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What does it imply?

Necessary, sufficient, necessary and sufficient conditions

Example 1

The team of the Timbertoes and the Shortlegs played a football match; there were no special happenings (there were no own goals, both teams took part, it did not finish earlier, etc.). Who won if the reporter said the following at the end of the commentary?

The Timbertoes scored one goal more than the Shortlegs did.
The Timbertoes scored.
One of the players of the Shortlegs was sent off.
The Timbertoes scored more goals than the Shortlegs did.

Solution (a)

If the Timbertoes scored one goal more than the Shortlegs did, then the Timbertoes won for sure. Thus scoring one goal more than the opponent is a sufficient condition for the Timbertoes to win.

But the Timbertoes can win in other ways too, not only by scoring one goal more, i.e. scoring one goal more than the opponent is not a necessary condition for the Timbertoes to win.

To summarise: scoring one goal more than the opponent is a sufficient but not necessary condition for the Timbertoes to win.

Solution (b)

If the Timbertoes scored, it is still possible that they lost the match if the opponent scored more goals. Thus the fact that the Timbertoes scored does not imply that they won. So scoring is not a sufficient condition for the Timbertoes to win.

But on the other hand: if the Timbertoes won, then they must have scored, otherwise they could not have won. So scoring is a necessary condition for the Timbertoes to win.

To summarise: scoring is a necessary but not sufficient condition for the Timbertoes to win.

Solution (c)

The fact that one of the players of the Shortlegs was sent off does not imply that the Timbertoes won, thus one player of the opponent being sent off is not a sufficient condition for the Timbertoes to win.

The fact that the Timbertoes won does not imply that one of the players of the Shortlegs was sent off, thus one player of the opponent being sent off is not a necessary condition for the Timbertoes to win.

To summarise: one of the players of the opponent being sent off is not a necessary and not a sufficient condition for the Timbertoes to win.

Solution (d)

If the Timbertoes scored more goals than the Shortlegs did, then they must have won; thus the fact that the Timbertoes scored more goals than the opponent implies that they won. So scoring more goals than the opponent is a sufficient condition for the Timbertoes to win.

If the Timbertoes won, it was only possible so that they scored more goals than the Shortlegs, i.e. the fact that the Timbertoes won implies that they scored more goals than the Shortlegs did. So scoring more goals than the opponent is a necessary condition for the Timbertoes to win.

To summarise: scoring more goals than the opponent is a necessary and sufficient condition for the Timbertoes to win.

Note: A condition can be necessary or not necessary, sufficient or not sufficient, there are a total of \latex{2\times2 = 4} options.

STATEMENT:

If the Timbertoes won, then the Timbertoes scored more goals than the opponent. (The Timbertoes won only if the Timbertoes scored more goals than the opponent.)
Scoring more Ugoals than the opponent is a necessary condition for the Timbertoes to win.

Notation: The Timbertoes won \latex{\Rightarrow} the Timbertoes scored more goals than the opponent.

THE CONVERSE OF THE STATEMENT:

If the Timbertoes scored more goals than the opponent, then the Timbertoes won.
Scoring more goals than the opponent is a sufficient condition for the Timbertoes to win.

Notation: The Timbertoes scored more goals than the opponent \latex{\Rightarrow} the Timbertoes won.

THE STATEMENT AND ITS CONVERSE:

The Timbertoes won if and only if the Timbertoes scored more goals than the opponent.
Scoring more goals than the opponent is a necessary and sufficient condition for the Timbertoes to win.

Notation: The Timbertoes won \latex{\lrArr} the Timbertoes scored more goals than the opponent.

Example 2

How can we decide whether a natural number \latex{n} is divisible by \latex{24}? Five students gave the following answers:

Ann: The natural number \latex{n} needs to be divisible by \latex{4}.
Ben: The natural number \latex{n} needs to be divisible by \latex{48}.
Chris: The natural number \latex{n} needs to be divisible by \latex{5}.

Dana: The natural number \latex{n} needs to be divisible by \latex{4} and \latex{6}.

Edward: The natural number \latex{n} is divisible by \latex{3} and \latex{8}.

