Broad word systematic problem

ОБПОУ 
«КЭМТ»

   УТВЕРЖДАЮ

                                                                                  
заместитель директора

    _____________ В.Н.Павленко

    «____»_______________2016 г.

Контрольные
задания

по учебной
дисциплине
ООПб.02 Иностранный язык

для
профессии:

15.01.05
Сварщик (ручной и частично механизированной сварки (наплавки)

для группы
 СВ-21

Преподаватель:  Саакян Ирина
Ивановна

Рассмотрено на
заседании

ПЦК преподавателей
иностранного языка

протокол № ____ от
___________ 2016 г.

председатель ПЦК
_____________ Е.А.Белошапкина

Контрольные задания

по учебной дисциплине  ООПб.02 Иностранный язык

по разделу: « Science and technology»
(«Наука и техника»)

для
группы CВ -21

EXERCISE
1. a) Read the words
following the rules of read­ing. Pronounce correctly. Practise aloud:

[e]
invention, investigate, attempt, interconnect, shelter,

engine,
measure, cell;

[æ]
fact,
examine, animal, act, national, overlap, satisfy;

[a:]
plants;

[^]
study,
subject, structure, become, discovery;

[o]
technology, quantity;

[u:]
true, computer, tool, prove, unity;

[i:]
field, deal, steam;

[i] 
contribute, digital, liquid, since;

[з:] 
search,
research, refer, word, universe;

[ei]
way, same, relationship, great, explain, complicated;

[au]
boundary;

[ou]
closely, process, social, grow;

[iə]
appear,
theory, clear;

[ai]
typewriter, wide, try, divide, provide;

[aiə]
science, scientist, variety

EXERCISE
2.

a)
Decide which word is the odd one out in each of the following groups of words:

natural,
social, digital, technical;

digital,
natural, analog, hybrid;

theories,
tools, machines, materials;

radio,
knowledge, telephone, television;

the
universe, plants, trees, animals

b)
Find the names for the groups of words. Fill in each of the spaces:

natural,
social, technical — _________________________________;

digital,
analog, hybrid —     _________________________________ ;

radio,
television, telephone — ________________________________;

tools,
machines, materials — _________________________________;

plants,
animals, trees — _____________________________________.

EXERCISE
3.

a) Translate the following definitions of
the words:

science
is
the study of knowledge which can be made into a system, and which usually
depends on seeing and interest­ing facts and stating general natural laws;

technology
is
a branch of knowledge dealing with scien­tific and industrial methods and their
practical use in in­dustry;

reseach
is
a serious and detailed study of a subject, that is aimed at learning new facts,
scientific laws, testing ideas, etc.;

a
shelter is a building or something of the kind
that gives protection;

a
typewriter is a
machine that prints letters.

b)
The following words also appear in the texts and dia­logues. Match each one
with its correct definition:

to
prove, to process, to unify, to explain, to appear

to
combine parts of something to form a single whole;

to
make clear or easy to understand, usually by speaking or writing;

to
become able to be seen, to come into sight;

to
show to be true by means of facts, documents, infor­mation, etc.;

to
put (information, numbers, etc.) into a computer for examination.

EXERCISE
4

Match
the words from the column A with the words from the column B.

1.
broad                                         a) word

2. systematic                               
b) problem

3. natural                                     
c) theory

4. Latin                                        
d) principle

5. different                                   
e) groups

6. general                                      
f) methods

7. major                                       
g) field

8. industrial                                  
h)objects

9. scientific                                   
i) technology

10. mathematical                      
    j) sciences

From Wikipedia, the free encyclopedia

In mathematics, especially in the area of abstract algebra known as combinatorial group theory, the word problem for a finitely generated group G is the algorithmic problem of deciding whether two words in the generators represent the same element. More precisely, if A is a finite set of generators for G then the word problem is the membership problem for the formal language of all words in A and a formal set of inverses that map to the identity under the natural map from the free monoid with involution on A to the group G. If B is another finite generating set for G, then the word problem over the generating set B is equivalent to the word problem over the generating set A. Thus one can speak unambiguously of the decidability of the word problem for the finitely generated group G.

