[tex]\huge \blue{ \bold { \pmb{ \fcolorbox{gray}{black} { {\ \ddag \: \ \red{QUESTION}\ \ddag \: }}} }}[/tex] Prove that √7 is an irrational number. About the author Iris
Answer: Given √7 To prove: √7 is an irrational number. Proof: Let us assume that √7 is a rational number. So it t can be expressed in the form p/q where p,q are co-prime integers and q≠0 √7 = p/q Here p and q are coprime numbers and q ≠ 0 Solving √7 = p/q On squaring both the side we get, => 7 = (p/q)2 => 7q2 = p2……………………………..(1) p2/7 = q2 So 7 divides p and p and p and q are multiple of 7. ⇒ p = 7m ⇒ p² = 49m² ………………………………..(2) From equations (1) and (2), we get, 7q² = 49m² ⇒ q² = 7m² ⇒ q² is a multiple of 7 ⇒ q is a multiple of 7 Hence, p,q have a common factor 7. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number √7 is an irrational number. Reply
To prove : [tex]\sf{\sqrt{7}}[/tex] is irrational. Proof : Assume that [tex]\sf{\sqrt{7}}[/tex] is rational. [tex]\sf{ \sqrt{7} = \dfrac{a}{b} \: – – – \gray{(a \: and \: b \: are \: coprime \: numbers)}}[/tex] [tex]\sf{\longrightarrow} \: \sqrt{7} = \dfrac{a}{b} [/tex] [tex]\sf{\longrightarrow} \: \sqrt{7}b = a[/tex] [tex]\sf{\longrightarrow} \: (\sqrt{7}b)^{2} = {a}^{2} – – \gray{(Squaring \: both \: the \: sides)}[/tex] [tex]\sf{\longrightarrow} \: {7b}^{2} = {a}^{2} – – -\gray{ (1)}[/tex] 7 and b² is a factor of a², which means 7 is also a factor of a. [tex]\sf{\longrightarrow} \:a = 7c – – -\gray{(2)}[/tex] Substitute Eq. 2 in Eq. 1 [tex]\sf{\longrightarrow} \: {7b}^{2} = {(7c)}^{2} [/tex] [tex]\sf{\longrightarrow} \:7 {b}^{2} = 49 {c}^{2} [/tex] [tex]\sf{\longrightarrow} \: {b}^{2} = {7c}^{2} [/tex] 7 and c² are factors of b². Which means 7 is a factor of b too. 7 is a common factor of a and b. But they were co prime numbers. This is a contradiction. The contradiction was arisen due to wrong assumption. Hence proved, [tex]\sf{\sqrt{7}}[/tex] is irrational. Reply
Answer:
Given √7
To prove: √7 is an irrational number.
Proof:
Let us assume that √7 is a rational number.
So it t can be expressed in the form p/q where p,q are co-prime integers and q≠0
√7 = p/q
Here p and q are coprime numbers and q ≠ 0
Solving
√7 = p/q
On squaring both the side we get,
=> 7 = (p/q)2
=> 7q2 = p2……………………………..(1)
p2/7 = q2
So 7 divides p and p and p and q are multiple of 7.
⇒ p = 7m
⇒ p² = 49m² ………………………………..(2)
From equations (1) and (2), we get,
7q² = 49m²
⇒ q² = 7m²
⇒ q² is a multiple of 7
⇒ q is a multiple of 7
Hence, p,q have a common factor 7. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number
√7 is an irrational number.
To prove : [tex]\sf{\sqrt{7}}[/tex] is irrational.
Proof :
Assume that [tex]\sf{\sqrt{7}}[/tex] is rational.
[tex]\sf{ \sqrt{7} = \dfrac{a}{b} \: – – – \gray{(a \: and \: b \: are \: coprime \: numbers)}}[/tex]
[tex]\sf{\longrightarrow} \: \sqrt{7} = \dfrac{a}{b} [/tex]
[tex]\sf{\longrightarrow} \: \sqrt{7}b = a[/tex]
[tex]\sf{\longrightarrow} \: (\sqrt{7}b)^{2} = {a}^{2} – – \gray{(Squaring \: both \: the \: sides)}[/tex]
[tex]\sf{\longrightarrow} \: {7b}^{2} = {a}^{2} – – -\gray{ (1)}[/tex]
7 and b² is a factor of a², which means 7 is also a factor of a.
[tex]\sf{\longrightarrow} \:a = 7c – – -\gray{(2)}[/tex]
Substitute Eq. 2 in Eq. 1
[tex]\sf{\longrightarrow} \: {7b}^{2} = {(7c)}^{2} [/tex]
[tex]\sf{\longrightarrow} \:7 {b}^{2} = 49 {c}^{2} [/tex]
[tex]\sf{\longrightarrow} \: {b}^{2} = {7c}^{2} [/tex]
7 and c² are factors of b². Which means 7 is a factor of b too.
7 is a common factor of a and b. But they were co prime numbers. This is a contradiction. The contradiction was arisen due to wrong assumption.
Hence proved, [tex]\sf{\sqrt{7}}[/tex] is irrational.