Question

so that 7 is not a cube of rational number

Answer 1

Assume that cube rt 7 is rational.
Then (7)^(1/3) = a/b where a and b are integers and a/b is reduced to lowest terms.
Then a=b[7^(1/3)]
Since a is a multiple of b and a is an integer, b divides a.
Since b divides a, a = nb and n is an integer.
Therefore 7^(1/3) = a/b = nb/b, so a/b is not reduced to lowest terms.


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Answer 2

here you go ☆☆

▪$let \: \sqrt{7} \: \ \: \: be \: \: any \: rationl \: n.$

▪$\frac{a}{b} = \sqrt{7}$ here a and b are co prime...

▪$b \sqrt{7} = a$________●

▪$squaring \: both \: sides \: \: we \: get \\ {a}^{2} = 7 {b}^{2}$

▪$it \: means \: {a}^{2} \: is \: divisble \: by \: \: 7 \\ ▪also \: a\: is \: divisble \: by \: 7$

let no. be c

a = 7c _______1

put 1 in ●

▪${(7c)}^{2} = 7 {b}^{2}$

▪$49 {c}^{2} = 7 {b}^{2} \\ 7 {c}^{2} = {b}^{2}$

▪it means b^2 is divisible by c

▪ALSO b is divisible by c

▪but a and b are co prime...

▪our assumption us wrong
▪there is contradiction.....


$so \sqrt{7} \: is \: irrational \: no..$


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