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# Each year for 4 years, a farmer increased the number of trees in a

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Each year for 4 years, a farmer increased the number of trees in a [#permalink]

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08 Jul 2012, 05:04
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Each year for 4 years, a farmer increased the number of trees in a certain orchard by 1/4 of the number of trees in the orchard of the preceding year. If all of the trees thrived and there were 6250 trees in the orchard at the end of 4 year period, how many trees were in the orchard at the beginning of the 4 year period.

A. 1250
B. 1563
C. 2250
D. 2560
E. 2752

[Reveal] Spoiler:
Can someone walk me through the logic behind this question. I am able to solve this by using options as well as by assuming the number of trees = x. However, had the question been, "If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period". then it would have been difficult to solve.

Thanks
[Reveal] Spoiler: OA

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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]

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08 Jul 2012, 05:35
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imhimanshu wrote:
Each year for 4 years, a farmer increased the number of trees in a certain orchard by 1/4 of the number of trees in the orchard of the preceding year. If all of the trees thrived and there were 6250 trees in the orchard at the end of 4 year period, how many trees were in the orchard at the beginning of the 4 year period.

A. 1250
B. 1563
C. 2250
D. 2560
E. 2752

Can someone walk me through the logic behind this question. I am able to solve this by using options as well as by assuming the number of trees = x. However, had the question been, "If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period". then it would have been difficult to solve.

Thanks

Say the number of trees at the beginning of the 4 year period was x, then:
At the end of the 1st year the number of trees would be $$x+\frac{1}{4}x=\frac{5}{4}*x$$;
At the end of the 2nd year the number of trees would be $$(\frac{5}{4})^2*x$$;
At the end of the 3rd year the number of trees would be $$(\frac{5}{4})^3*x$$;
At the end of the 4th year the number of trees would be $$(\frac{5}{4})^4*x$$;
At the end of the $$n_{th}$$ year the number of trees would be $$(\frac{5}{4})^n*x$$;

So, we have that $$(\frac{5}{4})^4*x=6,250$$ --> $$\frac{5^4}{4^4}*x=5^4*10$$ --> $$x=4^4*10=2,560$$.

If the question were "if all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period", then we would have that: $$(\frac{5}{4})^{15}*x=6,250$$ --> $$x\neq{integer}$$, so it would be a flawed question.

Hope it's clear.
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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]

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08 Jul 2012, 22:32
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imhimanshu wrote:
Each year for 4 years, a farmer increased the number of trees in a certain orchard by 1/4 of the number of trees in the orchard of the preceding year. If all of the trees thrived and there were 6250 trees in the orchard at the end of 4 year period, how many trees were in the orchard at the beginning of the 4 year period.

A. 1250
B. 1563
C. 2250
D. 2560
E. 2752

Can someone walk me through the logic behind this question. I am able to solve this by using options as well as by assuming the number of trees = x. However, had the question been, "If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period". then it would have been difficult to solve.

Thanks

The number of trees increases by 1/4 i.e. 25% every year. It is just a matter of thinking in terms of successive percentage changes e.g. population increase. Here, we are talking about the increase of tree population.

If x increases by 25%, how we denote it? (5/4)*x
If next year, this new number increases by 25% again, how do we denote it? (5/4)*(5/4)*x
and so on...

For more on this, check: http://www.veritasprep.com/blog/2011/02 ... e-changes/

So if we are taking into account 4 years, we simply get (5/4)^4 * x = 6250

As for your next question, the numbers given would be such that the calculation will not be tough.

Say, you have 8 years and 100% increase every year (population doubles every year). The final population will be divisible by 2^8 i.e. 256.
Something like 2^8 * x = 2560

and if you meant what you wrote (though I considered that the 4 was a typo because of the language of the question) "If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period", note that you still need to work with

(5/4)^4 * x = 6250
since you need the number of trees 4 yrs back only. The only thing is that the answer (2560) needs to be divisible by $$5^{11}$$ which it isn't so there is a problem in this question. If it were an actual question, the answer would be divisible by $$5^{11}$$ but you wouldn't really need to bother about it.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews SVP Status: The Best Or Nothing Joined: 27 Dec 2012 Posts: 1838 Location: India Concentration: General Management, Technology WE: Information Technology (Computer Software) Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink] ### Show Tags 25 Oct 2013, 00:26 4 This post received KUDOS 2 This post was BOOKMARKED Trees increase 1/4th every year, which means 100 trees become 125 after 1 yr & so on So, 125/100 = 5/4 is the resultant (after adding 1/4 as the growth) for 1 year So, for 4 yrs is 5^4/(4^4) From the condition given, inital trees were = 6250 x 4^4 / 5^4 = 2560 _________________ Kindly press "+1 Kudos" to appreciate Manager Joined: 12 Jan 2013 Posts: 212 Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink] ### Show Tags 27 Dec 2013, 09:31 Bunuel wrote: imhimanshu wrote: Each year for 4 years, a farmer increased the number of trees in a certain orchard by 1/4 of the number of trees in the orchard of the preceding year. If all of the trees thrived and there were 6250 trees in the orchard at the end of 4 year period, how many trees were in the orchard at the beginning of the 4 year period. A. 1250 B. 1563 C. 2250 D. 2560 E. 2752 Can someone walk me through the logic behind this question. I am able to solve this by using options as well as by assuming the number of trees = x. However, had the question been, "If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period". then it would have been difficult to solve. Thanks Say the number of trees at the beginning of the 4 year period was x, then: At the end of the 1st year the number of trees would be $$x+\frac{1}{4}x=\frac{5}{4}*x$$; At the end of the 2nd year the number of trees would be $$(\frac{5}{4})^2*x$$; At the end of the 3rd year the number of trees would be $$(\frac{5}{4})^3*x$$; At the end of the 4th year the number of trees would be $$(\frac{5}{4})^4*x$$; At the end of the $$n_{th}$$ year the number of trees would be $$(\frac{5}{4})^n*x$$; So, we have that $$(\frac{5}{4})^4*x=6,250$$ --> $$\frac{5^4}{4^4}*x=5^4*10$$ --> $$x=4^4*10=2,560$$. Answer: D. If the question were "if all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period", then we would have that: $$(\frac{5}{4})^{15}*x=6,250$$ --> $$x\neq{integer}$$, so it would be a flawed question. Hope it's clear. Isn't the question quite ambiguous, though? I mean the first scentence could be interpreted as "for the first year we have (4/4)x and for the second year (5/4)x and for the third..." etc.. With that reasoning one would have (5/4)^3 * x + x and then your approach doesnt work. Obviously, I understand that this was a flaw in my reasoning but I cannot understand how they - with that wording - will assume that we totally understand that at the end of year one he has (5/4)x.. Is there a straightforward "word translation" way in knowing how to interpret wordings like this? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7996 Location: Pune, India Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink] ### Show Tags 01 Jan 2014, 23:34 Expert's post 2 This post was BOOKMARKED aeglorre wrote: Bunuel wrote: imhimanshu wrote: Each year for 4 years, a farmer increased the number of trees in a certain orchard by 1/4 of the number of trees in the orchard of the preceding year. If all of the trees thrived and there were 6250 trees in the orchard at the end of 4 year period, how many trees were in the orchard at the beginning of the 4 year period. A. 1250 B. 1563 C. 2250 D. 2560 E. 2752 Can someone walk me through the logic behind this question. I am able to solve this by using options as well as by assuming the number of trees = x. However, had the question been, "If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period". then it would have been difficult to solve. Thanks Say the number of trees at the beginning of the 4 year period was x, then: At the end of the 1st year the number of trees would be $$x+\frac{1}{4}x=\frac{5}{4}*x$$; At the end of the 2nd year the number of trees would be $$(\frac{5}{4})^2*x$$; At the end of the 3rd year the number of trees would be $$(\frac{5}{4})^3*x$$; At the end of the 4th year the number of trees would be $$(\frac{5}{4})^4*x$$; At the end of the $$n_{th}$$ year the number of trees would be $$(\frac{5}{4})^n*x$$; So, we have that $$(\frac{5}{4})^4*x=6,250$$ --> $$\frac{5^4}{4^4}*x=5^4*10$$ --> $$x=4^4*10=2,560$$. Answer: D. If the question were "if all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period", then we would have that: $$(\frac{5}{4})^{15}*x=6,250$$ --> $$x\neq{integer}$$, so it would be a flawed question. Hope it's clear. Isn't the question quite ambiguous, though? I mean the first scentence could be interpreted as "for the first year we have (4/4)x and for the second year (5/4)x and for the third..." etc.. With that reasoning one would have (5/4)^3 * x + x and then your approach doesnt work. Obviously, I understand that this was a flaw in my reasoning but I cannot understand how they - with that wording - will assume that we totally understand that at the end of year one he has (5/4)x.. Is there a straightforward "word translation" way in knowing how to interpret wordings like this? Actually, it is not ambiguous. Read the statement: Each year a farmer increased the number of trees by 1/4. He did this for 4 years. (In GMAT Verbal and Quant are integrated. You need Verbal skills (slash and burn) in Quant and Quant skills (Data Interpretation) in Verbal. So in the first year, he increased it by 1/4 The next year, he again increased it by 1/4 (of preceding year) Next year, again the same. Next year, again the same. So he did it for a total of 4 years. So if initially the number of trees was x, in the first year he made them (5/4)x _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]

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07 Feb 2014, 18:28
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it can easily solved by compound interest formula
R = 25%
P = original number of trees
N = no of times of compounding within a year = 1
T = total number of years = 4

P+I at the end of compounding years = 6250
hence,

6250 = (1 + 0.25/1)^(4*1) * P
5^4*10 = 1.25^4 * P
(5/1/25)^4 *10 = P
4^4 * 10 = P
2560 = P

Another way to think is to pick ans choice which is exactly divisible by 4. A, B and C are not. Only D and E are.
Pick D and increase it by 25% for 4 times, you should get 6250.
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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]

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27 Jul 2014, 10:08
Hi Karishma,

I am not able to solve it by below method, where am I wrong:

6250*(3/4)*(3/4)*(3/4)

Regards,
Ravi
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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]

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27 Jul 2014, 22:08
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email2vm wrote:
Hi Karishma,

I am not able to solve it by below method, where am I wrong:

6250*(3/4)*(3/4)*(3/4)

Regards,
Ravi

If I increase A by 25% and get B, I will not get A if I reduce B by 25%. The bases are different in the two cases.

e.g. A = 80
25% of A is 20 so I increase A by 25% to get B = 100

Now if I decrease B by 25%, I will decrease B by 25 (25% of 100). This will give me 75 which is not the same as A (which is 80).

In the first step, I found 25% of 80 and added that. In the second step, I found 25% of 100 and subtracted that. These two numbers are different.

