Question

# 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, FB. T, FC. F, TD.T, T

Verified
145.8k+ views
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.

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.