Chapter
stringclasses 18
values | sentence_range
stringlengths 3
9
| Text
stringlengths 7
7.34k
|
|---|---|---|
1 | 18-21 | Solution x2 – 5x + 6 = 0 (given)
Appendix 1
PROOFS IN MATHEMATICS
Rationalised 2023-24
MATHEMATICS
188
⇒ (x – 3) (x – 2) = 0 (replacing an expression by an equal/equivalent expression)
⇒ x – 3 = 0 or x – 2 = 0 (from the established theorem ab = 0 ⇒ either a = 0 or
b = 0, for a, b in R)
⇒ x – 3 + 3 = 0 + 3 or x – 2 + 2 = 0 + 2 (adding equal quantities on either side of the
equation does not alter the nature of the
equation)
⇒ x + 0 = 3 or x + 0 = 2 (using the identity property of integers under addition)
⇒ x = 3 or x = 2 (using the identity property of integers under addition)
Hence, x2 – 5x + 6 = 0 implies x = 3 or x = 2 Explanation Let p be the given statement “x2 – 5x + 6 = 0” and q be the conclusion
statement “x = 3 or x = 2” From the statement p, we deduced the statement r : “(x – 3) (x – 2) = 0” by
replacing the expression x2 – 5x + 6 in the statement p by another expression (x – 3)
(x – 2) which is equal to x2 – 5x + 6 (i)There arise two questions:
How does the expression (x – 3) (x – 2) is equal to the expression x2 – 5x + 6 |
1 | 19-22 | Explanation Let p be the given statement “x2 – 5x + 6 = 0” and q be the conclusion
statement “x = 3 or x = 2” From the statement p, we deduced the statement r : “(x – 3) (x – 2) = 0” by
replacing the expression x2 – 5x + 6 in the statement p by another expression (x – 3)
(x – 2) which is equal to x2 – 5x + 6 (i)There arise two questions:
How does the expression (x – 3) (x – 2) is equal to the expression x2 – 5x + 6 (ii)
How can we replace an expression with another expression which is equal to
the former |
1 | 20-23 | From the statement p, we deduced the statement r : “(x – 3) (x – 2) = 0” by
replacing the expression x2 – 5x + 6 in the statement p by another expression (x – 3)
(x – 2) which is equal to x2 – 5x + 6 (i)There arise two questions:
How does the expression (x – 3) (x – 2) is equal to the expression x2 – 5x + 6 (ii)
How can we replace an expression with another expression which is equal to
the former The first one is proved in earlier classes by factorization, i |
1 | 21-24 | (i)There arise two questions:
How does the expression (x – 3) (x – 2) is equal to the expression x2 – 5x + 6 (ii)
How can we replace an expression with another expression which is equal to
the former The first one is proved in earlier classes by factorization, i e |
1 | 22-25 | (ii)
How can we replace an expression with another expression which is equal to
the former The first one is proved in earlier classes by factorization, i e ,
x2 – 5x + 6 = x2 – 3x – 2x + 6 = x (x – 3) –2 (x – 3) = (x – 3) (x – 2) |
1 | 23-26 | The first one is proved in earlier classes by factorization, i e ,
x2 – 5x + 6 = x2 – 3x – 2x + 6 = x (x – 3) –2 (x – 3) = (x – 3) (x – 2) The second one is by valid form of argumentation (rules of logic)
Next this statement r becomes premises or given and deduce the statement s
“ x – 3 = 0 or x – 2 = 0” and the reasons are given in the brackets |
1 | 24-27 | e ,
x2 – 5x + 6 = x2 – 3x – 2x + 6 = x (x – 3) –2 (x – 3) = (x – 3) (x – 2) The second one is by valid form of argumentation (rules of logic)
Next this statement r becomes premises or given and deduce the statement s
“ x – 3 = 0 or x – 2 = 0” and the reasons are given in the brackets This process continues till we reach the conclusion |
1 | 25-28 | ,
x2 – 5x + 6 = x2 – 3x – 2x + 6 = x (x – 3) –2 (x – 3) = (x – 3) (x – 2) The second one is by valid form of argumentation (rules of logic)
Next this statement r becomes premises or given and deduce the statement s
“ x – 3 = 0 or x – 2 = 0” and the reasons are given in the brackets This process continues till we reach the conclusion The symbolic equivalent of the argument is to prove by deduction that p ⇒ q
is true |
1 | 26-29 | The second one is by valid form of argumentation (rules of logic)
Next this statement r becomes premises or given and deduce the statement s