Let us examine these answers.

Solution (Ann)

Indeed, if a natural number is divisible by \latex{24}, then it is also divisible by \latex{4}. It means that divisibility by \latex{4} is a necessary condition of the divisibility by \latex{24}.

But there is a number divisible by \latex{4}, e.g. \latex{28}, which is not divisible by \latex{24}. Therefore if a number is divisible by \latex{4}, then it is not sure that it is divisible by \latex{24}, i.e. the divisibility by \latex{4} does not imply the divisibility by \latex{24}. In other words: the divisibility by \latex{4} is not a sufficient condition of the divisibility by \latex{24}.

To summarise: the divisibility by \latex{4} is a necessary but not sufficient condition of the divisibility by \latex{24}. (Figure 1)

Solution (Ben)

There is a number divisible by \latex{24}, e.g. \latex{72}, which is not divisible by \latex{48}. Therefore if a number is divisible by \latex{24}, then it is not sure that it is divisible by \latex{48}, i.e. the divisibility by \latex{24} does not imply the divisibility by \latex{48}. In other words: the divisibility by \latex{48} is not a necessary condition of the divisibility by \latex{24}.

However if a natural number is divisible by \latex{48}, then it is also divisible by \latex{24}. It means that the divisibility by \latex{48} is a sufficient condition of the divisibility by \latex{24}.

To summarise: the divisibility by \latex{48} is a sufficient but not necessary condition of the divisibility by \latex{24}. (Figure 2)

Solution (Chris)

There is a natural number divisible by \latex{24}, e.g. \latex{24}, which is not divisible by \latex{5}. Therefore if a number is divisible by \latex{24}, then it is not sure that it is divisible by \latex{5}, i.e. the divisibility by \latex{24} does not imply the divisibility by \latex{5}. It means that the divisibility by \latex{5} is not a necessary condition of the divisibility by \latex{24}.

Furthermore there is a natural number divisible by \latex{5}, e.g. \latex{25}, which is not divisible by \latex{24}. Therefore if a number is divisible by \latex{5}, then it is not sure that it is divisible by \latex{24}, i.e. the divisibility by \latex{5} does not imply the divisibility by \latex{24}.

In other words: the divisibility by \latex{5} is not a sufficient condition of the divisibility by \latex{24}.

To summarise: the divisibility by \latex{5} is not a necessary and not a sufficient condition of the divisibility by \latex{24}. (Figure 3)

Solution (Dana)

Indeed, if a natural number is divisible by \latex{24}, then it is also divisible by \latex{4} and \latex{6}. It means that the divisibility by \latex{4} and \latex{6} is a necessary condition of the divisibility by \latex{24}.

But there is a number divisible by \latex{4} and \latex{6}, e.g. \latex{12}, which is not divisible by \latex{24}. Therefore if a number is divisible by \latex{4} and \latex{6}, then it is not sure that it is divisible by \latex{24}, i.e. the divisibility by \latex{4} and \latex{6} does not imply the divisibility by \latex{24}.

In other words: the divisibility by \latex{4} and \latex{6} is not a sufficient condition of the divisibility by \latex{24}.

To summarise: the divisibility by \latex{4} and \latex{6} is a necessary but not a sufficient condition of the divisibility by \latex{24}. (Figure 4)

Note: In general it is not true that if \latex{a} and \latex{b} are natural numbers for which both \latex{a} and \latex{b} are divisors of \latex{n}, then \latex{a\times{b}} is a divisor of \latex{n}.

Solution (Edward)

Indeed, if a natural number is divisible by \latex{24}, then it is also divisible by \latex{3} and \latex{8}. It means that the divisibility by \latex{3} and \latex{8} is a necessary condition of the divisibility by \latex{24}.

If a natural number is divisible by \latex{3} and \latex{8}, then it is also divisible by \latex{24}. It means that the divisibility by \latex{3} and \latex{8} is a sufficient condition of the divisibility by \latex{24}.

To summarise: an arbitrary natural number \latex{n} is divisible by \latex{24} if and only if it is divisible by \latex{3} and \latex{8}. The divisibility by \latex{3} and \latex{8} is a necessary and sufficient condition of the divisibility by \latex{24}.