You will get the correct answer if you do 6250*(4/5)*(4/5)*(4/5)*(4/5)
4/5 is the inverse of 5/4 (which is 25% increase in x).
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 13 Feb 2014 Posts: 3 Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink] ### Show Tags 25 Oct 2014, 12:01 You can use the formula A=p(1 + r/100)^n with r = 25% You will get P as 2560. Intern Joined: 23 Aug 2014 Posts: 42 GMAT Date: 11-29-2014 Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink] ### Show Tags 30 Dec 2014, 10:59 Is it true that the number of trees in the beginning has to be a multiple of 4 (as additions are always 1/4 th of previous and will not be non-integers)? In that case I'm left with D or E. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7996 Location: Pune, India Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink] ### Show Tags 31 Dec 2014, 02:53 deeuk wrote: Is it true that the number of trees in the beginning has to be a multiple of 4 (as additions are always 1/4 th of previous and will not be non-integers)? In that case I'm left with D or E. Yes, that's correct. But you might still need to work with two options. On the other hand, working with 6250 will definitely yield the answer. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]

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06 Jan 2015, 10:39
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Trees increase by 1/4 the number of trees in preceding year. Hence, correct answer must be divisible by 4. Based on divisibility rules, if last 2 digits are divisible by 4 then the number is divisible by 4. Thus, we can eliminate A, B, C. The answer has to be D or E.

Again, trees increase by 1/4 the number of trees in preceding year. Hence, the number of trees increase by 5/4 times the number of trees the preceding year.

If x = initial number of trees = Answer D or E = 2560 or 2752
Year 1 = 5/4x
Year 2 = (5/4)(5/4)x
Year 3 = (5/4)(5/4)(5/4)x
Year 4 = (5/4)(5/4)(5/4)(5/4)x

Only for Answer D: (5/4)(5/4)(5/4)(5/4) 2560 = 6250

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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]

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02 Apr 2015, 03:36
@VeritasPrepKarishma
VeritasPrepKarishma wrote:
imhimanshu wrote:
Each year for 4 years, a farmer increased the number of trees in a certain orchard by 1/4 of the number of trees in the orchard of the preceding year. If all of the trees thrived and there were 6250 trees in the orchard at the end of 4 year period, how many trees were in the orchard at the beginning of the 4 year period.

A. 1250
B. 1563
C. 2250
D. 2560
E. 2752

Can someone walk me through the logic behind this question. I am able to solve this by using options as well as by assuming the number of trees = x. However, had the question been, "If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period". then it would have been difficult to solve.

Thanks

The number of trees increases by 1/4 i.e. 25% every year. It is just a matter of thinking in terms of successive percentage changes e.g. population increase. Here, we are talking about the increase of tree population.

If x increases by 25%, how we denote it? (5/4)*x
If next year, this new number increases by 25% again, how do we denote it? (5/4)*(5/4)*x
and so on...

For more on this, check: http://www.veritasprep.com/blog/2011/02 ... e-changes/

So if we are taking into account 4 years, we simply get (5/4)^4 * x = 6250

As for your next question, the numbers given would be such that the calculation will not be tough.

Say, you have 8 years and 100% increase every year (population doubles every year). The final population will be divisible by 2^8 i.e. 256.
Something like 2^8 * x = 2560

and if you meant what you wrote (though I considered that the 4 was a typo because of the language of the question) "If all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period", note that you still need to work with

(5/4)^4 * x = 6250
since you need the number of trees 4 yrs back only. The only thing is that the answer (2560) needs to be divisible by $$5^{11}$$ which it isn't so there is a problem in this question. If it were an actual question, the answer would be divisible by $$5^{11}$$ but you wouldn't really need to bother about it.

Karisma,

However, it still isn't clear to me why we are keeping the same base for the increase during the 4-year period time. To me, this seems to be an example of successive % in which the base is shifting every year. Indeed every year we have an increase of 25% more on the increase of the previous year.

Why is it incorrect to compute the increase of the successive year on the increased base of the previous year?

Thank you a lot for your help!

Regards,
Eli.
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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]

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02 Apr 2015, 19:43
elisabettaportioli wrote:
However, it still isn't clear to me why we are keeping the same base for the increase during the 4-year period time. To me, this seems to be an example of successive % in which the base is shifting every year. Indeed every year we have an increase of 25% more on the increase of the previous year.

Why is it incorrect to compute the increase of the successive year on the increased base of the previous year?

Thank you a lot for your help!

Regards,
Eli.

The base for each year changes.

Say, you have 100 trees and you increase them by 20% each year.

At the end of the first year, you will have 100 + (20/100)*100 = 100*(1 + 20/100) = 100*120/100 = 100*(6/5) = 120 trees.

So if you have to increase a number by 20%, you just need to multiply it by 6/5 every time no matter what the number is.

By the same logic, next year, you will have 120 * (6/5) = 144 trees when you increase them by 20%.

You can write 120 as 100 * (6/5) and then multiply it by another 6/5 to increase 120 by 20%.

This is the same as 100 * (6/5) * (6/5) = 144 trees.

So every time you multiply it by 6/5, you increase the base. The next time you multiply it by 6/5, you are increasing the new base by 20%. This is the concept of successive percentage changes and I have discussed it in the link I mentioned in my post above.
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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]

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02 Apr 2015, 20:15
elisabettaportioli wrote:

Why is it incorrect to compute the increase of the successive year on the increased base of the previous year?

Thank you a lot for your help!

Regards,
Eli.

hi,
you are correct that the increase is on the preceeding year and that is exactly what is happening when you are multiplying x by 5/4..
first year say it was x.. 2nd year it becomes 5/4*x.. now for next year when we write the value as 5/4*5/4*x, the increase is on 5/4*x and not x..
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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]

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13 Apr 2015, 08:38
1
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1/4 = 0.25 = 25%
using the formula of compound interest calculation
p(1+r)^n= A

p = principal amount (to be identified), r = rate (0.25), n = numbers years (4), A= amount at the end of term(6250 given)

p(1+0.25)^4=6250
p(1.25)^4= 25*25*10 = 5^2*5^2*10=5^4*10
p(125/100)^4 = 5^4 x 10
p(5/4)^4 = 5^4 x 10
p(5^4)/4^4 = 5^4 x 10
p= (5^4 x 10 x 4^4)/5^4
p= 10 x 4^4
p= 2560
Ans D
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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]

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19 May 2015, 12:37
we have here 4 (4 years) successive increases of 25% or *1,25 --> 1,25^4 * X = 6250, X = 2560

See MGMAT (Percents) for detailed explanation of such question types.....
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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]

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20 May 2015, 03:32
BrainLab wrote:
we have here 4 (4 years) successive increases of 25% or *1,25 --> 1,25^4 * X = 6250, X = 2560

See MGMAT (Percents) for detailed explanation of such question types.....

Dear BrainLab

Perfect logic but for easier calculation, you may want to work with ratio here (1/4 increase per annum) instead of percentages (25% increase per annum). Both convey the same thing but the equation

$$(\frac{5}{4})^4*X = 6250$$

will take lesser time to solve (especially if you know that $$5^4 = 625$$) than $$(1.25)^4*X = 6250$$

Hope this was useful!

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Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]

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In the rectangular coordinate system above, if the equation of m is y  [#permalink]

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15 Dec 2017, 02:09
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In the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?

(A) √2
(B) 1
(C) √2/2
(D) 1/2
(E) 1/4

Attachment:

2017-12-15_1259.png [ 4.85 KiB | Viewed 1084 times ]

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In the rectangular coordinate system above, if the equation of m is y  [#permalink]

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15 Dec 2017, 13:22
2
1
Bunuel wrote:

In the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?

(A) √2
(B) 1
(C) √2/2
(D) 1/2
(E) 1/4

Attachment:
The attachment 2017-12-15_1259.png is no longer available

With the formula (My diagram does not apply)
I did not use it, but to do so is standard. So I quote the formula from Bunuel HERE

Distance between two parallel lines $$y=mx+b$$ and $$y=mx+c$$ can be found by the formula:
$$D=\frac{|b-c|}{\sqrt{m^2+1}}$$

Equation for line $$m$$, $$y = x$$, to match the exact form of $$y = mx + b$$ can be written:
$$y =(1)x + 0$$ (all lines that pass through the origin have x- and y-intercepts of 0)
$$b = 0$$ = y-intercept, $$m = 1$$ = slope

Find the equation of line $$n$$, $$y = mx + c$$

Parallel lines have identical slopes. Line $$n$$ has slope $$m = 1$$
Use the point on line $$n$$ from the graph: (1, 0)
Plug its coordinates into slope-intercept form to find $$c$$
$$y = mx + c$$
$$(0) = 1(1) + c$$
$$-c = 1$$, so $$c = -1$$
Equation for line $$n: y = x - 1$$

Distance between parallel lines

$$D=\frac{|b-c|}{\sqrt{m^2+1}}$$

$$m = 1$$, $$b = 0$$, $$c = -1$$

$$D=\frac{|0-(-1)|}{\sqrt{1^2+1}}$$
$$D=\frac{1}{\sqrt{2}}$$
Does not match the answers. Rationalize the denominator; multiply by $$\frac{\sqrt2}{\sqrt2}$$

$$\frac{1}{\sqrt{2}}$$ * $$\frac{\sqrt2}{\sqrt2} = \frac{\sqrt{2}}{2}$$

Attachment:

linesmandn.png [ 18.05 KiB | Viewed 868 times ]

Without the formula see diagram

The shortest distance between parallel lines is a line perpendicular to both parallel lines
Draw a perpendicular line from point B to point A

That creates right isosceles ∆ ABO where OA = AB
-- The line y = x makes a 45° angle with both the x- and y-axes (as do all lines with slope 1 or -1)
-- ∠ BOA = 45° and ∠ at vertex A = 90° therefore ∠ ABO must = 45°
-- Sides opposite equal angles are equal: OA = AB

A right isosceles triangle
-- has angle measures 45-45-90 and
-- corresponding side lengths of $$x : x : x\sqrt{2}$$

The hypotenuse corresponds with $$x\sqrt{2} = 1$$: OB = 1
Find length of equal sides $$x$$ (one of which, AB, is the distance needed):

$$x\sqrt{2}= 1$$

$$x = \frac{1}{\sqrt{2}}$$
= AB = shortest distance between the parallel lines

That does not look like any of the answers.
Rationalize the denominator (clear the radical): Multiply by $$\frac{\sqrt2}{\sqrt2}$$
$$\frac{1}{\sqrt{2}}$$ * $$\frac{\sqrt2}{\sqrt2} = \frac{\sqrt{2}}{2}$$

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In the rectangular coordinate system above, if the equation of m is y  [#permalink]

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14 Apr 2018, 06:59
generis wrote:

In the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?