The related but different uniform word problem for a class K of recursively presented groups is the algorithmic problem of deciding, given as input a presentation P for a group G in the class K and two words in the generators of G, whether the words represent the same element of G. Some authors require the class K to be definable by a recursively enumerable set of presentations.

History[edit]

Throughout the history of the subject, computations in groups have been carried out using various normal forms. These usually implicitly solve the word problem for the groups in question. In 1911 Max Dehn proposed that the word problem was an important area of study in its own right,^[1] together with the conjugacy problem and the group isomorphism problem. In 1912 he gave an algorithm that solves both the word and conjugacy problem for the fundamental groups of closed orientable two-dimensional manifolds of genus greater than or equal to 2.^[2] Subsequent authors have greatly extended Dehn’s algorithm and applied it to a wide range of group theoretic decision problems.^[3][4][5]

It was shown by Pyotr Novikov in 1955 that there exists a finitely presented group G such that the word problem for G is undecidable.^[6] It follows immediately that the uniform word problem is also undecidable. A different proof was obtained by William Boone in 1958.^[7]

The word problem was one of the first examples of an unsolvable problem to be found not in mathematical logic or the theory of algorithms, but in one of the central branches of classical mathematics, algebra. As a result of its unsolvability, several other problems in combinatorial group theory have been shown to be unsolvable as well.

It is important to realize that the word problem is in fact solvable for many groups G. For example, polycyclic groups have solvable word problems since the normal form of an arbitrary word in a polycyclic presentation is readily computable; other algorithms for groups may, in suitable circumstances, also solve the word problem, see the Todd–Coxeter algorithm^[8] and the Knuth–Bendix completion algorithm.^[9] On the other hand, the fact that a particular algorithm does not solve the word problem for a particular group does not show that the group has an unsolvable word problem. For instance Dehn’s algorithm does not solve the word problem for the fundamental group of the torus. However this group is the direct product of two infinite cyclic groups and so has a solvable word problem.

A more concrete description[edit]

In more concrete terms, the uniform word problem can be expressed as a rewriting question, for literal strings.^[10] For a presentation P of a group G, P will specify a certain number of generators

x, y, z, …

for G. We need to introduce one letter for x and another (for convenience) for the group element represented by x⁻¹. Call these letters (twice as many as the generators) the alphabet for our problem. Then each element in G is represented in some way by a product

abc … pqr

of symbols from , of some length, multiplied in G. The string of length 0 (null string) stands for the identity element e of G. The crux of the whole problem is to be able to recognise all the ways e can be represented, given some relations.

The effect of the relations in G is to make various such strings represent the same element of G. In fact the relations provide a list of strings that can be either introduced where we want, or cancelled out whenever we see them, without changing the ‘value’, i.e. the group element that is the result of the multiplication.

For a simple example, take the presentation {a | a³}. Writing A for the inverse of a, we have possible strings combining any number of the symbols a and A. Whenever we see aaa, or aA or Aa we may strike these out. We should also remember to strike out AAA; this says that since the cube of a is the identity element of G, so is the cube of the inverse of a. Under these conditions the word problem becomes easy. First reduce strings to the empty string, a, aa, A or AA. Then note that we may also multiply by aaa, so we can convert A to aa and convert AA to a. The result is that the word problem, here for the cyclic group of order three, is solvable.

This is not, however, the typical case. For the example, we have a canonical form available that reduces any string to one of length at most three, by decreasing the length monotonically. In general, it is not true that one can get a canonical form for the elements, by stepwise cancellation. One may have to use relations to expand a string many-fold, in order eventually to find a cancellation that brings the length right down.

The upshot is, in the worst case, that the relation between strings that says they are equal in G is an Undecidable problem.

