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I was reading a book on elementary Algebra when I came across an article regarding recurring decimals, in it was a paragraph to find how many digits would recur in the decimal(Or I think it was for this reason).I could not understand the paragraph, so I'll just give the matter as it is:
Let n/d be any real fraction irreducible to lower terms.And suppose it were required to find how many decimal places to reduce it to before same terms return again.To determine this, suppose that 10n > d if not & if 100n or 1000n > d then we begin by reducing 10n or 100n / d to less terms or fraction n/d.
Therefore same number will return in quotient when n is again remainder.Suppose when this happens we have added s zeros & that q is the integral part of quotient, then abstracting from the point we have n x 10^s/d = q + n/d, therefore q = n x (10^s - 1)/d . Now for q to be an integer, we must find least integer number for 's' such that n x (10^s - 1)/d or 10^s - 1 / d is an integer, for this we consider 3 cases:
a: If d is factor 10, 100, 1000 etc. & so there will be no circulating fraction.
b: If d is odd & not a factor of 10 or its power, in which case value of 's' may rise to d - 1 but frequently it is less.
c: d is even but not factor of 10 etc. but nevertheless has a common divisor with one of those powers of 10, which can only be a number of the form 2^c, so that d/2^c = c, so the period will be same for fraction n/d but will not commence before figure represented by c.
I do not understand the meaning of this.
God Bless
I was reading a book on elementary Algebra when I came across an article regarding recurring decimals, in it was a paragraph to find how many digits would recur in the decimal(Or I think it was for this reason).I could not understand the paragraph, so I'll just give the matter as it is:
Let n/d be any real fraction irreducible to lower terms.And suppose it were required to find how many decimal places to reduce it to before same terms return again.To determine this, suppose that 10n > d if not & if 100n or 1000n > d then we begin by reducing 10n or 100n / d to less terms or fraction n/d.
Therefore same number will return in quotient when n is again remainder.Suppose when this happens we have added s zeros & that q is the integral part of quotient, then abstracting from the point we have n x 10^s/d = q + n/d, therefore q = n x (10^s - 1)/d . Now for q to be an integer, we must find least integer number for 's' such that n x (10^s - 1)/d or 10^s - 1 / d is an integer, for this we consider 3 cases:
a: If d is factor 10, 100, 1000 etc. & so there will be no circulating fraction.
b: If d is odd & not a factor of 10 or its power, in which case value of 's' may rise to d - 1 but frequently it is less.
c: d is even but not factor of 10 etc. but nevertheless has a common divisor with one of those powers of 10, which can only be a number of the form 2^c, so that d/2^c = c, so the period will be same for fraction n/d but will not commence before figure represented by c.
I do not understand the meaning of this.
God Bless
AB is a line
D is given angle
C is given triangle
We have to construct a parallelogram which has area equal to triangle C, with one side as AB & one angle equal to D
Now the way it's been proven is
A parallelogram BEFG is constructed with BE in line with AB & FG below it &angle D is equal to angle EBG & area of triangle C equal to area of this parallelogram(I know how to construct a parallelogram having area equal to a given triangle if they are on the same line but I don't know how to construct when they , as is shown in the figure which I couldn't upload, the trangle C is shown in a separate corner & not even near BEFG, in that case how do I construct the parallelogram having the same area is my question)
Coming back to the proof
Extend FG upto a certain point H & join AH such that AH is parallel to EF
Join H & B.
Now since HF is on 2 parallels AH & EF then angle AHF + angle HFE = 180 degrees
Also angle BHG + angle GFE(or HFE) is < 180 degrees ( as they intersect in that direction)
Since they must meet we extend HB & FE upwards where they meet at say K.
We draw KL || EA
Now extend HA & GB to meet KL at points L & say M respectively
Now the whole LKFH is a ||gram with HK as diagonal
Also inside it MKEB & ABGH are ||grams & LMBA & BEFG are complementary ||grams (on opposite side of the diagonal)
So bya previous proof complements on either side of a diagonal are equal in are
So Area LMBA = Area BEFG = Area triangle C
Angle GBE = Angle ABM = Angle D (since vertically opposite angles, if I am right)
So we have a parallelogram LMBA with Ab as its side, area same as C & an angle equal to D.
The remaining parts I have understood because there are proofs for all those etc, but only the step where I have noted is what I don't understand.
Thanking you again
* Earlier I had posted this for which the diagram given by bob was correct but I wherein he suggested that he could construct it if one of the side were parallel to the line but I still don't get how. Please help
AB is a line
D is given angle
C is given triangle
We have to construct a parallelogram which has area equal to triangle C, with one side as AB & one angle equal to D
Now the way it's been proven is
A parallelogram BEFG is constructed with BE in line with AB & FG below it &angle D is equal to angle EBG & area of triangle C equal to area of this parallelogram(I know how to construct a parallelogram having area equal to a given triangle if they are on the same line but I don't know how to construct when they , as is shown in the figure which I couldn't upload, the trangle C is shown in a separate corner & not even near BEFG, in that case how do I construct the parallelogram having the same area is my question)
Coming back to the proof
Extend FG upto a certain point H & join AH such that AH is parallel to EF
Join H & B.