(A) √2
(B) 1
(C) √2/2
(D) 1/2
(E) 1/4

Attachment:
2017-12-15_1259.png

Attachment:
linesmandn.png

Without the formula see diagram

The shortest distance between parallel lines is a line perpendicular to both parallel lines
Draw a perpendicular line from point B to point A

That creates right isosceles ∆ ABO where OA = AB
-- The line y = x makes a 45° angle with both the x- and y-axes (as do all lines with slope 1 or -1)
-- ∠ BOA = 45° and ∠ at vertex A = 90° therefore ∠ ABO must = 45°
-- Sides opposite equal angles are equal: OA = AB

A right isosceles triangle
-- has angle measures 45-45-90 and
-- corresponding side lengths of $$x : x : x\sqrt{2}$$

The hypotenuse corresponds with $$x\sqrt{2} = 1$$: OB = 1
Find length of equal sides $$x$$ (one of which, AB, is the distance needed):

$$x\sqrt{2}= 1$$

$$x = \frac{1}{\sqrt{2}}$$
= AB = shortest distance between the parallel lines

That does not look like any of the answers.
Rationalize the denominator (clear the radical): Multiply by $$\frac{\sqrt2}{\sqrt2}$$
$$\frac{1}{\sqrt{2}}$$ * $$\frac{\sqrt2}{\sqrt2} = \frac{\sqrt{2}}{2}$$

Great explanation, Kudos to you..
Did not quite get how you got that the triangle will be isosceles triangle. Could you or Bunuel help understand ?

thanks.
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Re: In the rectangular coordinate system above, if the equation of m is y  [#permalink]

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14 Apr 2018, 07:59
1
MT1988 wrote:
generis wrote:

In the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?

(A) √2
(B) 1
(C) √2/2
(D) 1/2
(E) 1/4

Attachment:
2017-12-15_1259.png

Attachment:
linesmandn.png

Without the formula see diagram

The shortest distance between parallel lines is a line perpendicular to both parallel lines
Draw a perpendicular line from point B to point A

That creates right isosceles ∆ ABO where OA = AB
-- The line y = x makes a 45° angle with both the x- and y-axes (as do all lines with slope 1 or -1)
-- ∠ BOA = 45° and ∠ at vertex A = 90° therefore ∠ ABO must = 45°
-- Sides opposite equal angles are equal: OA = AB

A right isosceles triangle
-- has angle measures 45-45-90 and
-- corresponding side lengths of $$x : x : x\sqrt{2}$$

The hypotenuse corresponds with $$x\sqrt{2} = 1$$: OB = 1
Find length of equal sides $$x$$ (one of which, AB, is the distance needed):

$$x\sqrt{2}= 1$$

$$x = \frac{1}{\sqrt{2}}$$
= AB = shortest distance between the parallel lines

That does not look like any of the answers.
Rationalize the denominator (clear the radical): Multiply by $$\frac{\sqrt2}{\sqrt2}$$
$$\frac{1}{\sqrt{2}}$$ * $$\frac{\sqrt2}{\sqrt2} = \frac{\sqrt{2}}{2}$$

Great explanation, Kudos to you..
Did not quite get how you got that the triangle will be isosceles triangle. Could you or Bunuel help understand ?

thanks.

Got it!!!! The Slope is 1 so the angle at the origin will be 45* , as for calculating distance we have to drop a perpendicular one angle will be 90 leaving the third angle as 45*.
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Re: In the rectangular coordinate system above, if the equation of m is y  [#permalink]

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14 Apr 2018, 16:33
1
Quote:
Bunuel wrote:
In the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?

(A) √2
(B) 1
(C) √2/2
(D) 1/2
(E) 1/4
Attachment:
2017-12-15_1259.png

MT1988 wrote:
MT1988 wrote:
generis wrote:
Without the formula see diagram

The shortest distance between parallel lines is a line perpendicular to both parallel lines
Draw a perpendicular line from point B to point A

That creates right isosceles ∆ ABO where OA = AB
-- The line y = x makes a 45° angle with both the x- and y-axes (as do all lines with slope 1 or -1)
-- ∠ BOA = 45° and ∠ at vertex A = 90° therefore ∠ ABO must = 45°
-- Sides opposite equal angles are equal: OA = AB

A right isosceles triangle
-- has angle measures 45-45-90 and
-- corresponding side lengths of $$x : x : x\sqrt{2}$$

The hypotenuse corresponds with $$x\sqrt{2} = 1$$: OB = 1
Find length of equal sides $$x$$ (one of which, AB, is the distance needed):

$$x\sqrt{2}= 1$$

$$x = \frac{1}{\sqrt{2}}$$
= AB = shortest distance between the parallel lines

That does not look like any of the answers.
Rationalize the denominator (clear the radical): Multiply by $$\frac{\sqrt2}{\sqrt2}$$
$$\frac{1}{\sqrt{2}}$$ * $$\frac{\sqrt2}{\sqrt2} = \frac{\sqrt{2}}{2}$$

Great explanation, Kudos to you..
Did not quite get how you got that the triangle will be isosceles triangle. Could you or Bunuel help understand ?

thanks.

Got it!!!! The Slope is 1 so the angle at the origin will be 45* , as for calculating distance we have to drop a perpendicular one angle will be 90 leaving the third angle as 45*.

BOOM. Excellent.
I came prepared with a diagram
to help, but behold, no need.

Here is a post that discusses
that create 45° angles.
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In the rectangular coordinate system above, if the equation of m is y  [#permalink]

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15 Apr 2018, 03:40
Bunuel wrote:

In the rectangular coordinate system above, if the equation of m is y = x, and m is parallel to n, what is the shortest distance between m and n?

(A) √2
(B) 1
(C) √2/2
(D) 1/2
(E) 1/4

Attachment:
The attachment 2017-12-15_1259.png is no longer available

My method to solve this:-
Attachments

File comment: for line y = x, the angle between x-axis and the line m will be 45 degrees.
The min distance will the perpendicular distance between the two lines.
In the image, by mistake I've written 1/root(2) = 2/root(2), it should be root(2)/2. Please excuse that mistake

IMG_20180415_160317.jpg [ 1.99 MiB | Viewed 347 times ]

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In the rectangular coordinate system above, if the equation of m is y &nbs [#permalink] 15 Apr 2018, 03:40
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# A company has two types of machines, type R and type S

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Re: A company has two types of machines, type R and type S  [#permalink]

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04 Nov 2016, 15:05
1
enigma123 wrote:
A company has two types of machines, type R and type S. Operating at a constant rate, a machine of type R does a certain job in 36 hrs and a machine of type S does the same job in 18 hours. If the company used the same number of each type of machine to do the job in 2 hours, how many machines of type R were used?

A. 3
B. 4
C. 6
D. 9
E. 12

We have a combined worker problem for which we can use the following formula:

work (1 machine) + work (2 machine) = total work completed

Since we are completing one job, we can say:

work (1 machine) + work (2 machine) = 1

We are given that a machine of type R does a certain job in 36 hours and a machine of type S does the same job in 18 hours.

Thus, the rates for the two machines are as follows:

rate of machine R = 1/36

rate of machine S = 1/18

We are also given that the company used the same number of each type of machine to do the job in 2 hours. If we let x = the number of each machine used, we can multiply each rate by x and we have:

rate of x number of R machines = x/36

rate of x number of S machines = x/18

Finally, since we know some number of R and S machines worked for two hours, and since work = rate x time, we can calculate the work done by each type of machine.

work done by x number of R machines = 2x/36 = x/18

work done by x number of S machines = 2x/18 = x/9

Now we can determine x using the combined worker formula:

work (machine R) + work (machine S) = 1

x/18 + x/9 = 1

x/18 + 2x/18 = 1

3x/18 = 1

x/6 = 1

x = 6

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Work/Rate problem- A company has..  [#permalink]

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01 Jan 2017, 06:10
A company has two types of machines, Type R and Type S. Operating at a certain rate, a machine of type R does a certain job in 36 hours and a machine of type S does the same job in 18 hours. If the company uses the same number of each type of machine to do the job in 2 hours, how many machines of Type R were used?

Options:
A. 3
B. 4
C. 6
D. 9
E. 12

Can anybody give me a quickest and simpler way to solve this? I took almost 3.5 min on this question to solve during a mock test. Still ended up guessing as I wasn't sure of the approach.
Thanks.
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Re: Work/Rate problem- A company has..  [#permalink]

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01 Jan 2017, 06:47
5
1
baalok88 wrote:
A company has two types of machines, Type R and Type S. Operating at a certain rate, a machine of type R does a certain job in 36 hours and a machine of type S does the same job in 18 hours. If the company uses the same number of each type of machine to do the job in 2 hours, how many machines of Type R were used?

Options:
A. 3
B. 4
C. 6
D. 9
E. 12

Can anybody give me a quickest and simpler way to solve this? I took almost 3.5 min on this question to solve during a mock test. Still ended up guessing as I wasn't sure of the approach.
Thanks.

We are given rates of each machine and that the number of machines of each type being used is the same = x.

$$\frac{x}{36} + \frac{x}{18} = \frac{1}{2}$$

$$\frac{3x}{36} = \frac{1}{2}$$

$$6x = 36$$

$$x = 6$$
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Re: A company has two types of machines, type R and type S  [#permalink]

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07 Jan 2017, 21:07
2
Rate(R) = $$\frac{1}{36}$$
Rate(S) = $$\frac{1}{18}$$
Combined rate of both machines ,
Rate(RS)=Rate(R) + Rate(S) = $$\frac{1}{12}$$

we have given time , T = 2hrs , Workdone = 1 (for single job)
Plug in the values :

[Rate] * [Time] * [No. of machines] = Workdone

$$\frac{1}{12}$$ * 2 * [No. of machines] = 1
[No. of machines] = 6

Ans : C
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Re: A company has two types of machines, type R and type S  [#permalink]

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18 Apr 2017, 05:05
Bunuel wrote:
enigma123 wrote:
A company has two types of machines, type R and type S. Operating at a constant rate, a machine of type R does a certain job in 36 hrs and a machine of type S does the same job in 18 hours. If the company used the same number of each type of machine to do the job in 2 hours, how many machines of type R were used?

A. 3
B. 4
C. 6
D. 9
E. 12

Rate of A - $$\frac{1}{36}$$ job/hour, rate of x machines of A - $$\frac{1}{36}x$$ job/hour;
Rate of B - $$\frac{1}{18}$$ job/hour, rate of x machines of B - $$\frac{1}{18}x$$ job/hour, (same number of each type);

Remember that we can sum the rates, hence combined rate of A and B is $$\frac{1}{36}x+\frac{1}{18}x=\frac{3}{36}x=\frac{1}{12}x$$ job/hour.

We are told that together equal number (x in our case) of machines A and B can do the job (1 job) in 2 hours --> $$Time*Rate=2*\frac{1}{12}x=1=Job$$ --> $$x=6$$.

When I am double checking, why am I not getting the right answer?
If 1 R machine is taking 36 hours, then 6 R machines will take 6 hours.
If 1 S machine is taking 18 hours, then 6 S machines will take 3 hours.
Together, they will take 6-3=3 hours.