Examples[edit]

The following groups have a solvable word problem:

• Automatic groups, including:
  • Finite groups
  • Negatively curved (aka. hyperbolic) groups
  • Euclidean groups
  • Coxeter groups
  • Braid groups
  • Geometrically finite groups
• Finitely generated free groups
• Finitely generated free abelian groups
• Polycyclic groups
• Finitely generated recursively absolutely presented groups,^[11] including:
  • Finitely presented simple groups.
• Finitely presented residually finite groups
• One relator groups^[12] (this is a theorem of Magnus), including:
  • Fundamental groups of closed orientable two-dimensional manifolds.
• Combable groups
• Autostackable groups

Examples with unsolvable word problems are also known:

• Given a recursively enumerable set A of positive integers that has insoluble membership problem, ⟨a,b,c,d | aⁿbaⁿ = cⁿdcⁿ : n ∈ A⟩ is a finitely generated group with a recursively enumerable presentation whose word problem is insoluble^[13]
• Every finitely generated group with a recursively enumerable presentation and insoluble word problem is a subgroup of a finitely presented group with insoluble word problem^[14]
• The number of relators in a finitely presented group with insoluble word problem may be as low as 14 ^[15] or even 12.^[16][17]
• An explicit example of a reasonable short presentation with insoluble word problem is given in Collins 1986:^[18][19]

Partial solution of the word problem[edit]

The word problem for a recursively presented group can be partially solved in the following sense:

Given a recursive presentation P = ⟨X|R⟩ for a group G, define:

then there is a partial recursive function f_P such that:

More informally, there is an algorithm that halts if u=v, but does not do so otherwise.

It follows that to solve the word problem for P it is sufficient to construct a recursive function g such that:

However u=v in G if and only if uv⁻¹=1 in G. It follows that to solve the word problem for P it is sufficient to construct a recursive function h such that:

Example[edit]

The following will be proved as an example of the use of this technique:

Theorem: A finitely presented residually finite group has solvable word problem.

Proof: Suppose G = ⟨X|R⟩ is a finitely presented, residually finite group.

Let S be the group of all permutations of N, the natural numbers, that fixes all but finitely many numbers then:

1. S is locally finite and contains a copy of every finite group.
2. The word problem in S is solvable by calculating products of permutations.
3. There is a recursive enumeration of all mappings of the finite set X into S.
4. Since G is residually finite, if w is a word in the generators X of G then w ≠ 1 in G if and only of some mapping of X into S induces a homomorphism such that w ≠ 1 in S.

Given these facts, algorithm defined by the following pseudocode:

For every mapping of X into S
    If every relator in R is satisfied in S
        If w ≠ 1 in S
            return 0
        End if
    End if
End for

defines a recursive function h such that:

This shows that G has solvable word problem.

Unsolvability of the uniform word problem[edit]

The criterion given above, for the solvability of the word problem in a single group, can be extended by a straightforward argument. This gives the following criterion for the uniform solvability of the word problem for a class of finitely presented groups:

To solve the uniform word problem for a class K of groups, it is sufficient to find a recursive function that takes a finite presentation P for a group G and a word in the generators of G, such that whenever G ∈ K:

Boone-Rogers Theorem: There is no uniform partial algorithm that solves the word problem in all finitely presented groups with solvable word problem.

In other words, the uniform word problem for the class of all finitely presented groups with solvable word problem is unsolvable. This has some interesting consequences. For instance, the Higman embedding theorem can be used to construct a group containing an isomorphic copy of every finitely presented group with solvable word problem. It seems natural to ask whether this group can have solvable word problem. But it is a consequence of the Boone-Rogers result that:

Corollary: There is no universal solvable word problem group. That is, if G is a finitely presented group that contains an isomorphic copy of every finitely presented group with solvable word problem, then G itself must have unsolvable word problem.