“ x – 3 = 0 or x – 2 = 0” and the reasons are given in the brackets This process continues till we reach the conclusion The symbolic equivalent of the argument is to prove by deduction that p ⇒ q
is true Starting with p, we deduce p ⇒ r ⇒ s ⇒ … ⇒ q |
1 | 27-30 | This process continues till we reach the conclusion The symbolic equivalent of the argument is to prove by deduction that p ⇒ q
is true Starting with p, we deduce p ⇒ r ⇒ s ⇒ … ⇒ q This implies that “p ⇒ q” is true |
1 | 28-31 | The symbolic equivalent of the argument is to prove by deduction that p ⇒ q
is true Starting with p, we deduce p ⇒ r ⇒ s ⇒ … ⇒ q This implies that “p ⇒ q” is true Example 2 Prove that the function f : R →
→
→
→
→ R
defined by f (x) = 2x + 5 is one-one |
1 | 29-32 | Starting with p, we deduce p ⇒ r ⇒ s ⇒ … ⇒ q This implies that “p ⇒ q” is true Example 2 Prove that the function f : R →
→
→
→
→ R
defined by f (x) = 2x + 5 is one-one Solution Note that a function f is one-one if
f (x1) = f (x2) ⇒ x1 = x2 (definition of one-one function)
Now, given that
f (x1) = f (x2), i |
1 | 30-33 | This implies that “p ⇒ q” is true Example 2 Prove that the function f : R →
→
→
→
→ R
defined by f (x) = 2x + 5 is one-one Solution Note that a function f is one-one if
f (x1) = f (x2) ⇒ x1 = x2 (definition of one-one function)
Now, given that
f (x1) = f (x2), i e |
1 | 31-34 | Example 2 Prove that the function f : R →
→
→
→
→ R
defined by f (x) = 2x + 5 is one-one Solution Note that a function f is one-one if
f (x1) = f (x2) ⇒ x1 = x2 (definition of one-one function)
Now, given that
f (x1) = f (x2), i e , 2x1+ 5 = 2x2 + 5
⇒
2x1+ 5 – 5 = 2x2 + 5 – 5 (adding the same quantity on both sides)
Rationalised 2023-24
PROOFS IN MATHEMATICS
189
⇒
2x1+ 0 = 2x2 + 0
⇒
2x1 = 2x2 (using additive identity of real number)
⇒
1
2
2 x =
2
2
2 x (dividing by the same non zero quantity)
⇒
x1 = x2
Hence, the given function is one-one |
1 | 32-35 | Solution Note that a function f is one-one if
f (x1) = f (x2) ⇒ x1 = x2 (definition of one-one function)
Now, given that
f (x1) = f (x2), i e , 2x1+ 5 = 2x2 + 5
⇒
2x1+ 5 – 5 = 2x2 + 5 – 5 (adding the same quantity on both sides)
Rationalised 2023-24
PROOFS IN MATHEMATICS
189
⇒
2x1+ 0 = 2x2 + 0
⇒
2x1 = 2x2 (using additive identity of real number)
⇒
1
2
2 x =
2
2
2 x (dividing by the same non zero quantity)
⇒
x1 = x2
Hence, the given function is one-one (ii) Mathematical Induction
Mathematical induction, is a strategy, of proving a proposition which is deductive in
nature |
1 | 33-36 | e , 2x1+ 5 = 2x2 + 5
⇒
2x1+ 5 – 5 = 2x2 + 5 – 5 (adding the same quantity on both sides)
Rationalised 2023-24
PROOFS IN MATHEMATICS
189
⇒
2x1+ 0 = 2x2 + 0
⇒
2x1 = 2x2 (using additive identity of real number)
⇒
1
2
2 x =
2
2
2 x (dividing by the same non zero quantity)
⇒
x1 = x2
Hence, the given function is one-one (ii) Mathematical Induction
Mathematical induction, is a strategy, of proving a proposition which is deductive in
nature The whole basis of proof of this method depends on the following axiom:
For a given subset S of N, if
(i)
the natural number 1 ∈ S and
(ii)
the natural number k + 1 ∈ S whenever k ∈ S, then S = N |
1 | 34-37 | , 2x1+ 5 = 2x2 + 5
⇒
2x1+ 5 – 5 = 2x2 + 5 – 5 (adding the same quantity on both sides)
Rationalised 2023-24
PROOFS IN MATHEMATICS
189
⇒
2x1+ 0 = 2x2 + 0
⇒
2x1 = 2x2 (using additive identity of real number)
⇒
1
2
2 x =
2
2
2 x (dividing by the same non zero quantity)
⇒
x1 = x2
Hence, the given function is one-one (ii) Mathematical Induction
Mathematical induction, is a strategy, of proving a proposition which is deductive in
nature The whole basis of proof of this method depends on the following axiom:
For a given subset S of N, if
(i)
the natural number 1 ∈ S and
(ii)
the natural number k + 1 ∈ S whenever k ∈ S, then S = N According to the principle of mathematical induction, if a statement “S(n) is true
for n = 1” (or for some starting point j), and if “S(n) is true for n = k” implies that “S(n)