Please tell me where am I wrong :/
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Re: A company has two types of machines, type R and type S  [#permalink]

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16 Aug 2017, 18:41
narendran1990 wrote:
Took 36 as the unit of work to be completed by R & S. [LCM of 36 & 18].

So, R does 1 unit per hour & S does 2 units per hour. Since an equal number of R & S machines have to be deployed, consider 'x' as the number of equipment. Additionally, the work is to be completed in 2 hours, so 18 units have to be manufactured.

Therefore, x+2x = 18, 3x = 18, x=6. [Correct Answer]

I was trying to solve using similar kind of method. But i could not understand how you arrived on 18? It will be great if you could elaborate.
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A company has two types of machines, type R and type S  [#permalink]

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09 Nov 2017, 21:26
There is no need to compute the combined rate in this case, explicitly that is.

Since the number of machines is the same, we can use a simple equation:

$$x*\frac{1}{36}+ x* \frac{1}{18} = \frac{1}{2}$$ (here x machines work at the respective rates for 2 hours)

Solving for x, we get 6.
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Re: A company has two types of machines, type R and type S  [#permalink]

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13 Nov 2018, 19:46
hannahkagalwala wrote:
Bunuel wrote:
enigma123 wrote:
A company has two types of machines, type R and type S. Operating at a constant rate, a machine of type R does a certain job in 36 hrs and a machine of type S does the same job in 18 hours. If the company used the same number of each type of machine to do the job in 2 hours, how many machines of type R were used?

A. 3
B. 4
C. 6
D. 9
E. 12

Rate of A - $$\frac{1}{36}$$ job/hour, rate of x machines of A - $$\frac{1}{36}x$$ job/hour;
Rate of B - $$\frac{1}{18}$$ job/hour, rate of x machines of B - $$\frac{1}{18}x$$ job/hour, (same number of each type);

Remember that we can sum the rates, hence combined rate of A and B is $$\frac{1}{36}x+\frac{1}{18}x=\frac{3}{36}x=\frac{1}{12}x$$ job/hour.

We are told that together equal number (x in our case) of machines A and B can do the job (1 job) in 2 hours --> $$Time*Rate=2*\frac{1}{12}x=1=Job$$ --> $$x=6$$.

When I am double checking, why am I not getting the right answer?
If 1 R machine is taking 36 hours, then 6 R machines will take 6 hours.
If 1 S machine is taking 18 hours, then 6 S machines will take 3 hours.
Together, they will take 6-3=3 hours.

Please tell me where am I wrong :/

It is not correct to subtracting time taken by one machine from that of other to calculate final time,

Time taken are for complete job, and if other machines are also performing the same job, then there will be less amount of work of each machine type and hence leaser time will be taken.

You are right till, 6 R machines will take 6 hours.
And 6 S machines will take 3 hours.

Now total time taken will be 1/(1/6 +1/3) = 1/(1/2) = 2 hours.

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Re: A company has two types of machines, type R and type S &nbs [#permalink] 13 Nov 2018, 19:46

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# Let $p$ and $q$ be two distinct primes. Pick the correct statements from

Let $p$ and $q$ be two distinct primes. Pick the correct statements from the following:

a. $Q(\sqrt p)$ is isomorphic to $Q(\sqrt q)$ as fields.

b. $Q(\sqrt p)$ is isomorphic to $Q(\sqrt{−q})$ as vector spaces over Q.

c. $[Q(\sqrt p,\sqrt q) : Q] = 4$.

d. $Q(\sqrt p,\sqrt q) = Q(\sqrt p + \sqrt q)$.

(c) & (d) are correct. (a) is not correct for fields but correct for vector spaces. (b) not sure. i think it is correct by the same arguement of (a).am i right?

-
Is this a homework question, or just one for personal development, as this may effect the type of response that people give? – David Ward Sep 17 '12 at 8:16
@DavidWard I don't know why it should affect the type of response. If one should not give a full answer to a homework question, many good questions might have no full answer in this site. I think this is against the policy of this site. – Makoto Kato Sep 17 '12 at 18:13
@MakotoKato My reasoning is that if it is a homework question, then the answer itself is of importance to the person asking the question. However, if the question is not homework, then for a question as above, I would feel happy to give the answer, but then lead someone through the reasoning on how to get there. – David Ward Sep 18 '12 at 8:26
@DavidWard I'm sorry that I don't understand your reasoning. Could you explain why you don't want to give the answer to a homework question? – Makoto Kato Sep 18 '12 at 8:43
@MakotoKato My reasoning is as follows: Firstly suppose that the question asked is from an assessed piece of work. Then as the question is a multiple-choice question, the actual answer will attract some if not all of the marks (depending on the desire of the assessor). Thus knowledge of the answer is of critical importance. On the other hand, if the question asked is purely for personal development, whilst leading the member towards the solution, it can be beneficial to actually know where you are being led. It just gives some motivation for where the reasoning is heading. – David Ward Sep 18 '12 at 15:26

a. Suppose $\mathbb{Q}(\sqrt p)$ is isomorphic to $\mathbb{Q}(\sqrt q)$ as fields. Then $\mathbb{Q}(\sqrt q)$ has an element $\alpha$ such that $\alpha^2 = p$. Let $\alpha = a + b\sqrt q$, where $a, b \in \mathbb{Q}$. We denote the conjugate of $\alpha$ by $\alpha'$. Since $\alpha + \alpha' = 0$, $a = 0$. Since $\alpha^2 = p$, $p = b^2 q$. Hence $p = q$. This is a contradiction. Hence $a$. is not true.

b. Since the both fields have dimension 2 as vector spaces over $\mathbb{Q}$, $b.$ is true.

c. As we see in the above, $\sqrt p$ is not contained in $\mathbb{Q}(\sqrt q)$. Hence $[\mathbb{Q}(\sqrt p,\sqrt q) : \mathbb{Q}(\sqrt q)] = 2$. Hence $[\mathbb{Q}(\sqrt p,\sqrt q) : \mathbb{Q}] = 4$. Thus $c.$ is true.

d. Let $K = \mathbb{Q}(\sqrt p,\sqrt q)$. Clearly $K/\mathbb{Q}$ is Galois. Let $\sigma \in Gal(K/\mathbb{Q})$. Then $\sigma(\sqrt p) = \sqrt p$ or $-\sqrt p$, $\sigma(\sqrt q) = \sqrt q$ or $-\sqrt q$. Since $|Gal(K/\mathbb{Q})| = 4$ by $c.$, $Gal(K/\mathbb{Q}) = \{1, \sigma_1, \sigma_2. \sigma_3\}$, where

$$\sigma_1(\sqrt p) = \sqrt p,\ \ \ \sigma_1(\sqrt q) = -\sqrt q$$

$$\sigma_2(\sqrt p) = -\sqrt p,\ \sigma_2(\sqrt q) = \sqrt q$$

$$\sigma_3(\sqrt p) = -\sqrt p,\ \ \ \sigma_3(\sqrt q) = -\sqrt q$$

Let $\alpha = \sqrt p + \sqrt q$.

Clearly $\alpha, \sigma_1(\alpha) = \sqrt p - \sqrt q, \sigma_2(\alpha) = -\sqrt p + \sqrt q, \sigma_3(\alpha) = -\sqrt p - \sqrt q$ are distinct. Hence $K = \mathbb{Q}(\alpha)$. Thus $d.$ is true.

-
please explain why α+α′=0? In the proof of a ? – GA316 Aug 9 '15 at 4:01


# How do I describe formally complexity of 2-sum problem algorithm?

I have algorithm that finds if there are two elements in sorted array that have sum zero.

1.ZeroSumPair(A[1..n]) // A[1..n] <-- sorted
2.    l <- 1, r <- n
3.    while(l < r)
4.        while(l < r or A[l] + A[r] > 0)
5.            r--
6.        if(A[l] + A[r] = 0)
7.            return true
8.        l++
9.    return false

Intuitively I know that this algorithm is $$O(n)$$, but how do I deduct it using proof with summations like in CLRS book?

I've also saw How to develop an $O(N)$ algorithm solve the 2-sum problem?, but I didn't see any formal proof.

• I would try putting in a counter t. Then increment t by 1 every time you do an operation. Then try to prove a loop invariant for the outer-while loop that claims $t = O(n)$. With the correct loop invariant on $t$, this can be proven by induction relatively easily. Perhaps something like $t \leq c(l + (n - r))$ for some constant $c$ might work. Then at termination (in the worst case) you know $l = r$, thus $t \leq cn \implies t = O(n)$. – ryan Mar 27 at 18:28

Consider this slightly modified algorithm where I've added an "operation counter" t. This will be incremented every time we do a comparison or assignment.

1. ZeroSumPair(A[1..n]) // A[1..n] <-- sorted
2.   l <- 1, r <- n
3.   t <- 3             // 2 for first two assignments and 1 for initial while check
4.   while(l < r)
5.     t++              // 1 for initial while check
6.     while(l < r or A[l] + A[r] > 0)
7.       t++            // 1 for decrementing r
8.       r--
9.       t++            // 1 for following while check
10.    t++              // 1 for if comparison
11.    if(A[l] + A[r] = 0)
12.      return true
13.    t++              // 1 for incrementing l
14.    l++
15.    t++              // 1 for following while check
16.  return false

This is a bit verbose, but it will work. Now we must simply prove that, at termination we have $$t = O(n)$$. We can do this inductively with a loop invariant.

Let's use the following loop invariant for the loop on line 4.

$$t = 3 + 4(l-1) + 2(n-r)$$

### Base Case

Initially $$t = 3$$, $$l = 1$$, and $$r = n$$. Thus we have:

$$3 = 3 + 4(1 - 1) + 2(n - n) = 3$$

### Inductive Case

Let $$t'$$, $$l'$$, and $$r'$$ be the values of $$t$$, $$l$$, and $$r$$ at the end of our previous iteration. At the end of our current iteration we have $$l = l' + 1$$, $$r = r' - k$$ for some $$k$$, and $$t = t' + 4 + 2k$$. Thus we have:

\begin{align*} t & = t' + 4 + 2k\\ & = 3 + 4(l' - 1) + 2(n - r') + 4 + 2k\\ & = 3 + 4(l - 2) + 2(n - (r + k)) + 4 + 2k\\ & = 3 + 4(l - 1) - 4 + 2(n - r) - 2k + 4 + 2k\\ & = 3 + 4(l - 1) + 2(n - r) & \square \end{align*}

Thus, we can conclude the loop invariant holds. At the end of the loop (in the worst case) we have $$l = r \leq n$$. We then have: \begin{align*} t & = 3 + 4(l - 1) + 2(n - r)\\ & = 3 + 4l - 4 + 2n - 2l\\ & = 2(n + l) - 1\\ & \leq 2(n + n) - 1\\ & = 4n - 1\\ & = O(n) & \square \end{align*}

Thus it is linear. There might be an easier way to do this, but this way makes sense to me pretty clearly.

