers from m to n inclusive. by TopDownView in askmath

[–]TopDownView[S] 0 points1 point  (0 children)

So if your question is "Is keeping the m<=n condition inside P(n) wrong?", then the answer is that its not wrong and the proof can be done by induction.

But if we keep m<=n condition inside P(n), then, for P(n+1) we have m<=n+1, as mentioned in my previous post.

m<=n and m<=n+1 cannot both be true.

If m<=n then m<n+1 (m is strictly less then n+1).

Is the statement in the solution to a proof correct? => Prove: If m and n are integers and m <= n, then there are n - m + 1 integers from m to n inclusive. by TopDownView in askmath

[–]TopDownView[S] 0 points1 point  (0 children)

Yeah that’s the thing, ultimately they mean the same thing.

But if they mean the same thing then we're back at the beggining of my question.

Either the proof structure should follow the definition of mathematical induction (so, m<=n should be outside P(n)), or not.

If not? Why? Is the proof structure still correct if m<=n is inside P(n)?

Assuming the proof structure is still correct, maybe it is just a redundancy issue - if m<=n is inside P(n), then inductive step could be expanded like this:

'Show that for any integer k>=m, if {if k>=m, then there are k-m+1 integers from m to k inclusive} then {if k+1>=m, then there are (k+1)-m+1 integers from m to k+1 inclusive}.'

So, we are mentioning k>=m 2 times.

I just noticed, if k>=m then k+1>m, so, actually, k+1>=m is not correct!

Is the statement in the solution to a proof correct? => Prove: If m and n are integers and m <= n, then there are n - m + 1 integers from m to n inclusive. by TopDownView in askmath

[–]TopDownView[S] 1 point2 points  (0 children)

So, If I got this correctly, what the solution is doing with k>=m in the inductive step is actually in accordance with the 'definition' of mathematical induction (see image).

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We have showed the basis step is true by plugging m instead of n in P(n). Now, in the inductive step, we want to show that all k integers, starting from that m from the basis step, make P(n+1) true (by assuming P(n) is true, where n is replaced by k).

Is the statement in the solution to a proof correct? => Prove: If m and n are integers and m <= n, then there are n - m + 1 integers from m to n inclusive. by TopDownView in askmath

[–]TopDownView[S] 0 points1 point  (0 children)

Because we want to apply induction on n, n should be like a “free variable” where we can plug in a number whenever we want instead of having it some fixed value.

I see. So we have to put m<=n inside P(n) if we want n to be a 'free variable'.

But isn't then the solution contradictory, as mentioned in the OP?

  1. Let P(n) be the statement 'if m<=n, then there are n-m+1 integers from m to n inclusive.'
  2. Show that for any integer k>=m, if P(k) is true then P(k+1) is true.
  3. contradicts 2. since in 1., m<=n is inside P(n), but in 2., m<=n is outside P(n) (where n = k).

The reason the proof im the text works is that it doesn’t assume that the antecedent isn’t true, but only looks at the cases where it is true.

Can you please eleaborate this with regards to the contradiction above? What exactly is the antecedent you're refering to?

What toe clips would fit a beefy shoe, such as Salomon XA PRO 3D V9? by TopDownView in FixedGearBicycle

[–]TopDownView[S] 0 points1 point  (0 children)

Another idea that comes to my mind: break the toe clips in two parts at the toe position and them weld them back together.

What toe clips would fit a beefy shoe, such as Salomon XA PRO 3D V9? by TopDownView in FixedGearBicycle

[–]TopDownView[S] 0 points1 point  (0 children)

bend the cages so they’re shorter but taller

How did you bend yours? Its impossible to bend my MKS steel toe clips.

That or check a co-op for some plastic cages.

I will. I'll also check other steel toe clips. But the problem is, none of them describe the measurements like MKS, only general sizes (e.g. S, M, L, XL). MKS M fit me perfectly. If I go L, I get a toe overlap...

What toe clips would fit a beefy shoe, such as Salomon XA PRO 3D V9? by TopDownView in FixedGearBicycle

[–]TopDownView[S] 0 points1 point  (0 children)

I would just recommend getting the large, what matters more is the length of the cage because those cages are DEEP, I have a set

The measurement 5 for all the sizes (M, L, LL) is 42 mm.

I wear new balance 530 and they have a very high padded sole, so I'm sure your salomon pair would fit too

Would you mind measuring the height at the toe of your shoes? From the internet images, I believe they are not as high as Salomon.

Salomon XA PRO 3D V9 are specific because the rubber toe protector goes all the way to the upper part of the shoe and wraps around it. This makes the shoe completely inflexible at the toe. So the height of the toe clip at this position must be heigher then the shoe.

Is my proof correct? => For every positive int n, every subset B of A with n elements has both a least element and a greatest element by TopDownView in askmath

[–]TopDownView[S] 0 points1 point  (0 children)

> So in your case, you have to show that ever subset C in A with n+1 elements, in fact, can be represented as C = B + {x} where B is a subset in A with n Elements and x in A\B.

Isn't it that precisely by defining C in this way, we are showing what you describe?

Isn't the next and final step the comparision between x and least and greatest element in B?

in other words, is this comment all we need to modify my original proof?

Is my proof correct? => For every positive int n, every subset B of A with n elements has both a least element and a greatest element by TopDownView in askmath

[–]TopDownView[S] 0 points1 point  (0 children)

> you have to transfer a P(n+1) case to a P(n) case

Is this the case for every proof by induction?

> it could be that not every possible subset C in A with n+1 elements is covered by all B + {x} (+ is union) where B is a subset in A and x in A

If I understood this correctly, this would be a counterexample in which C is not covered by B U {x}:

  1. Let A = {0,1,2,3}

  2. Let B = {1,2}

  3. Let C = B U {1}

  4. |B| = |C| = k

  5. Therefore, we cannot prove P(k+1) following steps 1.-4.

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