is true for n = k + 1” (whatever integer k ≥ j may be), then the statement is true for any
positive integer n, for all n ≥ j |
1 | 35-38 | (ii) Mathematical Induction
Mathematical induction, is a strategy, of proving a proposition which is deductive in
nature The whole basis of proof of this method depends on the following axiom:
For a given subset S of N, if
(i)
the natural number 1 ∈ S and
(ii)
the natural number k + 1 ∈ S whenever k ∈ S, then S = N According to the principle of mathematical induction, if a statement “S(n) is true
for n = 1” (or for some starting point j), and if “S(n) is true for n = k” implies that “S(n)
is true for n = k + 1” (whatever integer k ≥ j may be), then the statement is true for any
positive integer n, for all n ≥ j We now consider some examples |
1 | 36-39 | The whole basis of proof of this method depends on the following axiom:
For a given subset S of N, if
(i)
the natural number 1 ∈ S and
(ii)
the natural number k + 1 ∈ S whenever k ∈ S, then S = N According to the principle of mathematical induction, if a statement “S(n) is true
for n = 1” (or for some starting point j), and if “S(n) is true for n = k” implies that “S(n)
is true for n = k + 1” (whatever integer k ≥ j may be), then the statement is true for any
positive integer n, for all n ≥ j We now consider some examples Example 3 Show that if
A =
cos
sin
sin
cos
θ
θ
−
θ
θ
, then An =
cos
sin
sin
cos
n
n
n
n
θ
θ
−
θ
θ
Solution We have
P(n) : An =
cos
sin
sin
cos
n
n
n
n
θ
θ
−
θ
θ
We note that
P(1) : A1 =
cos
sin
sin
cos
θ
θ
−
θ
θ
Therefore, P(1) is true |
1 | 37-40 | According to the principle of mathematical induction, if a statement “S(n) is true
for n = 1” (or for some starting point j), and if “S(n) is true for n = k” implies that “S(n)
is true for n = k + 1” (whatever integer k ≥ j may be), then the statement is true for any
positive integer n, for all n ≥ j We now consider some examples Example 3 Show that if
A =
cos
sin
sin
cos
θ
θ
−
θ
θ
, then An =
cos
sin
sin
cos
n
n
n
n
θ
θ
−
θ
θ
Solution We have
P(n) : An =
cos
sin
sin
cos
n
n
n
n
θ
θ
−
θ
θ
We note that
P(1) : A1 =
cos
sin
sin
cos
θ
θ
−
θ
θ
Therefore, P(1) is true Assume that P(k) is true, i |
1 | 38-41 | We now consider some examples Example 3 Show that if
A =
cos
sin
sin
cos
θ
θ
−
θ
θ
, then An =
cos
sin
sin
cos
n
n
n
n
θ
θ
−
θ
θ
Solution We have
P(n) : An =
cos
sin
sin
cos
n
n
n
n
θ
θ
−
θ
θ
We note that
P(1) : A1 =
cos
sin
sin
cos
θ
θ
−
θ
θ
Therefore, P(1) is true Assume that P(k) is true, i e |
1 | 39-42 | Example 3 Show that if
A =
cos
sin
sin
cos
θ
θ
−
θ
θ
, then An =
cos
sin
sin
cos
n
n
n
n
θ
θ
−
θ
θ
Solution We have
P(n) : An =
cos
sin
sin
cos
n
n
n
n
θ
θ
−
θ
θ
We note that
P(1) : A1 =
cos
sin
sin
cos
θ
θ
−
θ
θ
Therefore, P(1) is true Assume that P(k) is true, i e ,
P(k) : Ak =
cos
sin
sin
cos
k
k
k
k
θ
θ
−
θ
θ
Rationalised 2023-24
MATHEMATICS
190
We want to prove that P(k + 1) is true whenever P(k) is true, i |
1 | 40-43 | Assume that P(k) is true, i e ,
P(k) : Ak =
cos
sin
sin
cos
k
k
k
k
θ
θ
−
θ
θ
Rationalised 2023-24
MATHEMATICS
190
We want to prove that P(k + 1) is true whenever P(k) is true, i e |
1 | 41-44 | e ,
P(k) : Ak =
cos
sin
sin
cos
k
k
k
k
θ
θ
−
θ
θ
Rationalised 2023-24
MATHEMATICS
190
We want to prove that P(k + 1) is true whenever P(k) is true, i e ,
P(k + 1) : Ak+1 =
cos (
1)
sin (
1)
sin(
1)
cos (
1 )
k
k
k
k
+
θ
+
θ
−
+
θ
+
θ
Now
Ak+1 = Ak |
1 | 42-45 | ,
P(k) : Ak =
cos
sin
sin
cos
k
k
k
k
θ
θ
−
θ
θ
Rationalised 2023-24
MATHEMATICS
190
We want to prove that P(k + 1) is true whenever P(k) is true, i e ,
P(k + 1) : Ak+1 =
cos (
1)
sin (
1)
sin(
1)
cos (
1 )
k
k
k
k
+
θ
+
θ
−
+
θ
+
θ
Now
Ak+1 = Ak A
Since P(k) is true, we have
Ak+1 =
cos
sin
sin
cos
k
k
k
k
θ
θ
−
θ
θ
cos
sin
sin
cos
θ
θ
−
θ
θ
=
cos
cos
sin
sin
cos
sin
sin
cos
sin
cos
cos
sin
sin
sin
cos
cos
k
k
k
k
k
k
k
k
θ
θ −
θ
θ
θ
θ +
θ
θ
−
θ
θ −
θ
θ
−
θ
θ +
θ
θ
(by matrix multiplication)
=
cos (
1)
sin (
1)
sin(
1)