• I see that this looks formal but I feel like it is “hacking”, amending the algorithm in order to prove its complexity, is this real thing in computer science? Not talking about practice here – kuskmen Mar 27 at 21:53
• @kuskmen you don't need $t$ explicitly in your code. I put it there for clarity. You can have an implicit "operation count" which is realistically what you're trying to determine. Then prove (through loop invariants) that the "operation count" is bounded by $n$. That's all I did here with $t$ being the operation counter. – ryan Mar 27 at 22:49
• I see, I have to admit I am bad in mathematical proofs or any sort of proofs as you can tell. Can you tell why $t=t'+ 4 + 2k$ in the inductive step? – kuskmen Mar 28 at 8:15
• Assume $t$ starts at $t'$ just before the loop executes. Now hop into the loop. Increment by 1 for the initial inner while loop check (line 5). Now assume the inner while loop executes $k$ times. There are 2 increments in the inner while loop (line 7 and line 9), so $t$ is then incremented $2k$ times. Next, exit the inner while loop, increment by 1 for the if comparison (line 10). Assume the if condition is false because we do worst case analysis, then increment by 1 for incrementing l (line 13). Increment 1 last time for the following while check (line 15). Add them all up and you get that. – ryan Mar 28 at 16:11
• @kuskmen, this may also help you: cs.stackexchange.com/a/23595/68251 – ryan Mar 28 at 17:45


# Show that the equilibrium point $(0,0)$ is asymptotically stable and an estimate of its basin of attraction

Consider the system \begin{aligned} \dot{x} &=-y-x^3+x^3y^2\\ \dot{y}&=x-y^3+x^2y^3\end{aligned} Show that the equilibrium point $$(0,0)$$ is asymptotically stable and an estimate of its attractiveness basin.

Clearly the point $$(0,0)$$ is a point of equilibrium and also $$D_f(x,y)=\begin{pmatrix}-3x^2+3y^4 & -1+2yx^3\\ 1+2xy^3 &-3y^2+3y^2x^2 \end{pmatrix}$$, so $$D_f(0,0)=\begin{pmatrix}0 & -1\\1 &0 \end{pmatrix}$$ and has eigenvalues $$\pm i$$ So, I can not conclude anything of stability for this point, I need a Liapunov function and I do not know what it is or how to find it, could someone help me please? Thank you very much.

• Always try the simplest one first. For example, quadratic. More precisely, $V(x, y) = x^2+y^2$. Oct 30 '17 at 5:44

In addition to what has been said by @Evgeny and @MrYouMath: the set $$M=\left\{ (x,y)\in\mathbb R^2 :\; x^2+y^2<2 \right\}$$ is a positively invariant set of the considered system since $\forall (x,y)\in M$ $$\dot V=-x^4-y^4+x^2y^2(x^2+y^2)<-x^4-y^4+2x^2y^2=-(x^2-y^2)^2\leq 0;$$ it is also a subset (guaranteed estimation) of the domain of attraction.

• Nice one! This estimate is even better than $\| (x, y) \|_{\infty} < 1$ . Oct 30 '17 at 18:06
• +1: Very nice observation. This might be useful in the future. Oct 31 '17 at 14:03
• How Do I show that $M$ is positively invariant with this fact?
– P.G
Nov 20 '20 at 13:44
• @P.G Just as it is done in the proof of the Lyapunov stability theorem
– AVK
Nov 21 '20 at 7:02

As @Evgeny suggested you could use the Lyapunov function candidate

$$V(x,y)=\dfrac{1}{2}\left[x^2+y^2\right].$$

$V(x,y)$ is clearly positive definite at the origin and radially unbounded (which we would need for assessing global asymptotic stability).

The time derivative of $V(x,y)$ is given by $$\dot{V}=x\dot{x}+y\dot{y}=x(-y-x^3+x^3y^2)+y(x-y^3+x^2y^3)$$ $$\dot{V}=-x^4-y^4+x^2y^2(x^2+y^2)=-(1-y^2)x^4-(1-x^2)y^4.$$

As the lower order terms $-x^4-y^4$ are negative definite we can conclude that the equilibrium point is asymptotically stable in a region around the origin. Using the comment given by @Evgeny we can see that this expression is negative semidefinite if $(x,y)$ lie inside the unit circle $D=\{(x,y)\in \mathbb{R}^2|x^2+y^2<1\}$. This is the basin of attraction (correction due to @Artem).

We cannot say if the origin is globally asymptotically stable because $\dot{V}$ is not negative definite for all regions around the origin. This does not mean that the origin cannot be globally asymptotically stable. It just means we can only show (local) asymptotic stability of the origin with $V(x)$ as our Lyapunov function candidate.

• Actually, you can do better if regroup terms correctly: $\dot{V} = -x^4(1-y^2)-y^4(1-x^2)$. This also gives a rough estimate of basin of attraction since this expression is negative when $\vert x \vert < 1$ and $\vert y \vert < 1$. Oct 30 '17 at 13:18
• @Evgeny: You are right. I felt that I should be able to factor this expression. Oct 30 '17 at 13:19
• Also, speaking about radial unboundedness: it's not a necessary condition really, see recent discussion in comments. Oct 30 '17 at 18:09
• It is not necessary for asymptotic stability but it is necessary for global asymptotic stability (see the comment in the brackets in my answer). I think that is what the example showed. Or do you mean something else? Oct 30 '17 at 18:14
• Even for the global asymptotic stability it is not necessary. You can prove the global asymptotic stability with a Lyapunov function which is not radially unbounded. Moreover, you can transform the image of any Lyapunov function such that it will become bounded but still will prove that something is globally asymptotic stable. Oct 30 '17 at 20:29

This problem can be handled with an optimization procedure, having in mind that generally is a non convex problem. The result depends on the test Lyapunov function used so we will generalize to a quadratic Lyapunov function

$$V(p) = p^{\dagger}\cdot M\cdot p = a x^2+b x y + c y^2,\ \ \ p = (x,y)^{\dagger}$$

and

$$f(p) = \{-y - x^3 + x^3 y^2, x - y^3 + x^2 y^3\}$$ with $$a>0,c>0, a b-b^2 > 0$$ to assure positivity on $$M$$. We will assure a set involving the origin $$Q_{\dot V}$$ such that $$\dot V(Q_{\dot V}) < 0$$. The optimization process will be used to guarantee a maximal $$Q_{\dot V}$$.

After determination of $$\dot V = 2 p^{\dagger}\cdot M\cdot f(p)$$ we follow with a change of variables

$$\cases{ x = r\cos\theta\\ y = r\sin\theta }$$

so $$\dot V = \dot V(a,b,c,r,\theta)$$. The next step is to make a sweep on $$\theta$$ calculating

$$S(a,b,c, r)=\{\dot V(a,b,c,r,k\Delta\theta\},\ \ k = 0,\cdots, \frac{2\pi}{\Delta\theta}$$

and then the optimization formulation follows as

$$\max_{a,b,c,r}r\ \ \ \ \text{s. t.}\ \ \ \ a > 0, c> 0, a c -b^2 > 0, \max S(a,b,c,r) \le -\gamma$$

with $$\gamma > 0$$ a margin control number.

Follows a MATHEMATICA script which implements this procedure in the present case.

f = {-y - x^3 + x^3 y^2, x - y^3 + x^2 y^3};
V = a x^2 + 2 b x y + c y^2;
dV = Grad[V, {x, y}].f /. {x -> r Cos[t], y -> r Sin[t]};
rest = Max[Table[dV, {t, -Pi, Pi, Pi/30}]] < -0.1;
rests = Join[{rest}, {r > 0, a > 0, c > 0, a c - b^2 > 0}];
sols = NMinimize[Join[{-r}, rests], {a, b, c, r}, Method -> "DifferentialEvolution"]
rest /. sols[[2]]

dV0 = Grad[V, p].f /. sols[[2]]
V0 = V /. sols[[2]]
r0 = 2;
rmax = r /. sols[[2]];
gr0 = StreamPlot[f, {x, -r0, r0}, {y, -r0, r0}];
gr1a = ContourPlot[dV0, {x, -r0, r0}, {y, -r0, r0}, ContourShading -> None, Contours -> 80];
gr1b = ContourPlot[dV0 == 0, {x, -r0, r0}, {y, -r0, r0}, ContourStyle -> Blue];
gr2 = ContourPlot[x^2 + y^2 == rmax^2, {x, -r0, r0}, {y, -r0, r0}, ContourStyle -> {Red, Dashed}];
Show[gr0, gr1a, gr1b, gr2]

Follows a plot showing in black the level sets $$Q_{\dot V}$$ an in blue the trace of $$\dot V = 0$$. In dashed red is shown the largest circular set $$\delta = 1.42486$$ defining the maximum attraction basin for the given test Lyapunov function's family.


# If $aH=bH \implies Ha=Hb$ for a subgroup H having *finite index*, then $gH=Hg$ for all $g \in G$?

Problem 2.5.9 of Herstein's Topics in Algebra asks us to prove that if $$H$$ is a subgroup of a group $$G$$ such that $$Ha \not = Hb \implies aH \not = bH$$, then $$gHg^{-1} \subset H$$ for all $$g \in G$$, which is equivalent to $$gH \subset Hg.$$ Suppose I have proved this (this result is of course true). I want to prove more, namely that, under the same premise, $$gH = Hg$$ for all $$g \in G$$. I think I can prove it if I further assume that the index of $$H$$ in $$G$$ is finite. Is this true? And if so, is this additional hypothesis necessary as well as sufficient? I.e. can you give an example of a subgroup $$H$$ of infinite index such that $$aH \subset Ha$$ but not vice versa? Thanks in advance. [DISCLAIMER: I am not yet familiar with normal subgroups]

Here's my attempt at a proof:

Suppose that $$|G:H|=n \in \mathbb{N}$$ and that $$aH \subset Ha$$ for all $$a \in G$$. Now, suppose that there is an element $$x$$ such that $$x \in Ha$$ but $$x \not \in aH$$. Then $$x$$ must be contained in a different left coset, say $$bH$$, because the left cosets form a partition of $$G$$. Then, by our hypothesis, $$x \in Hb$$, which implies $$Hb = Ha$$, because the right cosets form a partition of $$G$$ as well. So far we have shown that $$aH, bH \subset Ha=Hb$$. But now we can prove that there are more left cosets than right ones! In fact, the remaining $$n-1$$ right cosets distinct from $$Ha$$ must contain at least one left coset, by our hypothesis, but $$Ha$$ contains two left cosets. So there are at least $$n+1$$ left cosets, a contradiction. Therefore $$aH=Ha. \square$$

• Compare with this duplicate and its answers. – Dietrich Burde Mar 14 at 12:00
• @DietrichBurde The answers in your linked question only prove $gHg^{-1} \subset H$ and not equality. So I don't think they address the OP’s problem. – Claudius Mar 14 at 12:15
• @Claudius This is not true. They prove equality, e.g., see Deven's answer. – Dietrich Burde Mar 14 at 12:18
• @DietrichBurde Yes, Deven claims to have shown $gHg^{-1} =H$. But what he really did was to prove $gHg^{-1} \subset H$. At least, I don't see how he proved the other inclusion. – Claudius Mar 14 at 12:20

Your proof seems correct to me.

