Prove that each of the following numbers is irrational.
(i) √3 (ii) (2−√3)(iii) (3+√2) (iv) (2+√5)(v) (5+3√2) (vi) 3√7(vii) 3√5 (viii) (2−3√5)(ix) (√3+√5)
Prove that each of the following numbers is irrational.
(i) √3 (ii) (2−√3)(iii) (3+√2) (iv) (2+√5)(v) (5+3√2) (vi) 3√7(vii) 3√5 (viii) (2−3√5)(ix) (√3+√5)
Sol:
(i) Let 6–√6 =2–√2 X 3–√3 be rational.
Hence, 2–√2 , 3–√3 are both rational.
This contradicts the fact that 2–√2 , 3–√3 are irrational.
The contradiction arises by assuming 6–√6 is rational.
Hence, 6–√6 is irrational.
(ii) Let 2−3–√2−3 be rational.
Hence, 2 and 2−3–√2−3 are rational.
(2 – 2+3–√2+3) =3–√3 = rational [Therefore, Difference of two rational is rational]
This contradicts the fact that 3–√3 is irrational.
The contradiction arises by assuming 2−3–√2−3 is rational.
Hence, 2−3–√2−3 is irrational.
(iii) Let, 3+2–√3+2 be rational.
Hence, 3 and 3+2–√3+2 are rational.
3+2–√3+2 – 3 =2–√2 = rational [Therefore, Difference of two rational is rational]
This contradicts the fact that 2–√2 is irrational.
The contradiction arises by assuming 3 + 2–√2 is rational.
Hence, 3 + 2–√2 is irrational.
(iv) Let 2 + .VS be rational. Hence, 2 and .5 are rational. (2 + – 2 = 2 + f – 2 = f = rational [v Difference of two rational is rational] this contradicts the fact that VE is irrational. The contradiction arises by assuming 2 – is rational. Hence, 2 – is irrational.
(v) Let, 5 + 3f2 be rational. Hence, 5 and 5 + 3.V2 are rational. (5 + 3f – 5) = 3/2. = rational [v Difference of two rational is rational] x 30 = 1,/2 = rational Product of two rational is rational] This contradicts the fact that V2 is irrational. The contradiction arises by assuming 5 + 3f is rational. Hence, 5 3f is irrational.
(vi) Let 30 be rational. x 3f = 1/77 = rational [v Product of two rational is rational] This contradicts the fact that 1/7 is irrational. The contradiction arises by assuming 30 is rational. Hence, 30 are irrational.
(vii) Let 3 — be rational. 3 1 VS : – ,/5 3 x – = – = rational This contradicts the fact that 1 is irrational.
(viii) [Product of two rational is rational]
So, if is rational, then Nig is rational. 5 x 115 = f= rational Hence, 71/- is irrational.
[Product of two rational is rational]
3 The contradiction arises by assuming VS — is rational.
3 Hence, VS — is irrational
(viii) Let 2 — 3N5 be rational. Hence 2 and 2 — 3 f are rational. :. 2 — (2 — 3A/g = 2 — 2 + 3.‘,/g = 3.5 = rational [v Difference of two rational is rational] x 3 Vg = f = rational Product of two rational is rational] This contradicts the fact that -N5 is irrational. The contradiction arises by assuming 2 — 3, 5 is rational. Hence, 2 — 3, 73 is irrational.
(ix) Let 3–√+5–√3+5 be rational.
Therefore, 3–√+5–√3+5 = a, where a is rational
Therefore, 3–√3= a−5–√a−5…..(1)
On squaring both sides of equation (1), we get
3 = (a−5–√)2(a−5)2 = a2 + 5 — 2Vga a2 i 2 2a
This is impossible because right-hand side is rational, whereas the left-hand side is irrational.
This is a contradiction.
Hence, 3–√+5–√3+5 is irrational.
Sol:
(i) Let 6–√6 =2–√2 X 3–√3 be rational.
Hence, 2–√2 , 3–√3 are both rational.
This contradicts the fact that 2–√2 , 3–√3 are irrational.
The contradiction arises by assuming 6–√6 is rational.
Hence, 6–√6 is irrational.
(ii) Let 2−3–√2−3 be rational.
Hence, 2 and 2−3–√2−3 are rational.
(2 – 2+3–√2+3) =3–√3 = rational [Therefore, Difference of two rational is rational]
This contradicts the fact that 3–√3 is irrational.
The contradiction arises by assuming 2−3–√2−3 is rational.
Hence, 2−3–√2−3 is irrational.
(iii) Let, 3+2–√3+2 be rational.
Hence, 3 and 3+2–√3+2 are rational.
3+2–√3+2 – 3 =2–√2 = rational [Therefore, Difference of two rational is rational]
This contradicts the fact that 2–√2 is irrational.
The contradiction arises by assuming 3 + 2–√2 is rational.
Hence, 3 + 2–√2 is irrational.
(iv) Let 2 + .VS be rational. Hence, 2 and .5 are rational. (2 + – 2 = 2 + f – 2 = f = rational [v Difference of two rational is rational] this contradicts the fact that VE is irrational. The contradiction arises by assuming 2 – is rational. Hence, 2 – is irrational.
(v) Let, 5 + 3f2 be rational. Hence, 5 and 5 + 3.V2 are rational. (5 + 3f – 5) = 3/2. = rational [v Difference of two rational is rational] x 30 = 1,/2 = rational Product of two rational is rational] This contradicts the fact that V2 is irrational. The contradiction arises by assuming 5 + 3f is rational. Hence, 5 3f is irrational.
(vi) Let 30 be rational. x 3f = 1/77 = rational [v Product of two rational is rational] This contradicts the fact that 1/7 is irrational. The contradiction arises by assuming 30 is rational. Hence, 30 are irrational.
(vii) Let 3 — be rational. 3 1 VS : – ,/5 3 x – = – = rational This contradicts the fact that 1 is irrational.
(viii) [Product of two rational is rational]
So, if is rational, then Nig is rational. 5 x 115 = f= rational Hence, 71/- is irrational.
[Product of two rational is rational]
3 The contradiction arises by assuming VS — is rational.
3 Hence, VS — is irrational
(viii) Let 2 — 3N5 be rational. Hence 2 and 2 — 3 f are rational. :. 2 — (2 — 3A/g = 2 — 2 + 3.‘,/g = 3.5 = rational [v Difference of two rational is rational] x 3 Vg = f = rational Product of two rational is rational] This contradicts the fact that -N5 is irrational. The contradiction arises by assuming 2 — 3, 5 is rational. Hence, 2 — 3, 73 is irrational.
(ix) Let 3–√+5–√3+5 be rational.
Therefore, 3–√+5–√3+5 = a, where a is rational
Therefore, 3–√3= a−5–√a−5…..(1)
On squaring both sides of equation (1), we get
3 = (a−5–√)2(a−5)2 = a2 + 5 — 2Vga a2 i 2 2a
This is impossible because right-hand side is rational, whereas the left-hand side is irrational.
This is a contradiction.
Hence, 3–√+5–√3+5 is irrational.
