Answer :
Step-by-step explanation:
Let us assume to the contrary that
7
−
√
5
is rational number
i.e, we can find co-prime
a
&
b
where
b
≠
0
such that:-
⇒
7
−
√
5
=
a
b
⇒
7
−
a
b
=
√
5
by rearranging this equation we get
⇒
√
5
=
7
−
a
b
⇒
√
5
b
=
7
b
−
a
since
a
&
b
are integers,we get
7
b
−
a
b
is a rational number & so
√
5
is rational But, this contradicts the fact that
√
5
is irrational number.
∴
our assumption i.e,
7
−
√
5
is rational number is incorrect.
Hence,
7
−
√
5
is irrrational
Answer:
Proof that (\sqrt{5}) is irrational:Assumption: Assume that (\sqrt{5}) is rational. This means it can be expressed as a fraction of two integers ( \frac{a}{b} ) in simplest form, where (a) and (b) are coprime integers (i.e., their greatest common divisor is 1), and (b \neq 0).Equation: Since (\sqrt{5} = \frac{a}{b}), squaring both sides gives: [ 5 = \frac{a^2}{b^2} ] Multiplying both sides by (b^2) results in: [ 5b^2 = a^2 ]Implication: This equation implies that (a^2) is a multiple of 5. Consequently, (a) must also be a multiple of 5 (because the square of a non-multiple of 5 cannot be a multiple of 5). So we can write (a) as: [ a = 5k \quad \text{for some integer } k ]Substitution: Substituting (a = 5k) into the equation (5b^2 = a^2) gives: [ 5b^2 = (5k)^2 = 25k^2 ] Simplifying this, we get: [ b^2 = 5k^2 ]Implication: This equation implies that (b^2) is also a multiple of 5. Therefore, (b) must be a multiple of 5 as well.Contradiction: If both (a) and (b) are multiples of 5, then (a) and (b) are not coprime, which contradicts our initial assumption that (\frac{a}{b}) is in simplest form.Since our assumption that (\sqrt{5}) is rational leads to a contradiction, we conclude that (\sqrt{5}) must be irrational.Proof that (7\sqrt{5}) is irrational:Now that we have established that (\sqrt{5}) is irrational, we can easily prove that (7\sqrt{5}) is irrational.Assumption: Assume for contradiction that (7\sqrt{5}) is rational. This means it can be expressed as a fraction of two integers ( \frac{c}{d} ) in simplest form, where (c) and (d) are coprime integers, and (d \neq 0).Equation: Since (7\sqrt{5} = \frac{c}{d}), we can express (\sqrt{5}) as: [ \sqrt{5} = \frac{c}{7d} ]Implication: This implies that (\sqrt{5}) is a rational number because (\frac{c}{7d}) is a ratio of two integers.Contradiction: This contradicts our previous proof that (\sqrt{5}) is irrational.Therefore, our assumption that (7\sqrt{5}) is rational must be false, which means (7\sqrt{5}) is irrational.Thus, we have proved both that (\sqrt{5}) is irrational and that (7\sqrt{5}) is irrational.