If a+√ b=c+√ d, prove that a=c and b=d.

Suppose a, b, c, d are all rational number where √ b and √ d exist and are irrational. if If a+√ b=c+√ d, prove that a=c and b=d.

I have been thinking for a long time but I don't know where to start. The only thing i can think of is to try using proof by contradiction. I am really confused, please help.

Re: If a+√ b=c+√ d, prove that a=c and b=d.

Hey jwu.

Consider a + SQRT(b) = c + SQRT(d) then a - c = SQRT(b) - SQRT(d). Now if a and c are rational then a - c is also rational but if SQRT(b) - SQRT(d) is irrational (two numbers added or subtracted are irrational) then the only time when this can happen is when both sides are equal to 0.

Hence you get the result.

Re: If a+√ b=c+√ d, prove that a=c and b=d.

yes, but that proof lies subsumed under a proof that if b ≠ d, and √b, √d are irrational, √b - √d is irrational, which you have not proven.

Re: If a+√ b=c+√ d, prove that a=c and b=d.

When is the only case when both sides are rational? Also you have both numbers b and d being rational which you need to use.

The rationality means you can't get things like say (SQRT(2) + 1) - SQRT(2): this is a not a general statement but rather a statement where b and d are forced to be rational which has consequences on what the sums and differences are.

Re: If a+√ b=c+√ d, prove that a=c and b=d.

I respectfully suggest that a rigorous proof of the following statement hasn't yet been given:

Let b and d be rational numbers such that $\displaystyle \sqrt{b}$ and $\displaystyle \sqrt{d}$ are irrational. Then either $\displaystyle \sqrt{b} = \sqrt{d}$ or $\displaystyle \sqrt{b} - \sqrt{d}$ is irrational.

**Proof**: Suppose that $\displaystyle \sqrt{b} - \sqrt{d} = r \in \mathbb{Q}$ with $\displaystyle r \neq 0$ (If r = 0, then $\displaystyle \sqrt{b} = \sqrt{d}$). Then $\displaystyle \sqrt{b} = r + \sqrt{d}$. Squaring both sides and solving for $\displaystyle \sqrt{d}$, we get

$\displaystyle \sqrt{d} = \frac{-r^2 + b -d}{2r}$

which implies that $\displaystyle \sqrt{d} \in \mathbb{Q}$, contrary to the hypothesis. Therefore, $\displaystyle \sqrt{b} - \sqrt{d}$ is irrational.

Re: If a+√ b=c+√ d, prove that a=c and b=d.

Your problem is terribly stated. From what set do a, b, c, & d come from? Real numbers? Complex numbers? Rationals?

One could approach this problem from many different angles depending upon the "context" of the statements.

I would assume b and d are positive integers, having no common factors and square-free, a & c ordinary integers.

From there you can draw some elementary conclusions about a, b, c & d.

Re: If a+√ b=c+√ d, prove that a=c and b=d.

Quote:

Originally Posted by

**dave0147** Your problem is terribly stated. From what set do a, b, c, & d come from? Real numbers? Complex numbers? Rationals?

From the first post:

Quote:

Suppose a, b, c, d are all rational number where √ b and √ d exist and are irrational.

Seems properly stated to me.

Quote:

One could approach this problem from many different angles depending upon the "context" of the statements.

I would assume b and d are positive integers, having no common factors and square-free, a & c ordinary integers.

From there you can draw some elementary conclusions about a, b, c & d.

Re: If a+√ b=c+√ d, prove that a=c and b=d.

I'm sorry. I'm new here and didn't realize that the problem had been better stated in an earlier post.

I will present my solution once I have it in order.

Thanks

dave0147

Re: If a+√ b=c+√ d, prove that a=c and b=d.

Quote:

Originally Posted by

**dave0147** I'm sorry. I'm new here and didn't realize that the problem had been better stated in an earlier post.

I will present my solution once I have it in order.

Thanks

dave0147

Hey, no problem.

Re: If a+√ b=c+√ d, prove that a=c and b=d.

Quote:

Originally Posted by

**Petek** I respectfully suggest that a rigorous proof of the following statement hasn't yet been given:

Let b and d be rational numbers such that $\displaystyle \sqrt{b}$ and $\displaystyle \sqrt{d}$ are irrational. Then either $\displaystyle \sqrt{b} = \sqrt{d}$ or $\displaystyle \sqrt{b} - \sqrt{d}$ is irrational.

yes, this was the point i was trying to make earlier.

Quote:

**Proof**: Suppose that $\displaystyle \sqrt{b} - \sqrt{d} = r \in \mathbb{Q}$ with $\displaystyle r \neq 0$ (If r = 0, then $\displaystyle \sqrt{b} = \sqrt{d}$). Then $\displaystyle \sqrt{b} = r + \sqrt{d}$. Squaring both sides and solving for $\displaystyle \sqrt{d}$, we get

$\displaystyle \sqrt{d} = \frac{-r^2 + b -d}{2r}$

which implies that $\displaystyle \sqrt{d} \in \mathbb{Q}$, contrary to the hypothesis. Therefore, $\displaystyle \sqrt{b} - \sqrt{d}$ is irrational.

short, sweet and to the point.

Re: If a+√ b=c+√ d, prove that a=c and b=d.

Without loss of generality, one may assume that ALL rational numbers are fractions expressed in lowest terms.

Hence the set {(a-c), b^(1/2), -d^(1/2)} are linearly independent over the rational numbers (field), **Q**

Please excuse the clumsy notation (expressing square roots as ^(1/2)) but I am very new and haven't mastered the learning curve when it comes to using the notation of this platform)

Thus any linear combination A(a-c) + B(b^(1/2)) + C(-d^(1/2)) cannot equal zero unless each of A, B, & C are zero.

The original problem is: a+ b^(1/2) = c + d^(1/2) cannot be valid unless either A or (a - c) = 0 --> c = a as advertised! The rest follows easily.

Could someone please tell me where symbols like sqrt are?

Re: If a+√ b=c+√ d, prove that a=c and b=d.

Quote:

Originally Posted by

**dave0147** Could someone please tell me where symbols like sqrt are?

Here are some examples.

[tex] \cup [/tex] gives $\displaystyle \cup $ is union

[tex] \cap [/tex] gives $\displaystyle \cap $ is intersection

[tex] \subseteq [/tex] gives $\displaystyle \subseteq $ is subset

[tex] x^{\sqrt{x+1}} [/tex] gives $\displaystyle x^{\sqrt{x+1}} $

[tex] \in [/tex] gives $\displaystyle \in $ is element

[tex] \emptyset [/tex] gives $\displaystyle \emptyset $ is emptyset

[tex] \wedge [/tex] gives $\displaystyle \wedge $ , and

[tex] \vee [/tex] gives $\displaystyle \vee $ , or

[tex] \sqrt{x+1} [/tex] gives $\displaystyle \sqrt{x+1} $

[tex]x^{-i\theta}[/tex] gives $\displaystyle x^{-i\theta} $ instead of $\displaystyle x^-i \theta $.

The $\displaystyle \boxed{\Sigma}$ on the toolbar gives the [tex]..[/tex] wrap.