cos (
1 )
k
k
k
k
+
θ
+
θ
−
+
θ
+
θ
Thus, P(k + 1) is true whenever P(k) is true |
1 | 43-46 | e ,
P(k + 1) : Ak+1 =
cos (
1)
sin (
1)
sin(
1)
cos (
1 )
k
k
k
k
+
θ
+
θ
−
+
θ
+
θ
Now
Ak+1 = Ak A
Since P(k) is true, we have
Ak+1 =
cos
sin
sin
cos
k
k
k
k
θ
θ
−
θ
θ
cos
sin
sin
cos
θ
θ
−
θ
θ
=
cos
cos
sin
sin
cos
sin
sin
cos
sin
cos
cos
sin
sin
sin
cos
cos
k
k
k
k
k
k
k
k
θ
θ −
θ
θ
θ
θ +
θ
θ
−
θ
θ −
θ
θ
−
θ
θ +
θ
θ
(by matrix multiplication)
=
cos (
1)
sin (
1)
sin(
1)
cos (
1 )
k
k
k
k
+
θ
+
θ
−
+
θ
+
θ
Thus, P(k + 1) is true whenever P(k) is true Hence, P(n) is true for all n ≥ 1 (by the principle of mathematical induction) |
1 | 44-47 | ,
P(k + 1) : Ak+1 =
cos (
1)
sin (
1)
sin(
1)
cos (
1 )
k
k
k
k
+
θ
+
θ
−
+
θ
+
θ
Now
Ak+1 = Ak A
Since P(k) is true, we have
Ak+1 =
cos
sin
sin
cos
k
k
k
k
θ
θ
−
θ
θ
cos
sin
sin
cos
θ
θ
−
θ
θ
=
cos
cos
sin
sin
cos
sin
sin
cos
sin
cos
cos
sin
sin
sin
cos
cos
k
k
k
k
k
k
k
k
θ
θ −
θ
θ
θ
θ +
θ
θ
−
θ
θ −
θ
θ
−
θ
θ +
θ
θ
(by matrix multiplication)
=
cos (
1)
sin (
1)
sin(
1)
cos (
1 )
k
k
k
k
+
θ
+
θ
−
+
θ
+
θ
Thus, P(k + 1) is true whenever P(k) is true Hence, P(n) is true for all n ≥ 1 (by the principle of mathematical induction) (iii) Proof by cases or by exhaustion
This method of proving a statement p ⇒ q is possible only when p can be split into
several cases, r, s, t (say) so that p = r ∨ s ∨ t (where “∨ ” is the symbol for “OR”) |
1 | 45-48 | A
Since P(k) is true, we have
Ak+1 =
cos
sin
sin
cos
k
k
k
k
θ
θ
−
θ
θ
cos
sin
sin
cos
θ
θ
−
θ
θ
=
cos
cos
sin
sin
cos
sin
sin
cos
sin
cos
cos
sin
sin
sin
cos
cos
k
k
k
k
k
k
k
k
θ
θ −
θ
θ
θ
θ +
θ
θ
−
θ
θ −
θ
θ
−
θ
θ +
θ
θ
(by matrix multiplication)
=
cos (
1)
sin (
1)
sin(
1)
cos (
1 )
k
k
k
k
+
θ
+
θ
−
+
θ
+
θ
Thus, P(k + 1) is true whenever P(k) is true Hence, P(n) is true for all n ≥ 1 (by the principle of mathematical induction) (iii) Proof by cases or by exhaustion
This method of proving a statement p ⇒ q is possible only when p can be split into
several cases, r, s, t (say) so that p = r ∨ s ∨ t (where “∨ ” is the symbol for “OR”) If the conditionals
r ⇒ q;
s ⇒ q;
and
t ⇒ q
are proved, then (r ∨ s ∨ t) ⇒ q, is proved and so p ⇒ q is proved |
1 | 46-49 | Hence, P(n) is true for all n ≥ 1 (by the principle of mathematical induction) (iii) Proof by cases or by exhaustion
This method of proving a statement p ⇒ q is possible only when p can be split into
several cases, r, s, t (say) so that p = r ∨ s ∨ t (where “∨ ” is the symbol for “OR”) If the conditionals
r ⇒ q;
s ⇒ q;
and
t ⇒ q
are proved, then (r ∨ s ∨ t) ⇒ q, is proved and so p ⇒ q is proved The method consists of examining every possible case of the hypothesis |
1 | 47-50 | (iii) Proof by cases or by exhaustion
This method of proving a statement p ⇒ q is possible only when p can be split into
several cases, r, s, t (say) so that p = r ∨ s ∨ t (where “∨ ” is the symbol for “OR”) If the conditionals
r ⇒ q;
s ⇒ q;
and
t ⇒ q
are proved, then (r ∨ s ∨ t) ⇒ q, is proved and so p ⇒ q is proved The method consists of examining every possible case of the hypothesis It is
practically convenient only when the number of possible cases are few |
1 | 48-51 | If the conditionals
r ⇒ q;
s ⇒ q;
and
t ⇒ q
are proved, then (r ∨ s ∨ t) ⇒ q, is proved and so p ⇒ q is proved The method consists of examining every possible case of the hypothesis It is
practically convenient only when the number of possible cases are few Example 4 Show that in any triangle ABC,
a = b cos C + c cos B
Solution Let p be the statement “ABC is any triangle” and q be the statement
“a = b cos C + c cos B”
Let ABC be a triangle |
1 | 49-52 | The method consists of examining every possible case of the hypothesis It is
practically convenient only when the number of possible cases are few Example 4 Show that in any triangle ABC,
a = b cos C + c cos B
Solution Let p be the statement “ABC is any triangle” and q be the statement
“a = b cos C + c cos B”
Let ABC be a triangle From A draw AD a perpendicular to BC (BC produced if