In fact, you can remove the finite index hypothesis. If $$gHg^{-1} \subset H$$ for all $$g\in G$$, then for each $$g\in G$$ you also have $$H = g^{-1}(gHg^{-1})g \subset g^{-1}Hg = g^{-1}H(g^{-1})^{-1} \subset H,$$ so we must have equality throughout, i. e. $$H = g^{-1}Hg$$ (for each $$g\in G$$).

More generally, for any subgroup $$H$$ of $$G$$ the following holds: the set $$N:= \{ g\in G \mid gHg^{-1}\subset H\}$$ is a subgroup of $$G$$ if and only if $$gHg^{-1} = H$$ for all $$g\in N$$. (In that case $$N$$ is the normalizer of $$H$$ in $$G$$.)

• That's brilliant, thank you. Now I feel stupid for not having come up with that immediately! But why would Herstein keep this result from the reader and only ask to prove one inclusion? And what about this thread math.stackexchange.com/questions/217601/… in which there seems to be a counterexample? – The Footprint Mar 14 at 12:36
• In the answer of Brian M. Scott the set $\{g\in G\mid gHg^{-1} \subset H\}$ is not stable under inversion, hence not a group. It is only closed under composition. – Claudius Mar 14 at 12:40

You don't need any additional hypotheses. Multiplying both sides by $$g^{-1}$$ on the left and $$g$$ on the right yields

$$gHg^{-1} \subset H \implies H \subset g^{-1}Hg$$

Since this is true for all $$g \in G$$, we can substitute $$g$$ for $$g^{-1}$$, concluding

$$gHg^{-1} \subset H \subset gHg^{-1}$$

$$H = gHg^{-1}$$

This is equivalent to

$$gH = Hg$$


×

# A Calculus Problem!

The Problem:

Find $$I=\displaystyle\int_{0}^{2} (x^2+1)\,d\left \lfloor x\right \rfloor$$

My Doubt:

There seem two probable approaches to the question but both of them yield different results. Both seem to be mathematically correct and hence the confusion.

Approach 1:

We know that,

$$g(x). f'(x) + f(x). g'(x) = \large \frac{d\left(f(x). g(x)\right)}{dx}$$

Integrating the above expression we get:

$$\displaystyle\int_{a}^{b} g(x)\,df(x) + \displaystyle\int_{a}^{b} f(x)\,dg(x) = f(b).g(b)-f(a).g(a)$$

Using the above property to solve the integration we get the answer as $$\boxed{7}$$

Approach 2:

We can write

$$I=\displaystyle\int_{0}^{1} (x^2+1)\,d\left \lfloor x\right \rfloor +\displaystyle\int_{1}^{2}(x^2+1)\,d\left \lfloor x\right \rfloor$$

But $$\left \lfloor x\right \rfloor$$ assumes constant values of $$\left \lfloor x\right \rfloor=0$$ and $$\left \lfloor x\right \rfloor=1$$ in the respective intervals. And hence in both cases $$d\left \lfloor x\right \rfloor=0$$ and therefore $$I=0$$

I genuinely can't understand what's the correct method. I will really be grateful if someone can explain it to me.

Thank You!

Note by Miraj Shah
1 year, 9 months ago

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In Approach 2, the floor function fails to be constant on the two subintervals; consider the two endpoints! The first integral takes the value of 1 due to the step at 1, and the second one takes a value of 5. With this correction, both approaches work and yield the same result.

- 1 year, 9 months ago

First off, integration by parts doesn't work as the step function isn't differentiable.

Secondly, the integral is 2. I'll provide a proof of this later.

Btw, @Aditya Kumar The integral makes perfect sense and is defined.

- 1 year, 9 months ago

It would be really nice of you to put the proof here if possible! But I would also like to know then what's wrong with the Approach 1&2 illustrated above? It's a bit baffling

- 1 year, 9 months ago

Wait, the integral is 7.

We shall use the definition of integration as given in this book, with $$\alpha(x)=\lfloor x \rfloor$$ and $$f(x)=1+x^2$$.

Consider a partition $$P$$ of $$[0,2]$$ such that $$x_{r-1}\leq 1\leq x_{r}$$ where strict inequality holds in atleast one of the two inequalities. Also, by the definition of a partition, we have $$x_{n-1}<2=x_n$$. Then,

$U(P, f, \alpha)=f(x_r)+f(x_n) \\ L(P, f, \alpha)=f(x_{r-1})+f(x_{n-1}) \\\implies \inf U(P, f, \alpha)=\sup L(P, f, \alpha)=f(1)+f(2) \\\implies \int_0^2 f \, d\alpha = 7$

Note: I've only used the fact that $$f$$ is monotonic.

Plus, in the book mentioned above, in the exercises, a more generalised version of IBP, valid for even discontinuos functions is given. You've made use of this version in approach 1 (luckily, the step function is monotonic).

If the integrals are calculated using the definition in approach 2, you get the right answer, 7.

- 1 year, 9 months ago

- 1 year, 9 months ago

Oh yes! Really helpful @Deeparaj Bhat ! Sorry for replying a bit late. Busy studying for the big one which is to be held on 22nd May! Actually the above problem was from a test paper it self! Had a doubt in the question, so thought of taking help from the brilliant Brilliant Community as I was pretty sure that I will be able to get at least some insight!

Thank you!

- 1 year, 9 months ago

You're writing any other exams? (Except JEE ADVANCED, which is on 22nd)

- 1 year, 9 months ago

Actually yes. BITS and all! You have to give JEE this year?

- 1 year, 9 months ago

Yup. That, BITSAT and CMI in my case.

- 1 year, 9 months ago


# Number TheoryEuler's theorem for calculating mod

#### ATroelstein

##### New member
I am trying to use Euler's theorem for calculating the following:

$$8^{7} mod 187$$

I have determined that:

$$\phi(187) = \phi(11*17) = 160$$

and

$$8^{7} = 8^{4} * 8^{2} * 8^{1}$$

but am unfortunately confused about how to now proceed beyond this point. Thank you.

#### caffeinemachine

##### Well-known member
MHB Math Scholar
I am trying to use Euler's theorem for calculating the following:

$$8^{7} mod 187$$

I have determined that:

$$\phi(187) = \phi(11*17) = 160$$

and

$$8^{7} = 8^{4} * 8^{2} * 8^{1}$$

but am unfortunately confused about how to now proceed beyond this point. Thank you.
I don't think here Euler's theorem can be fruitfully used. But you can use Fermat's little in conjunction with Chinese Remainder to calculate this nicely.

Note that $8^7=2^{21}=x$ (say). Also, $187=11\times 17$. Now by Fermat we have $2^{10}\equiv 1 \pmod{11}$ and $2^{16}\equiv 1\pmod{17}$. These give $x\equiv 2\pmod{11}$ and $x\equiv 32\equiv -2\pmod{17}$. From here it's routine to use the Chinese remainder and get the remainder $x$ leaves mod $187$.

#### Deveno

##### Well-known member
MHB Math Scholar
As I have stated here many times before, I am pathetically lazy, so I look for the easy way out.

87 = (64)(8)(64)(8)(8) (mod 187)

Next, I calculate 64*8 the old-fashioned way:

(6*8*10) + (4*8) = 480 + 32 = 512.

I'm going to guess that 2 multiples of 187 are all we're going to pack into 512. So:

512 = 512 - 374 = 138 (mod 187).

So 87 = (138)(138)(8) (mod 187).

Did I mention I'm afraid of big numbers? Well, I am. So next I'm going to apply a trick that often works well in modular arithmetic:

Since 138 = -49 (mod 187),

87 = (138)(138)(8) = (-49)(-49)(8) = (49)(49)(8) (mod 187).

Here, I have an unpleasant choice to make: do I tackle 49*49 next, or 49*8?

The small numbers win!

Now 49*8 = (4*8*10) + (9*8) = 320 + 72 = 392. Well, that's pretty close to 374, which makes me happy. So:

87 = (49)(49)(8) = (49)(392) = (49)(18) (mod 187).

I've avoided "hard calculations" as much as possible, but I have no choice but to evaluate 49*18 now:

49*18 = (40+9)(10+8) = 400 + 320 + 90 + 72 = 882.

So I want to know what 882 (mod 187) is. Knowing that 180*5 = 900, I don't think I can get 5 multiples of 187 in 882, so 4 is my best guess:

187*4 = 400 + 320 + 28 = 748. So:

87 = 882 = 882 - 0 = 882 - 748 = 134 (mod 187)

Does this jibe with caffeinemachine's answer?

134 = 132 + 2 = 0 + 2 = 2 = (mod 11). Check.

134 = 119 + 15 = 0 + 15 = 15 = -2 (mod 17). Check.

#### Deveno

##### Well-known member
MHB Math Scholar
I'd like to point out here that caffeinemachine's solution is not so easy to carry out as it might seem.

Given that:

x = 2 (mod 11)
x = -2 (mod 17)

we seek integers k,m such that:

11k + 2 = 17m - 2

or, equivalently:

17m - 11k = 4

If we knew some integers r,s with:

17r - 11s = 1, we could take m = 4r, k = 4s.

Fortunately, the euclidean division algorithm gives us a way to find these two integers r and s:

17 = 11*1 + 6
11 = 6*1 + 5
6 = 5*1 + 1

Therefore:

1 = 6 - 5
5 = 11 - 6
6 = 17 - 11, so:

1 = 6 - 5 = 6 - (11 - 6) = 2*6 - 11 = 2*(17 - 11) - 11 = 2*17 - 3*11, so we have:

r = 2, s = 3, and in turn:

m = 8, k = 12. Hence:

11*12 + 2 = 17*8 - 2, evaluating either side gives us:

134, which is the desired power 87 (mod 187).

#### johng

##### Well-known member
MHB Math Helper
Hi,
I tend to agree with caffeinemachine that the easiest computation is via the Chinese remainder theorem.
Given m and n relatively prime, the unique x (mod mn) that satisfies x = a (mod m) and x = b (mod n) is of the form a + km for some k with 0 <= k < n.
(All of the given values are different for the range of k's and so one must be b (mod n)).

So I want -2 + 17k = 2 (mod 11)
6k = 4 (mod 11)
k = 8 (mod 11) -- the multiplicative inverse of 6 is 2 mod 11.
So x = -2 +8*17 = 136 - 2 = 134.