Remark: Suppose G = ⟨X|R⟩ is a finitely presented group with solvable word problem and H is a finite subset of G. Let H^* = ⟨H⟩, be the group generated by H. Then the word problem in H^* is solvable: given two words h, k in the generators H of H^*, write them as words in X and compare them using the solution to the word problem in G. It is easy to think that this demonstrates a uniform solution of the word problem for the class K (say) of finitely generated groups that can be embedded in G. If this were the case, the non-existence of a universal solvable word problem group would follow easily from Boone-Rogers. However, the solution just exhibited for the word problem for groups in K is not uniform. To see this, consider a group J = ⟨Y|T⟩ ∈ K; in order to use the above argument to solve the word problem in J, it is first necessary to exhibit a mapping e: Y → G that extends to an embedding e^*: J → G. If there were a recursive function that mapped (finitely generated) presentations of groups in K to embeddings into G, then a uniform solution of the word problem in K could indeed be constructed. But there is no reason, in general, to suppose that such a recursive function exists. However, it turns out that, using a more sophisticated argument, the word problem in J can be solved without using an embedding e: J → G. Instead an enumeration of homomorphisms is used, and since such an enumeration can be constructed uniformly, it results in a uniform solution to the word problem in K.

Proof that there is no universal solvable word problem group[edit]

Suppose G were a universal solvable word problem group. Given a finite presentation P = ⟨X|R⟩ of a group H, one can recursively enumerate all homomorphisms h: H → G by first enumerating all mappings h^†: X → G. Not all of these mappings extend to homomorphisms, but, since h^†(R) is finite, it is possible to distinguish between homomorphisms and non-homomorphisms, by using the solution to the word problem in G. «Weeding out» non-homomorphisms gives the required recursive enumeration: h₁, h₂, …, h_n, … .

If H has solvable word problem, then at least one of these homomorphisms must be an embedding. So given a word w in the generators of H:

Consider the algorithm described by the pseudocode:

Let n = 0
    Let repeatable = TRUE
        while (repeatable)
            increase n by 1
            if (solution to word problem in G reveals hn(w) ≠ 1 in G)
                Let repeatable = FALSE
output 0.

This describes a recursive function:

The function f clearly depends on the presentation P. Considering it to be a function of the two variables, a recursive function has been constructed that takes a finite presentation P for a group H and a word w in the generators of a group G, such that whenever G has soluble word problem:

But this uniformly solves the word problem for the class of all finitely presented groups with solvable word problem, contradicting Boone-Rogers. This contradiction proves G cannot exist.

Algebraic structure and the word problem[edit]

There are a number of results that relate solvability of the word problem and algebraic structure. The most significant of these is the Boone-Higman theorem:

A finitely presented group has solvable word problem if and only if it can be embedded in a simple group that can be embedded in a finitely presented group.

It is widely believed that it should be possible to do the construction so that the simple group itself is finitely presented. If so one would expect it to be difficult to prove as the mapping from presentations to simple groups would have to be non-recursive.

The following has been proved by Bernhard Neumann and Angus Macintyre:

A finitely presented group has solvable word problem if and only if it can be embedded in every algebraically closed group

What is remarkable about this is that the algebraically closed groups are so wild that none of them has a recursive presentation.

The oldest result relating algebraic structure to solvability of the word problem is Kuznetsov’s theorem:

A recursively presented simple group S has solvable word problem.

To prove this let ⟨X|R⟩ be a recursive presentation for S. Choose a ∈ S such that a ≠ 1 in S.

If w is a word on the generators X of S, then let:

There is a recursive function such that:

Write:

Then because the construction of f was uniform, this is a recursive function of two variables.

It follows that: is recursive. By construction:

Since S is a simple group, its only quotient groups are itself and the trivial group. Since a ≠ 1 in S, we see a = 1 in S_w if and only if S_w is trivial if and only if w ≠ 1 in S. Therefore:

The existence of such a function is sufficient to prove the word problem is solvable for S.

