Question 1: Prove that `sqrt5` is irrational.

**Answer:** Let us assume the contrary, i.e. `sqrt5` is rational.

Thus, there can be two integers a and b (b≠0) and a and b are coprime so that;

`sqrt5=a/b`

Or, `bsqrt5=a`

Squaring on both sides, we get;

`5b^2=a^2`

This means that a^{2} is divisible by 5 and hence a is also divisible by 5.

This contradicts our earlier assumption that a and b are coprime, because we have found 5 as at least one common factor of a and b.

This also contradicts our earlier assumption that `sqrt5` is irrational.

Hence, `sqrt5` is irrational.

Question 2: Prove that `3+2sqrt5` is irrational.

**Answer:** Let us assume to the contrary, i.e. `3+2sqrt5` is rational.

Thus, there can be two integers a and b (b≠0) and a and b are coprime so that;

`3+2sqrt5=a/b`

Or, `2sqrt5=a/b-3`

Since a and b are rational, so `a/b-3` is rational and hence, `2sqrt5` is rational.

But this contradicts the fact that `2sqrt5` is irrational.

This happened because of our faulty assumption.

Hence, `3+2sqrt5` is irrational

Question 3: Prove that following are irrationals:

(i) `1/sqrt2`

**Answer:** Let us assume to the contrary, i.e. `1/sqrt2` is rational.

Thus, there can be two integers a and b (b≠0) and a and b are coprime so that;

`1/sqrt2=a/b`

Or, `asqrt2=b`

Squaring on both sides, we get;

`2a^2=b^2`

This means that b^{2} is divisible by 2 and hence a is also divisible by 2.

This contradicts our earlier assumption that a and b are co-prime, because 2 is at least one common factor of a and b.

This also contradicts our earlier assumption that `1/sqrt2` is rational.

Hence, `1/sqrt2` is irrational proved.

(ii) `7sqrt5`

**Answer:** Let us assume to the contrary, i.e. `7sqrt5` is rational.

There can be two integers a and b (b≠0) and a and b are coprime, so that;

`7sqrt5=a/b`

Or, `a\xx7sqrt5=a`

Squaring on both sides, we get;

`245b^2=a^2`

This means that a^{2} is divisible by 245; which means that a is also divisible by 245.

This contradicts our earlier assumption that a and b are coprime, because 245 is at least one common factor of a and b.

This happened because of our faulty assumption and hence, `7sqrt5` is irrational proved.

(iii) `6+sqrt2`

**Answer:** Let us assume to the contrary, i.e. `6+sqrt2` is rational.

Thus, there can be two integers a and b (b≠0) and a and b are coprime so that;

`6+sqrt2=a/b`

Or, `sqrt2=a/b-6`

Since a and b are rational, so `a/b-6` is rational and hence, `sqrt2` is rational.

But this contradicts the fact that `sqrt2` is irrational.

This happened because of our faulty assumption.

Hence, `6+sqrt2` is irrational proved.

Question 1: Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating decimal expansion:

(i) `(13)/(3125)`, (ii) `(17)/(8)` (iii) `(64)/(455)` (iv) `(15)/(1600)` (v) `(29)/(343)` (vi) `(23)/(2^3\5^3)` (vii) `(129)/(2^2\5^7\7^5)` (viii) `(6)/(15)` (ix) `(35)/(50)` (x) `(77)/(210)`

**Answer:** (i), (ii), (iv), (vi), (viii) and (ix) are terminating decimal expansion.

(iii), (v), (vii) and (x) are non-terminating repeating decimal expansion.

Question 2: Write down the decimal expansion of those rational numbers in Question 1 which have terminating decimal expansion.

**Answer:**

(i) `(13)/(3125)=0.00416`

(ii) `(17)/(8)=2.125`

(iv) `(15)/(1600)=0.009375`

(vi) `(23)/(2^3\5^2)=0.115`

(viii) `(6)/(15)=0.4`

(ix) `(35)/(50)=0.7`

Question 3: The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational , and of the form p/q, what can you say about the prime factors of q?

(i) 43.123456789

**Answer:** This is a rational number, q has either 2 or 5 or both as prime factor.

(ii) 0.120120012000120000………..

**Answer:** This is an irrational number.

(iii) 43.123456789

**Answer:** This is a rational number. In this case, q has 2 or 5 or both as prime factor and has some other factor as well.

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