necessary) |
1 | 50-53 | It is
practically convenient only when the number of possible cases are few Example 4 Show that in any triangle ABC,
a = b cos C + c cos B
Solution Let p be the statement “ABC is any triangle” and q be the statement
“a = b cos C + c cos B”
Let ABC be a triangle From A draw AD a perpendicular to BC (BC produced if
necessary) As we know that any triangle has to be either acute or obtuse or right angled, we
can split p into three statements r, s and t, where
Rationalised 2023-24
PROOFS IN MATHEMATICS
191
r : ABC is an acute angled triangle with ∠ C is acute |
1 | 51-54 | Example 4 Show that in any triangle ABC,
a = b cos C + c cos B
Solution Let p be the statement “ABC is any triangle” and q be the statement
“a = b cos C + c cos B”
Let ABC be a triangle From A draw AD a perpendicular to BC (BC produced if
necessary) As we know that any triangle has to be either acute or obtuse or right angled, we
can split p into three statements r, s and t, where
Rationalised 2023-24
PROOFS IN MATHEMATICS
191
r : ABC is an acute angled triangle with ∠ C is acute s : ABC is an obtuse angled triangle with ∠ C is obtuse |
1 | 52-55 | From A draw AD a perpendicular to BC (BC produced if
necessary) As we know that any triangle has to be either acute or obtuse or right angled, we
can split p into three statements r, s and t, where
Rationalised 2023-24
PROOFS IN MATHEMATICS
191
r : ABC is an acute angled triangle with ∠ C is acute s : ABC is an obtuse angled triangle with ∠ C is obtuse t : ABC is a right angled triangle with ∠ C is right angle |
1 | 53-56 | As we know that any triangle has to be either acute or obtuse or right angled, we
can split p into three statements r, s and t, where
Rationalised 2023-24
PROOFS IN MATHEMATICS
191
r : ABC is an acute angled triangle with ∠ C is acute s : ABC is an obtuse angled triangle with ∠ C is obtuse t : ABC is a right angled triangle with ∠ C is right angle Hence, we prove the theorem by three cases |
1 | 54-57 | s : ABC is an obtuse angled triangle with ∠ C is obtuse t : ABC is a right angled triangle with ∠ C is right angle Hence, we prove the theorem by three cases Case (i) When ∠ C is acute (Fig |
1 | 55-58 | t : ABC is a right angled triangle with ∠ C is right angle Hence, we prove the theorem by three cases Case (i) When ∠ C is acute (Fig A1 |
1 | 56-59 | Hence, we prove the theorem by three cases Case (i) When ∠ C is acute (Fig A1 1) |
1 | 57-60 | Case (i) When ∠ C is acute (Fig A1 1) From the right angled triangle ADB,
BD
AB = cos B
i |
1 | 58-61 | A1 1) From the right angled triangle ADB,
BD
AB = cos B
i e |
1 | 59-62 | 1) From the right angled triangle ADB,
BD
AB = cos B
i e BD = AB cos B
= c cos B
From the right angled triangle ADC,
CD
AC = cos C
i |
1 | 60-63 | From the right angled triangle ADB,
BD
AB = cos B
i e BD = AB cos B
= c cos B
From the right angled triangle ADC,
CD
AC = cos C
i e |
1 | 61-64 | e BD = AB cos B
= c cos B
From the right angled triangle ADC,
CD
AC = cos C
i e CD = AC cos C
= b cos C
Now
a = BD + CD
= c cos B + b cos C |
1 | 62-65 | BD = AB cos B
= c cos B
From the right angled triangle ADC,
CD
AC = cos C
i e CD = AC cos C
= b cos C
Now
a = BD + CD
= c cos B + b cos C (1)
Case (ii) When ∠ C is obtuse (Fig A1 |
1 | 63-66 | e CD = AC cos C
= b cos C
Now
a = BD + CD
= c cos B + b cos C (1)
Case (ii) When ∠ C is obtuse (Fig A1 2) |
1 | 64-67 | CD = AC cos C
= b cos C
Now
a = BD + CD
= c cos B + b cos C (1)
Case (ii) When ∠ C is obtuse (Fig A1 2) From the right angled triangle ADB,
BD
AB = cos B
i |
1 | 65-68 | (1)
Case (ii) When ∠ C is obtuse (Fig A1 2) From the right angled triangle ADB,
BD
AB = cos B
i e |
1 | 66-69 | 2) From the right angled triangle ADB,
BD
AB = cos B
i e BD = AB cos B
= c cos B
From the right angled triangle ADC,
CD
AC = cos ∠ ACD
= cos (180° – C)
= – cos C
i |
1 | 67-70 | From the right angled triangle ADB,
BD
AB = cos B
i e BD = AB cos B
= c cos B
From the right angled triangle ADC,