Note: For all natural numbers \latex{a} and \latex{b}, which are relatively primes, it is true that both \latex{a} and \latex{b} are divisors of \latex{n} if and only if \latex{a\times{b}} is a divisor of \latex{n}.

Example 3

An \latex{8\times8} chessboard can be tiled by \latex{2\times1} dominoes in several ways. One such possible tiling can be seen in figure \latex{5}.

If one corner (\latex{1\times1} square), of the chessboard is removed, can the remaining board be tiled by dominoes?
If two adjacent squares are removed from the top left corner of the chessboard, can the remaining board be tiled by dominoes?

If two arbitrary but adjacent squares of the chessboard are removed, can the remaining board be tiled by dominoes?
If the two top corners of the chessboard are removed, can the remaining board be tiled by dominoes?
If two opposite corners of the chessboard are removed, can the remaining board be tiled by dominoes?
If two arbitrary and differently coloured squares of the chessboard are removed, can the remaining board be tiled by dominoes?

Solution (a)

It can be seen that with the help of \latex{2\times1} dominoes only an even number of squares can be tiled. After removing one corner of the chessboard \latex{{8\times8}-1 = 63} squares are left, which is an odd number, therefore it cannot be tiled by dominoes. Since originally the number of squares was even, in order to be left with an even number of squares also an even number of squares should be removed. (Figure 6)

The statements below, implied by the above, have the same meaning:

– If we remove an odd number of squares from the chessboard, the remaining board cannot be tiled by \latex{2\times1} dominoes.

– If the remaining board can be tiled by dominoes, then an even number of squares are left.

– An even number of remaining squares is a necessary condition for the remaining board to be able to be tiled by dominoes.

Solution (b)

After removing two adjacent squares from the top left corner of the chessboard the remaining board can be tiled by dominoes for example as seen in figure \latex{7}.

Solution (c)

If two arbitrary but adjacent squares of the chessboard are removed, then let us take a \latex{2\times8} rectangle in the chessboard according to figure \latex{8} (the longer side of this large rectangle should be perpendicular to the longer side of the rectangle created by the two squares removed). This rectangle can be tiled by dominoes for example so that the dominoes are parallel with the rectangle removed. The remaining whole rows or columns of the chessboard can be tiled by dominoes. So if two adjacent squares of the chessboard are removed, the remaining board can be tiled by dominoes.

The statements below, implied by the above, have the same meaning:

– If two adjacent squares of the chessboard are removed, the remaining board can be tiled by dominoes.
– The fact that two adjacent squares of the chessboard are removed implies that the remaining board can be tiled by dominoes.
– Removing two adjacent squares from the chessboard is a sufficient condition for the remaining board to be able to be tiled by dominoes.

Solution (d)

If the two top corners of the chessboard are removed, the remaining board can be tiled by dominoes for example as seen in figure \latex{9}.

The statements below, implied by the above, have the same meaning:

– It is not true that if the remaining board can be tiled by dominoes, then it is certain that two adjacent squares are removed.
– The fact that the remaining board can be tiled by dominoes does not imply that two adjacent squares are removed.
– Removing two adjacent squares from the chessboard is not a necessary condition for the remaining board to be able to be tiled by dominoes.

Therefore removing two adjacent squares from the chessboard is a sufficient but not necessary condition for the remaining board to be able to be tiled by dominoes.

Solution (e)

Let us try the tiling. We assume that it is not possible, but it seems difficult to prove, because the dominoes can be placed in many ways.

The following idea helps: one domino tile always covers one light and one dark coloured square on the chessboard, i.e. there should be the same number of light and dark coloured squares on the board part that can be tiled. Since the squares in the two opposite corners of the chessboard have the same colour, in figure \latex{10} there are \latex{30} light and \latex{32} dark coloured squares among the remaining squares, thus it cannot be tiled by dominoes.