Now since HF is on 2 parallels AH & EF then angle AHF + angle HFE = 180 degrees
Also angle BHG + angle GFE(or HFE) is < 180 degrees ( as they intersect in that direction)
Since they must meet we extend HB & FE upwards where they meet at say K.
We draw KL || EA
Now extend HA & GB to meet KL at points L & say M respectively
Now the whole LKFH is a ||gram with HK as diagonal
Also inside it MKEB & ABGH are ||grams & LMBA & BEFG are complementary ||grams (on opposite side of the diagonal)
So bya previous proof complements on either side of a diagonal are equal in are
So Area LMBA = Area BEFG = Area triangle C
Angle GBE = Angle ABM = Angle D (since vertically opposite angles, if I am right)
So we have a parallelogram LMBA with Ab as its side, area same as C & an angle equal to D.
The remaining parts I have understood because there are proofs for all those etc, but only the step where I have noted is what I don't understand.
Thanking you again
* Earlier I had posted this for which the diagram given by bob was correct but I wherein he suggested that he could construct it if one of the side were parallel to the line but I still don't get how. Please help
How do I do the construction?
The diagram is absolutely correct now I just want to know how to do the construction if you could tell me.
Thank you
I was reading a certain book in which certain divisibility rules for numbers were given, in it was a given that if any number is divided by 7 & if the fraction were recurring then the number of digits that would recur at maximum would be only six, I am unable to understand the reason behind this.
God bless
AB is a line
D is given angle
C is given triangle
We have to construct a parallelogram which has area equal to triangle C, with one side as AB & one angle equal to D
Now the way it's been proven is
A parallelogram BEFG is constructed with BE in line with AB & FG below it &angle D is equal to angle EBG & area of triangle C equal to area of this parallelogram(I know how to construct a parallelogram having area equal to a given triangle if they are on the same line but I don't know how to construct when they , as is shown in the figure which I couldn't upload, the trangle C is shown in a separate corner & not even near BEFG, in that case how do I construct the parallelogram having the same area is my question)
Coming back to the proof
Extend FG upto a certain point H & join AH such that AH is parallel to EF
Join H & B.
Now since HF is on 2 parallels AH & EF then angle AHF + angle HFE = 180 degrees
Also angle BHG + angle GFE(or HFE) is < 180 degrees ( as they intersect in that direction)
Since they must meet we extend HB & FE upwards where they meet at say K.
We draw KL || EA
Now extend HA & GB to meet KL at points L & say M respectively
Now the whole LKFH is a ||gram with HK as diagonal
Also inside it MKEB & ABGH are ||grams & LMBA & BEFG are complementary ||grams (on opposite side of the diagonal)
So bya previous proof complements on either side of a diagonal are equal in are
So Area LMBA = Area BEFG = Area triangle C
Angle GBE = Angle ABM = Angle D (since vertically opposite angles, if I am right)
So we have a parallelogram LMBA with Ab as its side, area same as C & an angle equal to D.
The remaining parts I have understood because there are proofs for all those etc, but only the step where I have noted is what I don't understand.
Thanking you again
Sir, How can I upload the image?
Thank you all for the answers & God bless you all
I don't want to sound very pestering but I still have a slight difficulty in understanding this:
I don't understand why do we take twice of the number? & also why do we take 10z x z where we also write z next to 5? I am not able understand that, sorry for the trouble but thanks again & God bless
I am referring to this algorithm http://www.basic-mathematics.com/square-root-algorithm.html
Also why do we take twice of that number?
Thank you so much & God bless you always & forever
We had to draw a parallelogram with a given line as its side & area equal
to a given triangle & also that one of the angles in the parallelogram
would be equal to a given angle
I understood everything in the proof except one that is that the given line
say AB was extended to E & that the parallelogram BEFG was constructed with
area equal to the given triangle, but the base of the triangle & the line
BE were not on the same line they were different, I know how to construct a
parallelogram with one of its sides in line with the base of a given
triangle & with their areas being equal but I don`t understand how to
construct it when the side of the parallelogram is not inline with the base
of the triangle.
Please help me
God bless
Why do we take 2 digits initially in the algorithm also why do we take 3 digits for cube roots etc?
God bless
How does the compound ratio system work i.e if it takes 20$ per post for one horse then how much will it take for 8 horses for 4.5 posts
I understand ratio of horses is 1:8
Ratio of posts i 1: 4.5 i.e 2:9
Taking compund ratio becomes 2:72
I.e 1:36
I.e it takes 720$ but how does compunding the ratio give the amount f money required. Plz help
God bless
I know it seems crazy but I want to know why a( x+ y). Is ax+ay, I know it is distributive law but how do you prove this law.plz help
God bless
I had this from Euler's Algebra proof to represent polygonal numbers of any kind but I don't understand what it means it goes like this:
Draw a polygon haing number of sides required n this number is constant for the whole series n equal to 2 + diff of arith progress
Then choose on of its angles n draw diagonals n the sides of he angle n the diagonal are to be indefinetly produced
After that I take these 2 sides n diagonals o he first polygon as I often as I choose n draw from corresponding points marked by compass lines parallel to first polygon n divide them in as many equal parts or as many points as there are actualy in the diagonals in the 2 sides produced.
Please give geometrical represantation of the proof if you can
God bless n thank you
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