CD
AC = cos ∠ ACD
= cos (180° – C)
= – cos C
i e |
1 | 68-71 | e BD = AB cos B
= c cos B
From the right angled triangle ADC,
CD
AC = cos ∠ ACD
= cos (180° – C)
= – cos C
i e CD = – AC cos C
= – b cos C
Fig A1 |
1 | 69-72 | BD = AB cos B
= c cos B
From the right angled triangle ADC,
CD
AC = cos ∠ ACD
= cos (180° – C)
= – cos C
i e CD = – AC cos C
= – b cos C
Fig A1 1
Fig A1 |
1 | 70-73 | e CD = – AC cos C
= – b cos C
Fig A1 1
Fig A1 2
Rationalised 2023-24
MATHEMATICS
192
Now
a = BC = BD – CD
i |
1 | 71-74 | CD = – AC cos C
= – b cos C
Fig A1 1
Fig A1 2
Rationalised 2023-24
MATHEMATICS
192
Now
a = BC = BD – CD
i e |
1 | 72-75 | 1
Fig A1 2
Rationalised 2023-24
MATHEMATICS
192
Now
a = BC = BD – CD
i e a = c cos B – ( – b cos C)
a = c cos B + b cos C |
1 | 73-76 | 2
Rationalised 2023-24
MATHEMATICS
192
Now
a = BC = BD – CD
i e a = c cos B – ( – b cos C)
a = c cos B + b cos C (2)
Case (iii) When ∠ C is a right angle (Fig A1 |
1 | 74-77 | e a = c cos B – ( – b cos C)
a = c cos B + b cos C (2)
Case (iii) When ∠ C is a right angle (Fig A1 3) |
1 | 75-78 | a = c cos B – ( – b cos C)
a = c cos B + b cos C (2)
Case (iii) When ∠ C is a right angle (Fig A1 3) From the right angled triangle ACB,
BC
AB = cos B
i |
1 | 76-79 | (2)
Case (iii) When ∠ C is a right angle (Fig A1 3) From the right angled triangle ACB,
BC
AB = cos B
i e |
1 | 77-80 | 3) From the right angled triangle ACB,
BC
AB = cos B
i e BC = AB cos B
a = c cos B,
and
b cos C = b cos 900 = 0 |
1 | 78-81 | From the right angled triangle ACB,
BC
AB = cos B
i e BC = AB cos B
a = c cos B,
and
b cos C = b cos 900 = 0 Thus, we may write
a = 0 + c cos B
= b cos C + c cos B |
1 | 79-82 | e BC = AB cos B
a = c cos B,
and
b cos C = b cos 900 = 0 Thus, we may write
a = 0 + c cos B
= b cos C + c cos B (3)
From (1), (2) and (3) |
1 | 80-83 | BC = AB cos B
a = c cos B,
and
b cos C = b cos 900 = 0 Thus, we may write
a = 0 + c cos B
= b cos C + c cos B (3)
From (1), (2) and (3) We assert that for any triangle ABC,
a = b cos C + c cos B
By case (i), r ⇒ q is proved |
1 | 81-84 | Thus, we may write
a = 0 + c cos B
= b cos C + c cos B (3)
From (1), (2) and (3) We assert that for any triangle ABC,
a = b cos C + c cos B
By case (i), r ⇒ q is proved By case (ii), s ⇒ q is proved |
1 | 82-85 | (3)
From (1), (2) and (3) We assert that for any triangle ABC,
a = b cos C + c cos B
By case (i), r ⇒ q is proved By case (ii), s ⇒ q is proved By case (iii), t ⇒ q is proved |
1 | 83-86 | We assert that for any triangle ABC,
a = b cos C + c cos B
By case (i), r ⇒ q is proved By case (ii), s ⇒ q is proved By case (iii), t ⇒ q is proved Hence, from the proof by cases, (r ∨ s ∨ t) ⇒ q is proved, i |
1 | 84-87 | By case (ii), s ⇒ q is proved By case (iii), t ⇒ q is proved Hence, from the proof by cases, (r ∨ s ∨ t) ⇒ q is proved, i e |
1 | 85-88 | By case (iii), t ⇒ q is proved Hence, from the proof by cases, (r ∨ s ∨ t) ⇒ q is proved, i e , p ⇒ q is proved |
1 | 86-89 | Hence, from the proof by cases, (r ∨ s ∨ t) ⇒ q is proved, i e , p ⇒ q is proved Indirect Proof Instead of proving the given proposition directly, we establish the proof
of the proposition through proving a proposition which is equivalent to the given
proposition |
1 | 87-90 | e , p ⇒ q is proved Indirect Proof Instead of proving the given proposition directly, we establish the proof
of the proposition through proving a proposition which is equivalent to the given
proposition (i)
Proof by contradiction (Reductio Ad Absurdum) : Here, we start with the
assumption that the given statement is false |
1 | 88-91 | , p ⇒ q is proved Indirect Proof Instead of proving the given proposition directly, we establish the proof
of the proposition through proving a proposition which is equivalent to the given
proposition (i)
Proof by contradiction (Reductio Ad Absurdum) : Here, we start with the
assumption that the given statement is false By rules of logic, we arrive at a
conclusion contradicting the assumption and hence it is inferred that the assumption
