10th Maths 1.3
NCERT Class 10th solution of Exercise 1.1
NCERT Class 10th solution of Exercise 1.2
NCERT Class 10th solution of Exercise 1.4
Exercise 1.3
Q1. Prove that `sqrt5` is irrational.
Sol. :
Let `sqrt5` is a rational number. Then we can take `a` and `b` two co-prime non-zero integers.
Such that
`sqrt5=frac(a)(b)`, where `bne0`
`bsqrt5=a`
`5b^2=a^2`
Therefore, `5` divides `a^2` so we can take `a=5c`, where `c` is an integer.
`5b^2=(5c)^2`
`5b^2=25c^2`
`b^2=5c^2`
`b^2`, as well as `b`, also is divisible by `5`.
`a` and `b` both have at least one common factor.
But this is contradiction arisen because of our incorrect assumption that `sqrt5` is a rational number.
Thus, we conclude that `sqrt5` is irrational.
Proved.
Q2. Prove that `3+2sqrt5` is irrational.
Sol. :
Let `3+2sqrt5` is a rational number. Then we can take `a` and `b` two coprime nonzero integers.
Such that
`3+2sqrt5=frac(a)(b)`, where `bne0`
`2sqrt5=frac(a)(b)-3`
`sqrt5=frac(a)(2b)-frac(3)(2)`
Since `a` and `b` are two integers where `bne0`, so `(frac(a)(2b)-frac(3)(2))` is rational, so `sqrt5` is rational.
But this contradiction has arisen because of our incorrect assumption that `3+2sqrt5` is rational.
So, we conclude that `3+2sqrt5` is irrational.
Proved.
Q3. Prove that following are irrationals :
i) `frac(1)(sqrt2)` ii) `7sqrt5` iii) `6+sqrt2`
Sol. :
i) Let `frac(1)(sqrt2)` is a rational number.
Then we can take `a` and `b` two co-prime non-zero integers
such that
`frac(1)(sqrt2)=frac(a)(b)`, where `bne0`
`sqrt2=frac(b)(a)`,
since `b` and `a` are non-zero integers
`b/a` is a rational number and so `sqrt2` is also a rational number.
But it is a contradiction to the fact that `sqrt2` is an irrational number. It is arisen due to our incorrect assumption.
So, we conclude that `frac(1)(sqrt2)` is an irrational number.
Proved.
ii) Let `7sqrt5` is a rational number.
Then we can take `a` and `b` two co-prime non-zero integers
such that
`7sqrt5=frac(a)(b)`, where `bne0`
`sqrt5=frac(b)(7a)`,
since `a, b` and `7` are non-zero integers
`b/(7a)` is a rational number and so `sqrt5` is also a rational number.
But it is a contradiction to the fact that `sqrt5` is an irrational number. It is arisen due to our incorrect assumption.
So, we conclude that `7sqrt5` is an irrational number.
Proved.
iii) Let `6+sqrt2` is a rational number.
Then we can take `a` and `b` two co-prime non-zero integers
such that
`6+sqrt2=frac(a)(b)`, where `bne0`
`sqrt2=frac(b)(a)-6`,
since `a, b` and `6` are non-zero integers
`b/a-6` is a rational number and so `sqrt2` is also a rational number.
But it is a contradiction to the fact that `sqrt2` is an irrational number. It is arisen due to our incorrect assumption.
So, we conclude that `6+sqrt2` is an irrational number.
Proved.
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