The statements below, implied by the above, have the same meaning:

– If two squares with the same colour are removed, the remaining board cannot be tiled by dominoes.
– If the remaining board can be tiled by dominoes, then equal number of light and dark coloured squares is left.
– Equal number of light and dark coloured squares remaining is a necessary condition for the remaining board to be able to be tiled by dominoes.

Notes:

The idea above, where we solved the problem by colouring the squares, can also be useful when solving other exercises.
Another approach, which can also be used often: Let us place the \latex{8\times8} chessboard in a coordinate system so that its lower left corner is in the \latex{\left(0;0\right)} point of the coordinate system. Let the coordinates of a square be the coordinates of its lower left corner. The sum of the coordinates of all the squares of the chessboard is even, since every \latex{x}- and every \latex{y}-coordinate appears eight times in the sum. (Figure 11) If we remove the squares with the coordinates \latex{\left(0;0\right)} and \latex{\left(7;7\right)}, the sum of the coordinates of the remaining squares stays even. The coordinates of the two squares belonging to one domino are either \latex{(x; y)} and \latex{(x; y + 1)}, or \latex{(x; y)} and \latex{(x + 1; y)}. The sum of the four coordinates in both cases is \latex{2x + 2y + 1}, which is odd for sure. The remaining \latex{62} squares could be tiled by \latex{31} dominoes, since their number is odd; the sum of the coordinates of the \latex{31} dominoes is also odd. The tiling cannot be done, because the sum of the coordinates of the squares to tile is even, while the sum of the coordinates of the dominoes is odd.

Solution (f)

While removing two differently coloured squares in several different ways let us try to tile the remaining board. We assume that the tiling is always possible, but because of the large number of cases it seems to be difficult to prove.

After stringing the squares of the chessboard according to figure \latex{12} the chessboard can be tiled by dominoes along the string. If we remove two squares, the string is cut into two pieces, and since the squares removed are differently coloured, these pieces consist of even number of squares. So both pieces can be tiled by dominoes along the string.

The statements below, implied by the above, have the same meaning:

– If one dark and one light coloured square of the chessboard are removed, the remaining board can be tiled by dominoes.
– The fact that one dark and one light coloured square of the chessboard are removed implies that the remaining board can be tiled by dominoes.
– Removing one dark and one light coloured square of the chessboard is a sufficient condition for the remaining board to be able to be tiled by dominoes.

Items e) and f) together have the following meaning:

– By removing two squares of the chessboard the remaining board can be tiled by dominoes if and only if one dark and one light coloured square is removed.
– Removing one dark and one light coloured square of the chessboard is a necessary and sufficient condition for the remaining board (after removing two squares) to be able to be tiled by dominoes.

Note: In figure \latex{13} it can be seen that two light and two dark coloured squares of the chessboard can be removed so that the remaining board cannot be tiled by \latex{2\times1} dominoes.

Proof by giving an example, by giving a counter-example, by deduction. Proof by contradiction

Example 4

The following proposition can be read in the Biology course book: Every zebra is striped. “What does it imply?” – asked the teacher. Four students gave the answers below:

Adam: If an animal is a zebra, then it is striped.

Barbara: If an animal is striped, then it needs to be a zebra.
Celia: If an animal is striped, then it is possible that it is a zebra.
Dolly: If an animal is not striped, then it cannot be a zebra.

Let us decide whose answer is correct and why.

Solution (Adam)

Adam’s “If an animal is a zebra, then it is striped.” statement is true, because every zebra is striped.

The proposition of the course book implies Adam's proposition.

Solution (Barbara)

Barbara’s “If an animal is striped, then it needs to be a zebra.” statement is false, because there is an animal which is striped but not a zebra, for example a clownfish.

We verified that Barbara's statement was false with the help of a counterexample.

Solution (Celia)

Celia’s “If an animal is striped, then it is possible that it is a zebra.” statement is true, because for example a zebra is striped and is indeed a zebra.

We verified that Celia's statement was true with the help of an example.

Solution (Dolly)

Dolly’s “If an animal is not striped, then it cannot be a zebra.” statement is true; we are going to prove it as follows.

Let us assume that the statement is false.

