Question

prove that 2root3 plus root 5 is irrational​

Answer 1

Answer:

To Prove :  2√3 + √5 is irrational

we know that √3 & √5 are irrational numbers.

Now we use result which states that

Product of a rational and an irrational no. is an irrational no.

⇒ 2√3 is irrational no.

Now. according to closure property of irrational no.

Sum of two irrational no. is also an irrational no.

⇒ 2√3 + √5 is an irrational no.

Step-by-step explanation:

Your Answer

BRAINLIST

Answer 2

Answer:

To Prove

$2 \sqrt{3} + \sqrt{5}$

as irrational.

Proof

$let \: 2 \sqrt{3} + \sqrt{5} \: is \: a \: rational \: number \\ 2 \sqrt{3} + \sqrt{5} = \frac{p}{q} (where \: p \: and \: q \: are \: co - prime \: and \: q \: is \: not \: \\ equal \: to \: zero)$

squaring both sides

${(2 \sqrt{3} + \sqrt{5}) }^{2} = { (\frac{p}{q} )}^{2} \\ \\ 12 + 5 + 4 \sqrt{15} = \frac{ {p}^{2} }{ {q}^{2} } \\ \\ 17 + 4 \sqrt{15} = \frac{ {p}^{2} }{ {q}^{2} } \\ \\ 4 \sqrt{15} = \frac{ {p}^{2} }{ {q}^{2} } - 17 \\ \\ 4 \sqrt{15} = \frac{ {p}^{2} - 17 {q}^{2} }{ {q}^{2} } \\ \\ \sqrt{15} = \frac{ {p}^{2} - 17 {q}^{2} }{ 4{q}^{2} }$

• Here LHS is product of two irrational numbers is also irrational whereas RHS is a rational number.
• This is not possible.
• Our assumption is wrong.
• Therefore, it is an irrational number.

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