p: Every square is a rectangle.
q: Every rhombus is a kite.
The truth values of $p\to q\ and\ p\leftrightarrow q$ are _____ and _____ respectively.
A. F, F
B. T, F
C. F, T
D.T, T
Answer
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Hint: Consider the two statements p and q. Prove that they both are true. if p and q are true then $p\to q$ also becomes true. If both are true $p\leftrightarrow q$ also becomes true i.e. p implies q and q implies p.
Complete step-by-step answer:
p: Every square is a rectangle.
(i) A rectangle is a quadrilateral with all 4 angles $90{}^\circ $.
From this definition, you can prove that the opposite sides are parallel and of the same lengths. A rectangle can be tall & thin, short & fat or all the sides can have the same length.
(ii) A square is a quadrilateral with all 4 angles right angles and all 4 sides of same length.
So a square is a special kind of rectangle, it is one where all the sides have the same length. Thus every square is a rectangle because it is a quadrilateral with all 4 angles right angles. However not every rectangle is a square, to be a square its sides must have the same length.
$\therefore $ p: Every square is a rectangle is true.
q: Every rhombus is a kite.
When all the angles are also $90{}^\circ $ the kite will be square. The sum of its side, opposite angles are equal. Therefore the rhombus is symmetrical about each of its diagonals. So a rhombus can become a kite. The difference between a rhombus and kite is that a kite doesn’t always have 4 equal sides or 2 pairs of parallel sides like rhombus.
$\therefore $ Every rhombus is a kite, but every kite is not a rhombus.
$\therefore $ q is true.
So we found out that both p and q are true.
$p\to q$ is T, it means that p implies q.
Also we know that $T\leftrightarrow T$ is true.
$\therefore $ $p\leftrightarrow q$ is true, it means that p implies q and q implies p.
Hence the truth values of $p\to q$ and $p\leftrightarrow q$ is T, T.
Note: If it was given,
p: Every rectangle is square.
q: Every kite is rhombus.
Then $p\to q$ would have been F.
As both the statements are wrong.
So, $p\leftrightarrow q$ is false.
So, the value of $p\to q$ and $p\leftrightarrow q$ becomes F, F or from the table of logical implication here p and q are T.
$\therefore $ $p\to q$ is T and $q\to p$is T i.e. case 1 is true.
Complete step-by-step answer:
p: Every square is a rectangle.
(i) A rectangle is a quadrilateral with all 4 angles $90{}^\circ $.
From this definition, you can prove that the opposite sides are parallel and of the same lengths. A rectangle can be tall & thin, short & fat or all the sides can have the same length.
(ii) A square is a quadrilateral with all 4 angles right angles and all 4 sides of same length.
So a square is a special kind of rectangle, it is one where all the sides have the same length. Thus every square is a rectangle because it is a quadrilateral with all 4 angles right angles. However not every rectangle is a square, to be a square its sides must have the same length.
$\therefore $ p: Every square is a rectangle is true.
q: Every rhombus is a kite.
When all the angles are also $90{}^\circ $ the kite will be square. The sum of its side, opposite angles are equal. Therefore the rhombus is symmetrical about each of its diagonals. So a rhombus can become a kite. The difference between a rhombus and kite is that a kite doesn’t always have 4 equal sides or 2 pairs of parallel sides like rhombus.
$\therefore $ Every rhombus is a kite, but every kite is not a rhombus.
$\therefore $ q is true.
So we found out that both p and q are true.
$p\to q$ is T, it means that p implies q.
Also we know that $T\leftrightarrow T$ is true.
$\therefore $ $p\leftrightarrow q$ is true, it means that p implies q and q implies p.
Hence the truth values of $p\to q$ and $p\leftrightarrow q$ is T, T.
| p | q | $p\to q$ | $q\to p$ |
| T | T | T | T |
| T | F | F | T |
| F | T | T | F |
| F | F | T | T |
Note: If it was given,
p: Every rectangle is square.
q: Every kite is rhombus.
Then $p\to q$ would have been F.
As both the statements are wrong.
So, $p\leftrightarrow q$ is false.
So, the value of $p\to q$ and $p\leftrightarrow q$ becomes F, F or from the table of logical implication here p and q are T.
$\therefore $ $p\to q$ is T and $q\to p$is T i.e. case 1 is true.
Last updated date: 19th Sep 2023
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