Question

prove that √7 is irrational along that 3+√7 is also irrational. pls write down all the steps . steps are d most important .


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Answer 1

Step-by-step explanation:

let √7 be an irrational no. in the form of p/q, here , p&q are co-prime integers, q not equal to zero.

so,

√7=p/q

squaring both sides

7=p^2/q^2

p^2=7q^2

therefore

7 is a factor of p^2 as well as p- - -1

as p is a factor of p

p=7r

p^2=49r^2

but

p^2=7q^2

so, 7q^2=49r^2

therefore

7 is a factor of q^2 as well as q- - -2

from 1&2

7 is a factor of both p&q

so p&q are not co-prime

so our assumption was wrong and

√7 is an irrational no.

let 3+√7 is an rational no.

3+√7=p/q here , p&q are co-prime integers, q not equal to zero.

so,

3+√7=p/q

√7=p/q-3=rational (rational- rational= rational)

but √7 is irrational

so irrational cannot be equal to rational.

this is not possible.

so our assumption was wrong

so

3+√7 is an irrational no.