I.e. there is an animal, which is not striped and is still a zebra, so there is a zebra which is not striped. Since every zebra is striped according to the Biology course book, this statement is false.

It is a contradiction, thus the original statement is true.

⯁ ⯁ ⯁

We verified the last statement of the example with the help of proof by contradiction. The steps of the proof by contradiction:

Let us assume the negative of the statement.
Let us draw conclusions.
We end up with a contradiction.
Thus the original statement is true.

Example 5

Which statements below are true and which are false?

There is a rhombus that is a rectangle.
There is a kite in which all the angles are of different measure.
If the diagonals of a quadrilateral are perpendicular, then it is a square.
If the diagonals of a quadrilateral are not of equal length, then it is not a rectangle.

Solution (a)

The statement “There is a rhombus that is a rectangle.” is true. (Figure 15)

It can be proven with the help of an example: a square is such a rhombus which is also a rectangle. A square is a rhombus, because all of its sides are of equal length, and it is also a rectangle, because all of its angles are of equal measure.

Solution (b)

The statement “There is a kite in which all the angles are of different measure.” is false.

It can be verified by deduction. A kite is an axially symmetric quadrilateral about one of its diagonals. So it has two opposite angles, which are of equal measure, therefore it is not possible that all its angles are of different measure.

Solution (c)

The statement “If the diagonals of a quadrilateral are perpendicular, then it is a square.” is false.

It can be verified by a counter-example. For example the kite, in which not all the sides are of equal length, is such a quadrilateral in which the diagonals are perpendicular to each other, but it still not a square.

Solution (d)

The statement “If the diagonals of a quadrilateral are not of equal length, then it is not a rectangle.” is true.

It can be verified by proof by contradiction.

Let us assume the negative of the statement. There is a quadrilateral in which the diagonals are not of equal length, and it is still a rectangle. About rectangles we know that the diagonals are of equal length, so this is not possible.

It is a contradiction, thus the statement is true.

Figure 16

Exercises

{{exercise_number}}. Write down the converse of the following statements, and decide whether they are true or false.

If it is raining in the town, then the road is wet.
If I leave home, then I lock the door.
If he is Winnie-the-Pooh, then he is a bear.
If I win the national competition of secondary schools, then I am admitted to the university.
If I have a ticket, I can see the play in the theatre.

{{exercise_number}}. Write down the converse of the following theorems, and decide whether they are true or false.

If a number is divisible by \latex{4}, then it is divisible by \latex{2}.
If a number is a terminating decimal, then it is a rational number.
If a triangle is a right-angled triangle, then the square of its longest side is equal to the sum of the squares of the other two sides.
If at least one of two numbers is \latex{0}, then the product of the two numbers is \latex{0}.

{{exercise_number}}. What type of condition is \latex{\bold{\color{bc0822}{B}}} of \latex{\bold{\color{356ab5}{A}}}? (Necessary but not sufficient; necessary and sufficient; sufficient but not necessary; not necessary and not sufficient.)

\latex{\bold{\color{356ab5}{A:}}} Human life,

\latex{\bold{\color{bc0822}{B:}}} air;

\latex{\bold{\color{256bb6}{A:}}} I can withdraw money from an ATM,

\latex{\bold{\color{bd0822}{B:}}} I have a debit card;

\latex{\bold{\color{256bb6}{A:}}} it is noon in London,

\latex{\bold{\color{bd0822}{B:}}} Big Ben chimes;

\latex{\bold{\color{256bb6}{A:}}} I will be admitted to the university,

\latex{\bold{\color{bd0822}{B:}}} I will pass my graduation exams;

\latex{\bold{\color{256bb6}{A:}}} Thomas wins the Olympic Games B: Thomas runs world record \latex{ 100 } \latex{ m } race,

\latex{\bold{\color{bd0822}{B:}}} Thomas runs world record in the Olympic final of \latex{ 100 } \latex{ m } race.

{{exercise_number}}. What type of condition is \latex{\bold{\color{bc0822}{B}}} of \latex{\bold{\color{356ab5}{A}}}? (Necessary but not sufficient; necessary and sufficient; sufficient but not necessary; not necessary and not sufficient.)

