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# Prove that any real number of the form a+bsqrt(2) is not equal to 0, where a and b are rational numbers not?

both 0. Let G = {a+bsqrt(2) | a,b rational, not both 0}. Prove that G is a group under ordinary multiplication.

Steps please!! I have no clue where to start.

### 2 Answers

- ?Lv 710 years agoFavorite Answer
ok, we want to show that if a,b ≠ 0, then a + b√2 ≠ 0.

let's try to prove this by contradiction, so we will suppose that a and b are not both 0.

but that a + b√2 = 0. so

(a + b√2)(a - b√2) = 0(a - b√2) = 0

a^2 + 2b^2 = 0.

now if one of a or b (or both) is non-zero, then a^2 + 2b^2 > 0,

a contradiction.

so let's see if we can prove G is a group.

first, we need to show closure, that is:

we need to show that (a + b√2)(c + d√2) is in G.

now (a + b√2)(c + d√2) = ac + ad√2 + bc√2 + 2bd

= (ac + 2bd) + (ad + bc)√2.

certainly ac + 2bd, ad + bc are both rational numbers,

but we need to show that at least one of these must be non-zero.

there are 4 cases to check:

a ≠ 0, c ≠ 0

a ≠ 0, d ≠ 0

b ≠ 0, c ≠ 0

b ≠ 0, d ≠ 0.

1) suppose a ≠ 0, c ≠ 0. then ac ≠ 0. now if ac + 2bd = 0,

ac = -2bd, so neither b nor d can be 0,

in which case ad ≠ 0 and bc ≠ 0. now the only way

ad + bc = 0 would be if ad = -bc. from:

ac = -2bd

ad = -bc, we get:

abc = -2b^2d

a^2d = -abc, so

a^2d = 2b^2d. now we know d ≠ 0, so we have:

a^2 = 2b^2, so (a/b)^2 = 2, so a/b = √2,

which is impossible, since that would mean √2 is rational.

so if ac + 2bd is 0, ad + bc cannot be 0.

2) if a ≠ 0, d ≠ 0, then ad ≠ 0. so if ad + bc = 0,

then ad = -bc, so that b ≠ 0, c ≠ 0. now if ac + 2bd = 0

as well, we proceed as in case 1, to show (a/b) = √2

which is again impossible, so if ad + bc = 0,

we cannot have ac + 2bc = 0 as well.

the other two cases are proved similarly.

this shows closure (which is actually the hardest part).

multiplication is associative, since multiplication is associative

in the real numbers (and all of these numbers are real).

alternatively: [(a + b√2)(c + d√2)](g + h√2) =

(ac + 2bd + (ad + bc)√2)(g + h√2)

= (ac+2bd)g + 2(ad+bc)h + [(ac+2bd)h + (ad+bc)g]√2

= acg + 2bdg + 2adh + 2bch + (ach + 2bdh + adg + bcg)√2

= a(cg+2dh) + 2b(ch+dg) + [a(ch+dg) + b(cg+2bh)]√2

= (a + b√2)(cg + 2dh + (ch + dg)√2) = (a + b√2)[(c + d√2)(g + h√2)].

clearly 1 = 1 + 0√2 serves as an identity.

so all that's left to show is the existence of inverses.

define (a + b√2)^-1 to be a/(a^2 + 2b^2) + (-b/(a^2 + 2b^2))√2,

it is a straight-forward calculation to prove

(a + b√2)(a/(a^2 + 2b^2) + (-b/(a^2 + 2b^2))√2) =

(a^2/(a^2+b^2) + 2b^2/(a^2 + 2b^2)) + (-ab/(a^2 + 2b^2) + ab/(a^2 + 2b^2))√2

= (a^2 + 2b^2)/(a^2 + 2b^2) + ((-ab + ab)/(a^ + 2b^2))√2

= 1 + 0√2 = 1.

thus, G is a group.

- bayleyLv 44 years ago
For any c, x in |R (0 + c)x = 0x + cx ... (distributive assets) cx = 0x + cx ... (identity assets) -cx + cx = -cx + (0x + cx) -cx + cx = -cx + (cx + 0x) ... (commutative assets) -cx + cx = (-cx + cx) + 0x ... (associative assets) 0 = 0 + 0x ... (inverse assets) 0x = 0 ... (identity assets)