#### Deveno

##### Well-known member
MHB Math Scholar
Hi,
I tend to agree with caffeinemachine that the easiest computation is via the Chinese remainder theorem.
Given m and n relatively prime, the unique x (mod mn) that satisfies x = a (mod m) and x = b (mod n) is of the form a + km for some k with 0 <= k < n.
(All of the given values are different for the range of k's and so one must be b (mod n)).

So I want -2 + 17k = 2 (mod 11)
6k = 4 (mod 11)
k = 8 (mod 11) -- the multiplicative inverse of 6 is 2 mod 11.
So x = -2 +8*17 = 136 - 2 = 134.
Yes, but....

The ease of this is "hidden" in the step:

6k = 4 --> k = 8.

In this particular case, one can guess rather easily that [6-1]11 = 2 by trial-and-error. In general, finding a (multiplicative) inverse mod n (if it exists, which it assuredly will if n is prime, and the element inverted is not a multiple of n = p) involves exactly the same steps I outlined above. One often does not have a discrete log table lying around to read off which power of a generator a particular element is, and one HAS to resort to the division algorithm to find the inverse (finding primitive elements is not always a trivial task, although it IS easier for small numbers).

Don't misunderstand me, I find the application of the CRT a beautiful and elegant solution. At its heart, the CRT states that:

If (m,n) = 1, then $\Bbb Z_{mn} \cong \Bbb Z_m \times \Bbb Z_n$

However, while the ring-homomorphism one way is trivial to display, the inverse homomorphism is nontrivial (and amounts to actually solving the congruence pair). There is a slight difference between knowing a solution exists and exhibiting that solution.

Your calculations and mine are, of course, the same, as can be seen by taking the equation:

11*12 + 2 = 17*8 - 2 (mod 11).

I also do not know if the original poster has seen a PROOF of the CRT, with an understanding of what it means for "compound" moduli. In all fairness (both to you and caffeinemachine) I admit sometimes an understanding of more advanced methods makes problems like these easier to deal with. Do I run the risk of under-estimating the original poster's sophistication? Of course. I am the first to admit the laziest proof is not always the shortest.

If the exponent has been larger, my way would undeniably be more cumbersome. I certainly mean no disrespect to either of your fine answers. Knowing we can reduce an exponent b in ab (mod n), by taking b (mod φ(n)), is something well worth remembering, if one has been exposed to Euler's theorem (which we can assume the OP has, by the thread title), and I note in passing that Fermat's "Little Theorem" (which caffeinemachine DOES invoke) is just a special case of Euler's Theorem.

One hopes that the original poster gains more from our discussion of solution techniques here than just the answer to his problem.

#### johng

##### Well-known member
MHB Math Helper
Yes, but....

The ease of this is "hidden" in the step:

6k = 4 --> k = 8.

In this particular case, one can guess rather easily that [6-1]11 = 2 by trial-and-error. In general, finding a (multiplicative) inverse mod n (if it exists, which it assuredly will if n is prime, and the element inverted is not a multiple of n = p) involves exactly the same steps I outlined above. One often does not have a discrete log table lying around to read off which power of a generator a particular element is, and one HAS to resort to the division algorithm to find the inverse (finding primitive elements is not always a trivial task, although it IS easier for small numbers).

Given the parameters of the CRT, namely m, n are relatively prime, I'm trying to solve a + mk = b for k. So of course m does have an inverse mod n. Granted, the finding of such inverse requires a little work, but it's not exorbitant. A slight extension of the Euclidean algorithm for gcd's produces x and y with mx + ny = 1, and so x is the inverse of m. I find the extension a bit cumbersome to do by hand, but it's a Programming I exercise to encode (maybe toward the end of the course). Even for large integers, the gcd algorithm is "usually" quite fast; as you probably know the worst case is for consecutive Fibonacci numbers. But even here, it's logarithmic.

#### Deveno

##### Well-known member
MHB Math Scholar
Yes, that is what I was getting at. If the exponent is large-ish (say > 1000 for example), using a gcd algorithm is going to be MUCH faster than direct computation.

But with an exponent of only 7, direct computation can be done straight-forwardly, without involving more abstract results. Is this preferable? It depends on your point of view, I suppose. One thing that CAN be said about your solution/caffeinemachine's solution is that it is more widely applicable to a greater range of problems


# Mathematical "inclusion" and "exclusion" of elements from a given set $A$?

If I have a set A, comprising of numbers from 1 to 10: $$A = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$$. Let's say I want to make another set by "including" all even numbers:

$$\{2, 4, 6, 8, 10\}$$

Or I wanted to make a different set by "excluding" all odd numbers:

$$\{2, 4, 6, 8, 10\}$$

These sets are of course the same. So is it correct to say that inclusion/exclusion are synonymous when it comes to set theory, as they're just different ways of building a set?

This might sound trivial, but I have a reason for asking: I want to understand if inclusion and exclusion are "commutative" properties, i.e. it doesn't matter in which order you apply them.

For example, let's say I make an operation to "filter" my set, by including all even numbers as we did before, producing set B

$$B = \{2, 4, 6, 8, 10\}$$

And then a separate operation to "exclude" any numbers less than 6 from set B, resulting in set C:

$$C = \{6, 8, 10\}$$

What if I started with A and applied the operations the other way around? First remove all numbers less than 6:

$$B = \{6, 7, 8, 9, 10\}$$

Then filter B to "include" only even numbers:

$$C = \{6, 8, 10\}$$

It seems intuitively to me that the result will always be the same no matter which order you apply the operation. Is this true for all cases no matter the set, however? Is there a way to prove that applying "filters" to a set (I'm not sure of the proper term) will always be commutative?

So in summary, there are two questions here:

1. Are the notions of "inclusion" and "exclusion" really synonymous from the point of view of applying an operation to a set to produce a subset?
2. Will applying these operations to produce a subset of a set always be commutative, i.e. produce the same result?
• It looks lile you are taking repeated intersections of sets. Intersection is both associative and commutative. May 26 at 10:25
• @CrackedBauxite I've had a think about this. Is it intersection, or is it a set difference / relative complement? If I've got the set $\{1, 2, 3, 4, 5\}$ and I exclude all odd numbers, I've got a set of odd numbers $\{1, 3, 5\}$. It's my original set $\{1, 2, 3, 4, 5\} - \{1, 3, 5\} = \{2, 4\}$
– Lou
May 26 at 16:45
• But if I were to do "include all even numbers" from the set $\{1, 2, 3, 4, 5\}$, then I have a set of even numbers $\{2, 4\}$. $\{1, 2, 3, 4, 5\} \cap \{2, 4\} = {2, 4}$ I think - but please correct me if my logic is wrong
– Lou
May 26 at 16:46
• So I think "include" as an operation represents an intersection, but "exclude" represents a set difference?
– Lou
May 26 at 16:48
• Set differences are also intersections. $A\setminus B = A\cap B^c$, where $B^c$ is the complement of $B$. May 27 at 8:38

@KevinS offers an excellent answer from a logic point of view. Here's another that relies on the idea of a filter. That's a concept useful in programming (particularly in lisp). You pass the items in your set through a filter that lets some through and blocks others.

In this sense "inclusion" and "exclusion" are really different ways to describe the same result. You can specify what you keep or what you reject. "Keep (just) the odds" is the same as "reject (only) the evens".

If you have two filters each of which is described independently of the other and refers only to properties of the things you are filtering then you can filter in either order.

The independence matters. If you think that the wine must match the main course then your options will depend on whether you see the wine list or the menu first.

• You've hit upon my actual use case - I'm doing filtering in Python and trying to understand if "inclusion" and "exclusion" for my use case are order-sensitive, or commutative operations. You've all helped me understand that they are commutative and therefore order should not matter.
– Lou
May 26 at 10:52

This is perhaps easier to see in the math logic. When you apply conditions to a set, you get subsets that satisfy those conditions (possibly the emptyset if the conditions aren't met). In your example, define: $$A := \{n\in\mathbb{N}\text{ }|\text{ }1\leq n\leq 10\}.$$ Then $$B$$ and $$C$$ are had by adding conditionals in the set definition: $$B := \{n\in A\text{ }|\text{ }\exists k\in\mathbb{N}: n=2*k\}$$ and $$C:= \{n\in B\text{ }|\text{ }n\geq 6\}.$$ $$\implies C = \bigg\{n\in \mathbb{N}\text{ }\bigg|\text{ }(1\leq n\leq 10)\wedge(\exists k\in\mathbb{N}: (n=2*k))\wedge(n\geq 6)\bigg\}.$$ From this viewpoint, the reductions made were simply conjuctions (a logical operation). Conjunction is certainly commutative.

Also, the term "exclusion" is synonymous with set difference: $$X-Y:= X\cap Y^c,$$ whereas "inclusion" usually refers to an injective map: $$\iota: S\hookrightarrow X$$ which can be used as an identifier of a subset. I wouldn't say the two notions are synonymous.

• I'm a beginner in set theory, would you mind clarifying what you mean about an "injective map"? I understand the term set difference.
– Lou
May 26 at 14:21
• It seems as though the terms used might have multiple meanings as they apply to math and programming. An injection is a 1-1 function. The math notion I gave generalizes to categories (using equivalence classes of monomorphisms to identify sub-objects). This is not inherently programmable and is best used for theory. May 26 at 21:34

To your first question, consider using the set-builder notation to build a subset. That is, define $$B\subseteq A$$ such that $$B = \{a\in A : \Phi(a)\}$$, where $$\Phi$$ is the logical formula which dictates what how we pick the values of $$A$$.

Now say that $$A$$ is some collection of numbers, and $$\Phi$$ is the rule "include all even numbers". Then some other rule $$\Psi$$ which says "do not include all not even numbers" is completely logically equivalent to $$\Phi$$.

And this generalises nicely by considering that, if we impose a condition like "even number", any element of a set will satisify that condition, or not satisfy that condition. We cannot have an element which does neither.

The second question follows as others have pointed out, by writing the conditions as operations which are commutative.


# Chance of each player getting an ace when the dealer has 12 cards?

4 players are playing a card game wherein each of them gets 10 cards, and the dealer gets 12 cards. What is the probability that each player gets an ace?

I want the use the $p = \dfrac{n_A}{N}$ method, where $n_A$ equals the favourable outcomes and $N$ equals all possible outcomes.

Starting with $N$, I figured we could consider the dealer to be a fifth player, and considering we don't care about the order of the players we'd get:

$$N = \dfrac{52!}{(10!)^4\times12!} \times \dfrac{1}{5!}$$

Now for $n_A$, the aces can be divided among the players in $4!$ ways, and each of the players would still get 9 other cards from a total of 48, with the dealer getting the remaining twelve, thus giving us: $$n_A = 4! \times \dfrac{48!}{(9!)^412!} \times \dfrac{1}{5!}$$

But if we calculate $p$ this way we get a probability of $\approx 3\%$, which is just intuitively orders of magnitude too large to be correct, so I am sure I made a mistake somewhere. Can anyone help me spot it and then explain what I did wrong?

