Question

prove that√7 is an irrational number

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Answer 1

Step-by-step explanation:

let us assume that √7 be rational.

then it must in the form of p / q.

As definition of rational number says.. P is whole number q is non zero whole number.. And p and q is simplest ratio which is expressed.. That means there exists no prime factor common in p and q.

√7 = p / q

√7 x q = p

squaring on both sides

7q² = p² ------1.

p is divisible by 7

p = 7c [c is a positive integer] [squaring on both sides ]

p²= 49c²

subsitute p² in eqn(1) we get

7q² = 49 c²

q² = 7c²

q is divisble by 7

thus q and p have a common factor 7.

there is a contradiction to our assumption

as our assumsion p & q are co prime but it has a common factor.

so that √7 is an irrational.

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Answer 2

Answer:

$\mathcal{\underline{\blue{answer = \sqrt{7} \: is \: a \: irrational \: number}}}$

Step-by-step explanation:

Let us assume that root 7 is rational

$(0 \leqslant b < r)$

Then , and

p/q=root 7 ( p and q are the common factor)

$p = \sqrt{7}q$

Squaring both sides

p²=7q²

7 divides 7q²

7 divides p²

7 divides p

Now let

p=7m

squaring both sides

p²=49m²

p²=7q² So ,now

7q²=49m²

q²=7m²

7 Divides 7m²

7 Divides q²

7 Divides q

This means that 7 is the common factor of p and q ,This is the contradiction of our assumption ,This means that our assumption is wrong

$thus \: \sqrt{7} \: is \: a \: irrational \: number$