Question

a) Check whether 6n can end with the digit 0 for any natural number n.
b)Prove that 3 + 2√5 is irrational.​

Answer 1

$\huge\underline\mathrm{Solution}:-$

I) If the number 6n ends with the digit 0 then it should be divisible by 2 and we know that any number with unit place as 0 or 5 is divisible by 5.

So, Prime Factorisation of 6n = (2 × 3)n

Therefore, the prime factorization of 6n doesn't contain prime number 5.

Hence, it is clear that for any natural number n, 6n is not divisible by 5 and thus it proves that 6n can not end with digit 0 for any natural number n.

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ii) let us assume that 3+2√5 is rational.

Then we can find co prime x and y where (y ≠ 0)such that 3 + 2√5 = x/y⋅

Rearranging, we get,

2√5= x/y – 3

√5=1/2(x/y-3)

Since x and y are integers,

thus, 1/2(x/y-3) is a rational number.

Therefore, √5 is also a rational number. But this contradicts the fact that √5 is irrational.

So, we conclude that 3 + 2√5 is irrational.

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Answer 2

$\large\underline\mathfrak{Question\:1-}$

Check whether 6ⁿ can end with the digit 0 for any natural number n.

$\large\underline\mathfrak{Solution-}$

Prime factorisation of 6ⁿ = (2×3)ⁿ

If any number ends with 0, it should be divisible by 10 i.e the both 2 and 5 should be present in the prime factorisation.

Now in this case, we observe that 5 is missing in the prime factorisation of 6 which is a contradiction.

Therefore, 6ⁿ can't end with the digit 0 for any natural number n.

$\rule{200}2$

$\large\underline\mathfrak{Question\:2-}$

Prove that 3 + 2√5 is irrational.

$\large\underline\mathfrak{Solution-}$

Let 3 + 2√5 = a [ where a is a rational number ]

$\implies$ 2√5 = a - 3

$\implies$ √5 = $\dfrac{a-3}{2}$ ______(A)

Let √5 is a rational Number.

so it can be written in the form of p/q.

$\implies$ √5 = p/q

Squaring both sides,

$\implies$ (√5)² = p²/q²

$\implies$ 5 = p²/q²

$\implies$ 5q² = p² ________(1)

$\implies$ q² = p²/5

So, p² is divisible by 5, and therefore p is a factor or 5.

$\rule{100}3$

Let p = 5m

Squaring both sides,

$\implies$ p² = 25m² ________(2)

From (1) and (2),

5q² = 25m²

$\implies$ m² = 5q²/25

$\implies$ m² = q²/5

So, q² is divisible by 5 and therefore q is a factor of 5.

Here, we get, p and q both are the factors of 5. But it isn't possible. Hence, our supposition is wrong.

√5 is an irrational number.

$\rule{100}3$

Now, from (A), we see that, (a-3)/2 is a rational number.

An irrational number can't be equal to a rational number.

Hence, 3 + 2√5 is an irrational number.