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

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Nov 5th 2012, 08:21 PM
jwu

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.
Nov 6th 2012, 02:06 AM
chiro

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.
Nov 6th 2012, 11:17 AM
Deveno

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.
Nov 6th 2012, 04:37 PM
chiro

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.
Nov 7th 2012, 12:15 PM
Petek

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 $\sqrt{b}$ and $\sqrt{d}$ are irrational. Then either $\sqrt{b} = \sqrt{d}$ or $\sqrt{b} - \sqrt{d}$ is irrational.

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

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

which implies that $\sqrt{d} \in \mathbb{Q}$ , contrary to the hypothesis. Therefore, $\sqrt{b} - \sqrt{d}$ is irrational.
Nov 7th 2012, 03:07 PM
dave0147

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.
Nov 7th 2012, 03:23 PM
Petek

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.
Nov 7th 2012, 03:41 PM
dave0147

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
Nov 7th 2012, 06:34 PM
Petek

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.
Nov 7th 2012, 07:33 PM
Deveno

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 $\sqrt{b}$ and $\sqrt{d}$ are irrational. Then either $\sqrt{b} = \sqrt{d}$ or $\sqrt{b} - \sqrt{d}$ is irrational.

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

Quote:

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

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

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

short, sweet and to the point.
Nov 11th 2012, 01:51 PM
dave0147

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?
Nov 11th 2012, 01:59 PM
Plato

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 $\cup$ is union
[tex] \cap [/tex] gives $\cap$ is intersection
[tex] \subseteq [/tex] gives $\subseteq$ is subset
[tex] x^{\sqrt{x+1}} [/tex] gives $x^{\sqrt{x+1}}$
[tex] \in [/tex] gives $\in$ is element
[tex] \emptyset [/tex] gives $\emptyset$ is emptyset
[tex] \wedge [/tex] gives $\wedge$ , and
[tex] \vee [/tex] gives $\vee$ , or
[tex] \sqrt{x+1} [/tex] gives $\sqrt{x+1}$
[tex]x^{-i\theta}[/tex] gives $x^{-i\theta}$ instead of $x^-i \theta$ .

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