Question

prove √7+√3 is irrational​

Answer 1

Answer:

Step-by-step explanation:

To prove :-

√3-√7 is irrational

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Lets assume that √3-√7 is rational.

√3-√7 = r , where r is a rational number.

Squaring both the sides ,

[√3-√7 ]² = r²

3 - 2√21 +7 = r²

10 - 2√21 = r²

- 2√21 = r² - 10

√21 = r² - 10 / -2

Here ,

RHS is purely rational .

But , on the other hand , LHS is irrational [ √ 21 ]

This is a contradiction.

Hence , our assumption was wrong.

Thus , √3-√7 is irrational.

Hope this Helps You !!!!

Answer 2

ANSWER✔

$\large\underline\bold{GIVEN,}$

$\sf\dashrightarrow \sqrt{7}+ \sqrt{3}$

$\large\underline\bold{TO\:PROVE,}$

$\rm{\boxed{\bf{\green{ \star\:\:\sqrt{7}+ \sqrt{3}\:IS\: IRRATIONAL\:NUMBER. \:\: \star}}}}$

$\large\underline\bold{SOLUTION,}$

$\sf\therefore assuming\:\sqrt{7}+ \sqrt{3} \:as \:a\:rational\:number.$

where,

$\sf\dashrightarrow \sqrt{7}+ \sqrt{3}= \dfrac{a}{b} \:---[ where\:a \:and\:b\:are\:coprime\:and\:integers\:and\: b \neq 0$

$\sf\implies \sqrt{7}+ \sqrt{3}= \dfrac{a}{b}$

$\sf\implies \sqrt{7} = \dfrac{a}{b}-\sqrt{3}$

$\sf\implies \sqrt{7} = \dfrac{a-\sqrt{3}b}{b}$

$\sf\therefore \sqrt{7}\:is\:an\:irrational\:number$

$\sf\therefore \dfrac{a-\sqrt{3}b}{b}\:is\:also\:an\:irrational\:number$

$\sf\therefore it\:contradicts\:the\:fact\:that\: \sqrt{7}+ \sqrt{3} \:is\:an\:irrational\:number.$

$\sf\therefore\red{these\: contradiction\:has\:occured\:due\:to\:our\:wrong\:assumption\:that\: \sqrt{7}+ \sqrt{3} \:as \:a\:rational\:number}$

HENCE,

$\large{\boxed{\bf{ \star\:\: \sqrt{7}+ \sqrt{3} \:IS\:AN\:IRRATIONAL\:NUMBER.\:\: \star}}}$

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