• Why the "$\pm$" in $\pm3\%$? – Barry Cipra Oct 10 '16 at 23:09
• @BarryCipra I meant "approximately equal to" but I couldn't find the sign for that on my keyboard, so I thought ± would be the next best thing – YakSal Tafri Oct 11 '16 at 7:01
• The way to get $\approx$ is to type $\approx$. Changed it. – BruceET Oct 11 '16 at 7:54

I wouldn't consider the dealer as a fifth player but instead, let me guide you through another way to get the answer using combinatorics.

We should start by counting $N$ as ${52 \choose 10}{42 \choose 10}{32 \choose 10}{22 \choose 10}$ for the number of ways the dealer can give ten cards from 52 to each of the four players.

$N = {52 \choose 10}{42 \choose 10}{32 \choose 10}{22 \choose 10} \approx 971089585681469963688868551062400$

Now for $n_A$ we will consider all four aces given in $4!$ and count for the number of ways nine cards from the remaining 48 can be given to each of the four players as ${48 \choose 9}{39 \choose 9}{30 \choose 9}{21 \choose 9}$.

$n_A = 4!{48 \choose 9}{39 \choose 9}{30 \choose 9}{21 \choose 9} \approx 35869963456698493441273194240000$

Hence

$p = {n_A \over N} = {4!{48 \choose 9}{39 \choose 9}{30 \choose 9}{21 \choose 9}\over {52 \choose 10}{42 \choose 10}{32 \choose 10}{22 \choose 10}} ={400 \over 10829} \approx 0.036$

Comment: Because 0.036 doesn't seem to match the answer you anticipated, and because there was at least one false start towards a combinatorial answer, I decided to simulate the 'deal' a million times in R statistical software, and see what proportion of deals gave one ace to each player. I got $0.037 \pm 0.0004.$ So I think @AlfredoLozano's method is correct (but note that $400/10829 = 0.03693785 \approx 0.037).$

My deck has 1's for Aces and 0's for all other cards for simplicity counting results, but the sample function treats each 'card' as distinct. The m-vector each.1 is 'logical' with elements TRUE and FALSE; its mean is the proportion of its TRUEs.

m = 10^6; each.1 = logical(m)
deck=c(1,1,1,1, rep(0,48))
for (i in 1:m) {
d = sample(deck, 40)
each.1[i] = (sum(d[1:10]==1)&sum(d[11:20]==1)&sum(d[21:30]==1)&sum(d[31:40]==1))
}
mean(each.1)
## 0.037124

Here is another way to get the answer. Imagine the dealer deals the first $10$ cards to player A, the next $10$ to player B, the next $10$ to player C, the next $10$ to player D, and keeps the rest for himself. Clearly the cards can be shuffled in any of $52!$ ways. The question becomes, in how many ways can the dealer "stack" the deck so that each player gets an Ace?

First, remove the Aces from the deck and shuffle the remaining $48$ cards, which can be done in $48!$ ways. Then shuffle the $4$ Aces, in $4!$ ways. Then insert the first Ace so it's one of the first $10$ cards in the deck, the second Ace so it's one of the next $10$ cards, and so forth. All this can be done in $48!\times4!\times10^4$ ways. So the the probability of each player getting an Ace is

$${48!\times4!\times10^4\over52!}={24\times10^4\over52\times51\times50\times49}=24\left(10\over52\right)\left(10\over51\right)\left(10\over50\right)\left(10\over49\right)\approx25\left(1\over5\right)^4=0.04\%$$

The exact answer, as a reduced fraction, is ${400\over10829}=0.0369378\ldots$


# Partial orders - Question

• Mar 11th 2010, 12:38 PM
gate13
Partial orders - Question
Good day to all,

We have just started relations and I came across the following question in my textbook:

Let X = {1, 2}. List all the partial orders that can be defined on X.

My solution:

I began by computing X x X (Cartesian product) which gave me:

X x X ={(1,1), (1,2), (2,1), (2,2)}

Since we are looking for partial orders, the relation has to be simultaneously reflexive, antisymmetric and transitive.

If the relation R on X is reflexive then it must contain (1,1) and (2,2)

The ordered pairs (1,2) and (2,1) cannot belong to R for if they did and R is also antisymmetric then that would imply 1=2, which is false.

Therefore I concluded that the list of partial orders is the set:
{(1,1), (2,2)}

I was wondering if my logic is flawed and if so what errors have I committed?

Finally, is it possible in this problem to determine the number of partial orders (cardinality)?

Any advice would be greatly appreciated.

Kindest regards
• Mar 11th 2010, 12:44 PM
Plato
Would $\{(1,1),(1,2),(2,2)\}$ also work?
Why or why not?
• Mar 11th 2010, 12:58 PM
gate13
Thank you Plato for your quick response.

I am not sure. The definition of antisymmetric states:

for every a,b in X ((a,b) belongs to R and (b,a) belongs to R implies a=b)

If (1,2) belongs to R and (2,1) does not belong to R (based on the set you listed) then the hypothesis of the implication is false which means that the implication is true. If this is a valid reasoning could we not say the same for the set: {(1,1), (2,1), (2,2)}

Slightly confused!
• Mar 11th 2010, 01:06 PM
Plato
But $(2,1)\notin\{(1,1),(1,2),(2,2)\}$. Is it?
Antisymmetric says that in both $(a,b)\in R~\&~(b,a)\in R$ then it must be true that $a=b$.
• Mar 11th 2010, 01:35 PM
gate13
There is something that I am obviously not understanding. (2,1) does not belong to the set you listed.

I believe the set http://www.mathhelpforum.com/math-he...26e4344e-1.gif is antisymmetric, since for it not to be antisymmetric we would have to have: http://www.mathhelpforum.com/math-he...0853db63-1.gif and a different than b. Therefore the set you provided is a partial order as well (reflexive, transitive and antisymmetric).

I apologize as I realize this may be obvious to many.
• Mar 11th 2010, 01:38 PM
Plato
Quote:

Originally Posted by gate13
I believe the set http://www.mathhelpforum.com/math-he...26e4344e-1.gif is antisymmetric, since for it not to be antisymmetric we would have to have: http://www.mathhelpforum.com/math-he...0853db63-1.gif and a different than b. Therefore the set you provided is a partial order as well (reflexive, transitive and antisymmetric).

That is correct. And there is one more p.o. on the set.
What is it?
• Mar 11th 2010, 01:43 PM
gate13
I believe the other partial order would be :

{(1,1), (2,1), (2,2)} for the same reasons listed previously.
• Mar 11th 2010, 02:01 PM
gate13
I just wanted to thank you Plato for your time and explanations. Many thanks.


### Welcome to our community

#### dwsmith

##### Well-known member
So on the recent graduate problem of the week, I saw that $\int_0^{\infty}\frac{\sin x}{x}dx = \frac{\pi}{2}$, but so does, $\int_0^{\infty}\frac{\sin^2 x}{x^2}dx = \frac{\pi}{2}$.
How can they both be the same?

#### Chris L T521

##### Well-known member
Staff member
So on the recent graduate problem of the week, I saw that $\int_0^{\infty}\frac{\sin x}{x}dx = \frac{\pi}{2}$, but so does, $\int_0^{\infty}\frac{\sin^2 x}{x^2}dx = \frac{\pi}{2}$.
How can they both be the same?
Let us use integration by parts to compute $\displaystyle\int_0^{\infty}\frac{\sin^2 x}{x^2}\,dx$. At the end, we will need to use the fact that $\displaystyle\int_0^{\infty}\frac{\sin x}{x}\,dx=\frac{\pi}{2}$

Let $u=\sin^2x$ and $\,dv=\dfrac{\,dx}{x^2}$. Then $\,du=2\sin x\cos x\,dx=\sin(2x)\,dx$ and $v=-\dfrac{1}{x}$. Therefore,
$\int_0^{\infty}\frac{\sin^2 x}{x^2}\,dx = \left[-\frac{\sin^2 x}{x}\right]_0^{\infty}+\int_0^{\infty}\frac{\sin(2x)}{x}\,dx=\int_0^{\infty}\frac{\sin(2x)}{x}\,dx.$
(We note that $|\sin x|\leq 1\implies |\sin^2 x|\leq 1$ and thus $\displaystyle\lim_{x\to\infty} \frac{\sin^2 x}{x}\sim \lim_{x\to\infty} \frac{1}{x}=0$; We also note that $\displaystyle\lim_{x\to 0}\frac{\sin^2 x}{x}=\lim_{x\to 0}\frac{\sin x}{x}\cdot\lim_{x\to 0}\sin x=0$. Hence, that's why the $\displaystyle\left[-\frac{\sin^2 x}{x}\right]_0^{\infty}$ term goes to zero.)

Now let $t=2x\implies\,dt=2\,dx$. Therefore,
$\int_0^{\infty}\frac{\sin(2x)}{x}\,dx\xrightarrow{t=2x}{} \int_0^{\infty}\frac{\sin t}{t/2}\frac{\,dt}{2}=\int_0^{\infty}\frac{\sin t}{t}=\frac{\pi}{2}.$
And thus, we also have that $\displaystyle\int_0^{\infty}\frac{\sin^2 x}{x^2}\,dx =\frac{\pi}{2}$.

I hope this makes sense!

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
$$F(a)=\int^{\infty}_0\frac{\sin^2(ax)}{x^2}$$

Differentiate w.r.t a :

$$F'(a)=\int^{\infty}_0 \frac{\sin(2ax)}{x}$$

Let 2ax=t

$$F'(a)=\int^{\infty}_0 \frac{\sin(t)}{t}=\frac{\pi}{2}$$

$$F(a)=\frac{\pi}{2}a+C$$

Putting a =0 we get C = 0 hence

$$\int^{\infty}_0\frac{\sin^2(ax)}{x^2}=\frac{\pi \cdot a}{2}$$

So for a =1 we get our result :

$$\int^{\infty}_0\frac{\sin^2(x)}{x^2}=\frac{\pi}{2}$$

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
If your question is why such thing happen , then I don't know , to me it is pretty strange !

If you see the graph of both functions , then you have no indications ...

#### dwsmith

##### Well-known member
I just thought it was strange. When I took Theory of Complex Variables, I had the $\int_0^{\infty}\frac{\sin^2x}{x^2}dx = \frac{\pi}{2}$ exercise so I was surprised to see that $\frac{\sin x}{x}$ lead to the same conclusion.

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
In complex analysis $$\int^{\infty}_0 \dfrac{1-\cos(x)}{x^2}$$ and $$\int^{\infty}_0 \frac{\sin(x)}{x}$$ are conventional exercises to solve by contour integration ....