is wrong and hence the given statement is true |
1 | 89-92 | Indirect Proof Instead of proving the given proposition directly, we establish the proof
of the proposition through proving a proposition which is equivalent to the given
proposition (i)
Proof by contradiction (Reductio Ad Absurdum) : Here, we start with the
assumption that the given statement is false By rules of logic, we arrive at a
conclusion contradicting the assumption and hence it is inferred that the assumption
is wrong and hence the given statement is true Let us illustrate this method by an example |
1 | 90-93 | (i)
Proof by contradiction (Reductio Ad Absurdum) : Here, we start with the
assumption that the given statement is false By rules of logic, we arrive at a
conclusion contradicting the assumption and hence it is inferred that the assumption
is wrong and hence the given statement is true Let us illustrate this method by an example Example 5 Show that the set of all prime numbers is infinite |
1 | 91-94 | By rules of logic, we arrive at a
conclusion contradicting the assumption and hence it is inferred that the assumption
is wrong and hence the given statement is true Let us illustrate this method by an example Example 5 Show that the set of all prime numbers is infinite Solution Let P be the set of all prime numbers |
1 | 92-95 | Let us illustrate this method by an example Example 5 Show that the set of all prime numbers is infinite Solution Let P be the set of all prime numbers We take the negation of the statement
“the set of all prime numbers is infinite”, i |
1 | 93-96 | Example 5 Show that the set of all prime numbers is infinite Solution Let P be the set of all prime numbers We take the negation of the statement
“the set of all prime numbers is infinite”, i e |
1 | 94-97 | Solution Let P be the set of all prime numbers We take the negation of the statement
“the set of all prime numbers is infinite”, i e , we assume the set of all prime numbers
to be finite |
1 | 95-98 | We take the negation of the statement
“the set of all prime numbers is infinite”, i e , we assume the set of all prime numbers
to be finite Hence, we can list all the prime numbers as P1, P2, P3, |
1 | 96-99 | e , we assume the set of all prime numbers
to be finite Hence, we can list all the prime numbers as P1, P2, P3, , Pk (say) |
1 | 97-100 | , we assume the set of all prime numbers
to be finite Hence, we can list all the prime numbers as P1, P2, P3, , Pk (say) Note
that we have assumed that there is no prime number other than P1, P2, P3, |
1 | 98-101 | Hence, we can list all the prime numbers as P1, P2, P3, , Pk (say) Note
that we have assumed that there is no prime number other than P1, P2, P3, , Pk |
1 | 99-102 | , Pk (say) Note
that we have assumed that there is no prime number other than P1, P2, P3, , Pk Now consider N = (P1 P2 P3…Pk) + 1 |
1 | 100-103 | Note
that we have assumed that there is no prime number other than P1, P2, P3, , Pk Now consider N = (P1 P2 P3…Pk) + 1 (1)
N is not in the list as N is larger than any of the numbers in the list |
1 | 101-104 | , Pk Now consider N = (P1 P2 P3…Pk) + 1 (1)
N is not in the list as N is larger than any of the numbers in the list N is either prime or composite |
1 | 102-105 | Now consider N = (P1 P2 P3…Pk) + 1 (1)
N is not in the list as N is larger than any of the numbers in the list N is either prime or composite Fig A1 |
1 | 103-106 | (1)
N is not in the list as N is larger than any of the numbers in the list N is either prime or composite Fig A1 3
Rationalised 2023-24
PROOFS IN MATHEMATICS
193
If N is a prime, then by (1), there exists a prime number which is not listed |
1 | 104-107 | N is either prime or composite Fig A1 3
Rationalised 2023-24
PROOFS IN MATHEMATICS
193
If N is a prime, then by (1), there exists a prime number which is not listed On the other hand, if N is composite, it should have a prime divisor |
1 | 105-108 | Fig A1 3