This proof does not prove the existence of a uniform algorithm for solving the word problem for this class of groups. The non-uniformity resides in choosing a non-trivial element of the simple group. There is no reason to suppose that there is a recursive function that maps a presentation of a simple groups to a non-trivial element of the group. However, in the case of a finitely presented group we know that not all the generators can be trivial (Any individual generator could be, of course). Using this fact it is possible to modify the proof to show:

The word problem is uniformly solvable for the class of finitely presented simple groups.

See also[edit]

• Combinatorics on words
• SQ-universal group
• Word problem (mathematics)
• Reachability problem
• Nested stack automata (have been used to solve the word problem for groups)

Notes[edit]

References[edit]

• Boone, W.W.; Cannonito, F.B.; Lyndon, Roger C. (1973). Word problems : decision problems and the Burnside problem in group theory. Studies in logic and the foundations of mathematics. Vol. 71. North-Holland. ISBN 9780720422719.
• Boone, W. W.; Higman, G. (1974). «An algebraic characterization of the solvability of the word problem». J. Austral. Math. Soc. 18: 41–53. doi:10.1017/s1446788700019108.
• Boone, W. W.; Rogers Jr, H. (1966). «On a problem of J. H. C. Whitehead and a problem of Alonzo Church». Math. Scand. 19: 185–192. doi:10.7146/math.scand.a-10808.
• Borisov, V. V. (1969), «Simple examples of groups with unsolvable word problem», Akademiya Nauk SSSR. Matematicheskie Zametki, 6: 521–532, ISSN 0025-567X, MR 0260851
• Collins, Donald J. (1969), «Word and conjugacy problems in groups with only a few defining relations», Zeitschrift für Mathematische Logik und Grundlagen der Mathematik, 15 (20–22): 305–324, doi:10.1002/malq.19690152001, MR 0263903
• Collins, Donald J. (1972), «On a group embedding theorem of V. V. Borisov», Bulletin of the London Mathematical Society, 4 (2): 145–147, doi:10.1112/blms/4.2.145, ISSN 0024-6093, MR 0314998
• Collins, Donald J. (1986), «A simple presentation of a group with unsolvable word problem», Illinois Journal of Mathematics, 30 (2): 230–234, doi:10.1215/ijm/1256044631, ISSN 0019-2082, MR 0840121
• Collins, Donald J.; Zieschang, H. (1990), Combinatorial group theory and fundamental groups, Springer-Verlag, p. 166, MR 1099152
• Dehn, Max (1911), «Über unendliche diskontinuierliche Gruppen», Mathematische Annalen, 71 (1): 116–144, doi:10.1007/BF01456932, ISSN 0025-5831, MR 1511645, S2CID 123478582
• Dehn, Max (1912), «Transformation der Kurven auf zweiseitigen Flächen», Mathematische Annalen, 72 (3): 413–421, doi:10.1007/BF01456725, ISSN 0025-5831, MR 1511705, S2CID 122988176
• Kuznetsov, A.V. (1958). «Algorithms as operations in algebraic systems». Izvestia Akad. Nauk SSSR Ser Mat. 13 (3): 81.
• Miller, C.F. (1991). «Decision problems for groups — survey and reflections». Algorithms and Classification in Combinatorial Group Theory. Mathematical Sciences Research Institute Publications. Vol. 23. Springer. pp. 1–60. doi:10.1007/978-1-4613-9730-4_1. ISBN 978-1-4613-9730-4.
• Nyberg-Brodda, Carl-Fredrik (2021), «The word problem for one-relation monoids: a survey», Semigroup Forum, 103 (2): 297–355, arXiv:2105.02853, doi:10.1007/s00233-021-10216-8
• Rotman, Joseph (1994), An introduction to the theory of groups, Springer-Verlag, ISBN 978-0-387-94285-8
• Stillwell, J. (1982). «The word problem and the isomorphism problem for groups». Bulletin of the AMS. 6: 33–56. doi:10.1090/s0273-0979-1982-14963-1.