\latex{\bold{\color{256bb6}{A:}}} The sum of two numbers is even,

\latex{\bold{\color{bd0822}{B:}}} both numbers are even;

\latex{\bold{\color{256bb6}{A:}}} the difference of two numbers is \latex{\\} divisible by \latex{3},

\latex{\bold{\color{bd0822}{B:}}} the two numbers leave the same remainder when divided by \latex{3};

\latex{\bold{\color{256bb6}{A:}}} a quadrilateral is a square,

\latex{\bold{\color{bd0822}{B:}}} all sides of a quadrilateral are of equal length;

\latex{\bold{\color{256bb6}{A:}}} the product of two numbers is even,

\latex{\bold{\color{bd0822}{B:}}} one of the numbers is even, the other one is odd;

\latex{\bold{\color{256bb6}{A:}}} a quadrilateral is a parallelogram,

\latex{\bold{\color{bd0822}{B:}}} the diagonals of a quadrilateral are perpendicular to each other.

{{exercise_number}}. When rounding \latex{x} to the nearest thousand we get \latex{4000}. What type of conditions of it are the following?

\latex{3,400\lt {x} \leq 4,550}

\latex{3,890\leq {x} \lt 4,390}

\latex{3,500\leq {x} \lt 4,499}

\latex{3,500\leq {x} \lt 4,500}

\latex{3,999\lt {x} \lt 4,999}

\latex{3,501\leq {x} \leq 4,500}

{{exercise_number}}. \latex{10} exercises should be solved at an exam; each of them is worth \latex{10} points. The exam is failed if someone gets less than \latex{40}% of the total points, otherwise it is passed. Formulate necessary but not sufficient; sufficient but not necessary; necessary and sufficient; not necessary and not sufficient conditions for the exam to be passed.

{{exercise_number}}. Find necessary but not sufficient; sufficient but not necessary; necessary and sufficient conditions in everyday life and in Mathematics.

{{exercise_number}}. There is a snail resting on every square of a \latex{7\times7} chessboard. For an indication every snail crawls to an adjacent square (adjacent squares have a common side). Is it possible that after this there is again exactly one snail on every square?

{{exercise_number}}. We want to tile a \latex{7\times7} square by \latex{3\times1} triminoes (without overlaps and gaps) so that we can leave out \latex{2} squares at most. Find a necessary, a sufficient, a necessary and sufficient condition of tiling.

{{exercise_number}}. A rectangular forest can be seen in the figure; every point represents a tree. A woodpecker is sitting on the tree \latex{ A }, a robin is sitting on the tree \latex{ B }. Both of the birds fly to the nearest tree in the direction north-south or east-west in every minute. Is it possible that both of them are sitting on the same tree at one time? Find a necessary, a sufficient, a necessary and sufficient condition of the meeting with respect to the initial position of the birds.

BAÉ

{{exercise_number}}. Can the edges of a cube be numbered by numbers \latex{1;\, 2;\, 3;\, 4;\, 5;\, …;\, 11;\, 12} so that we get the same sums for every vertex when adding up the numbers written on the three edges starting from the vertex? Find a necessary condition for the conductibility of the correct numbering with respect to the numbers written on the edges. Is it also sufficient? Find a sufficient condition.

{{exercise_number}}. We coloured each of the \latex{36} squares of a \latex{6\times6} square yellow or blue. After that we have the right to change the colour of every square in any row or column at once (yellow to blue, blue to yellow). We can repeat this step any number of times. Is it possible after any of the initial situations to colour every square yellow? Try the initial situations shown in the figure.

Find a necessary, a sufficient, a necessary and sufficient condition with respect to the initial situation so that if it is satisfied, then following the correct steps it is possible to colour every square yellow.

Puzzle

We want to decide whether the following statement is true or false in the case of each of the four envelopes in the figure: If an envelope is stuck down, then there is a \latex{40} euro cent stamp on it. At least how many envelopes need to be turned round to decide? (If we can see the front of an envelope, we do not know whether it is stuck down; if we can see the back of it and it is turned down, then it is also stuck down.)

Mathematics 10.

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