Does sqr(2) + sqr(6) = sqr(8) . . . prove it?

Answer 1

No
sqrt(2) + sqrt(6) = sqrt(8)
sqrt(2) + sqrt(2*3) = sqrt(2*4)
sqrt(2) + sqrt(2)sqrt(3) = 2 sqrt(2)

Divide both sides by sqrt(2)
1 + sqrt(3) = 2
Subtract both sides by 1:
sqrt(3) = 1
Square both sides:
3 = 1
Not true.

Answer 2

For nonsquares A,B,C can we have an equation (#):

sqrtA + sqrtB = sqrtC ? If so, then all three have a common

factor or else 4AB = (C - (A+B))^2 renders both A and B

perfect squares since (A,B)=1, a contradiction.


With the GCD = d, we set A = ad, B = bd, C = cd and

divide eqn(#) by sqrt(d) to get

sqrt(a) + sqrt(b) = sqrt(c) ($)

Again, 4ab = (c - (a+b))^2 and a and b are perfect squares

and by $, c is a perfect square. Setting a=u^2, b=v^2

gives c= (u+v)^2 and finally,

A = (d) u^2

B = (d) v^2

C= (d)(u + v)^2

is the general solution to eqn (#) and in the example provided

must have d=2, u = 1, yet v^2 = 3, a contradiction, so the

example fails.

Answer 3

No.

square both sides to get:
(sqr(2) + sqr(6) )(sqr(2) + sqr(6) ) = 8
=> 2 +2sqr(2)sqr(6) + 6 = 8
so 8 + a pos value = 8 a contradiction

This is a trivial problem.

Answer 4

Claim: sqrt(2) + sqrt(6) != sqrt(8)

( != meaning "not equal")

Proof (by contradiction); assume that they are equal; that is, assume

sqrt(2) + sqrt(6) = sqrt(8). Then, if we square both sides, we get

[ sqrt(2) + sqrt(6) ]^2 = 8

FOILing it out, we get

2 + 2sqrt(2)sqrt(6) + 6 = 8
2sqrt(2)sqrt(6) + 8 = 8
2sqrt(12) + 8 = 8
2sqrt(12) = 0
sqrt(12) = 0

If it's not obvious at this point, then squaring both sides again will make it obvious.

12 = 0

BUT, this contradicts the fact that 12 != 0.
Therefore, by contradiction, sqrt(2) + sqrt(6) != sqrt(8)

Answer 5

Hey there!

Here's the answer.

sqrt(2)+sqrt(6)=sqrt(8) --> Write the problem.
sqrt(2)+sqrt(2*3)=sqrt(8) --> Rewrite 6 as 2*3.
sqrt(2)+sqrt(2)sqrt(3)=sqrt(8) --> Use the formula sqrt(ab)=sqrt(a)*sqrt(b).
sqrt(2)(1+sqrt(3))=sqrt(8) --> Factor out sqrt(2).
sqrt(2)(1+sqrt(3))=sqrt(4*2) --> Rewrite 8 as 4*2.
sqrt(2)(1+sqrt(3))=sqrt(4)sqrt(2) --> Use the formula sqrt(ab)=sqrt(a)*sqrt(b).
1+sqrt(3)=sqrt(4) --> Divide sqrt(2) on both sides of the equation.
1+sqrt(3)=2 --> Rewrite sqrt(4) as 2.
sqrt(3)=1 --> Subtract 1 on both sides of the equation.
3=1 Square both sides of the equation.

Since 3 does not equal to 1, the statement is false.

Hope it helps!

Answer 6

No it doesn't. Just square both sides and you'll find that you have 8 on the right side of the equation and 8 + other stuff on the left side, so the two sides are not equal.