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

• November 5th 2012, 07:21 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.
• November 6th 2012, 01:06 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.
• November 6th 2012, 10:17 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.
• November 6th 2012, 03:37 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.
• November 7th 2012, 11:15 AM
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.
• November 7th 2012, 02:07 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.
• November 7th 2012, 02:23 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.
• November 7th 2012, 02:41 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
• November 7th 2012, 05:34 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.
• November 7th 2012, 06:33 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.
• November 11th 2012, 12:51 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?
• November 11th 2012, 12:59 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.

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

The $\boxed{\Sigma}$ on the toolbar gives the $$..$$ wrap.