Solving Systems of Equations
Real World Problems

Wow! You have learned many different strategies for solving systems of equations! First we started with Graphing Systems of Equations.
Then we moved onto solving systems using the Substitution Method. In our last lesson we used the
Linear Combinations or Addition Method to solve systems of equations.

Now we are ready to apply these strategies to solve real world
problems! Are you ready? First let’s look at some guidelines for
solving real world problems and then we’ll look at a few examples.

Ok… let’s look at a few examples. Follow along with me. (Having a calculator will make it easier for you to follow along.)

Example 1: Systems Word Problems

You are running a concession stand at a basketball game. You are selling hot dogs and sodas. Each hot dog costs $1.50 and each soda costs $0.50. At the end of the night you made a total of $78.50. You sold a total of 87 hot dogs and sodas combined. You must report the number of hot dogs sold and the number of sodas sold. How many hot dogs were sold and how many sodas were sold?

Solution

1.  Let’s start by identifying the important information:

• hot dogs cost $1.50
• Sodas cost $0.50
• Made a total of $78.50
• Sold 87 hot dogs and sodas combined

2.  Define your variables.

• Ask yourself, «What am I trying to solve for? What don’t I know?

In this problem, I don’t know how many hot dogs or sodas were sold. So this is what each variable will stand for. (Usually the question at the end will give you this information).

Let x = the number of hot dogs sold

Let y = the number of sodas sold

3. Write two equations.

One equation will be related to the price and one equation will be related to the quantity (or number) of hot dogs and sodas sold.

1.50x + 0.50y = 78.50    (Equation related to cost)

 x + y = 87   (Equation related to the number sold)

4.  Solve! 

We can choose any method that we like to solve the system of equations. I am going to choose the substitution method since I can easily solve the 2nd equation for y.

5. Think about what this solution means.

x is the number of hot dogs and x = 35. That means that 35 hot dogs were sold.

y is the number of sodas and y = 52. That means that 52 sodas were sold.

6.  Write your answer in a complete sentence.

35 hot dogs were sold and 52 sodas were sold.

7.  Check your work by substituting.

1.50x + 0.50y = 78.50

1.50(35) + 0.50(52) = 78.50

52.50 + 26 = 78.50

AND

x + y = 87

35 + 52 = 87

Since both equations check properly, we know that our answers are correct!

That wasn’t too bad, was it? The hardest part is writing the equations.
From there you already know the strategies for solving. Think
carefully about what’s happening in the problem when trying to write the
two equations.

Example 2: Another Word Problem

You and a friend go to Tacos Galore for lunch. You order three soft tacos and three burritos and your total bill is $11.25. Your friend’s bill is $10.00 for four soft tacos and two burritos. How much do soft tacos cost? How much do burritos cost?

1.  Let’s start by identifying the important information:

• 3 soft tacos + 3 burritos cost $11.25
  
• 4 soft tacos + 2 burritos cost $10.00

2.  Define your variables.

• Ask yourself, «What am I trying to solve for? What don’t I know?

In this problem, I don’t know the price of the soft tacos or the price of the burritos.

Let x = the price of 1 soft taco

Let y = the price of 1 burrito

3. Write two equations.

One equation will be related your lunch and one equation will be related to your friend’s lunch.

3x + 3y = 11.25  (Equation representing your lunch)

4x + 2y = 10   (Equation representing your friend’s lunch)

4.  Solve! 

We can choose any method that we like to solve the system of equations. I am going to choose the combinations method.

5. Think about what the solution means in context of the problem.

x = the price of 1 soft taco and x = 1.25.

That means that 1 soft tacos costs $1.25.

y = the price of 1 burrito and y = 2.5.

That means that 1 burrito costs $2.50.

Yes, I know that word problems can be intimidating, but this is the
whole reason why we are learning these skills. You must be able to
apply your knowledge!

If you have difficulty with real world problems, you can find more examples and practice problems in the Algebra Class E-course.

Take a look at the questions that other students have submitted:

1. Home

>

2. System of Equations

>

3. 
   Systems Word Problems