Rationalised 2023-24
PROOFS IN MATHEMATICS
193
If N is a prime, then by (1), there exists a prime number which is not listed On the other hand, if N is composite, it should have a prime divisor But none of the
numbers in the list can divide N, because they all leave the remainder 1 |
1 | 106-109 | 3
Rationalised 2023-24
PROOFS IN MATHEMATICS
193
If N is a prime, then by (1), there exists a prime number which is not listed On the other hand, if N is composite, it should have a prime divisor But none of the
numbers in the list can divide N, because they all leave the remainder 1 Hence, the
prime divisor should be other than the one in the list |
1 | 107-110 | On the other hand, if N is composite, it should have a prime divisor But none of the
numbers in the list can divide N, because they all leave the remainder 1 Hence, the
prime divisor should be other than the one in the list Thus, in both the cases whether N is a prime or a composite, we ended up with
contradiction to the fact that we have listed all the prime numbers |
1 | 108-111 | But none of the
numbers in the list can divide N, because they all leave the remainder 1 Hence, the
prime divisor should be other than the one in the list Thus, in both the cases whether N is a prime or a composite, we ended up with
contradiction to the fact that we have listed all the prime numbers Hence, our assumption that set of all prime numbers is finite is false |
1 | 109-112 | Hence, the
prime divisor should be other than the one in the list Thus, in both the cases whether N is a prime or a composite, we ended up with
contradiction to the fact that we have listed all the prime numbers Hence, our assumption that set of all prime numbers is finite is false Thus, the set of all prime numbers is infinite |
1 | 110-113 | Thus, in both the cases whether N is a prime or a composite, we ended up with
contradiction to the fact that we have listed all the prime numbers Hence, our assumption that set of all prime numbers is finite is false Thus, the set of all prime numbers is infinite ANote Observe that the above proof also uses the method of proof by cases |
1 | 111-114 | Hence, our assumption that set of all prime numbers is finite is false Thus, the set of all prime numbers is infinite ANote Observe that the above proof also uses the method of proof by cases (ii)
Proof by using contrapositive statement of the given statement
Instead of proving the conditional p ⇒ q, we prove its equivalent, i |
1 | 112-115 | Thus, the set of all prime numbers is infinite ANote Observe that the above proof also uses the method of proof by cases (ii)
Proof by using contrapositive statement of the given statement
Instead of proving the conditional p ⇒ q, we prove its equivalent, i e |
1 | 113-116 | ANote Observe that the above proof also uses the method of proof by cases (ii)
Proof by using contrapositive statement of the given statement
Instead of proving the conditional p ⇒ q, we prove its equivalent, i e ,
~ q ⇒ ~ p |
1 | 114-117 | (ii)
Proof by using contrapositive statement of the given statement
Instead of proving the conditional p ⇒ q, we prove its equivalent, i e ,
~ q ⇒ ~ p (students can verify) |
1 | 115-118 | e ,
~ q ⇒ ~ p (students can verify) The contrapositive of a conditional can be formed by interchanging the conclusion
and the hypothesis and negating both |
1 | 116-119 | ,
~ q ⇒ ~ p (students can verify) The contrapositive of a conditional can be formed by interchanging the conclusion
and the hypothesis and negating both Example 6 Prove that the function f : R →
→
→
→
→ R defined by f (x) = 2x + 5 is one-one |
1 | 117-120 | (students can verify) The contrapositive of a conditional can be formed by interchanging the conclusion
and the hypothesis and negating both Example 6 Prove that the function f : R →
→
→
→
→ R defined by f (x) = 2x + 5 is one-one Solution A function is one-one if f (x1) = f (x2) ⇒ x1 = x2 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.