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Say, we have $a,b,c,d\in \mathbb{N}$ such that $a + \sqrt{b} = c + \sqrt{d}.$

Our teachers compare both sides and magically conclude that $a=c$ and $b=d.$ This is intuitve for complex numbers, but feels odd for irrational numbers like square roots. Please give a nice and formal proof of this, or help me understand it intuitively.

ryang
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Ahan Sasmal
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    The implication $$ a+\sqrt b = c + \sqrt d \implies a=c \text{ and } b = d $$ definitely doesn't hold in general. For instance, $$ 2 + \sqrt 4 = 4 + \sqrt 0 \text{ but } 2 \ne 0 \text{ and } 4 \ne 0 $$ The most relevant thing you're looking for is probably $$ a + b \sqrt k = c + d \sqrt k \implies a = c \text{ and } b = d $$ whenever $a,b,c,d,k$ are integers and $k$ is not a perfect square, which is perhaps more obvious. That follows as $$ a + b \sqrt k - \Big( c + d \sqrt k \Big) = (a-c) + (b-d) \sqrt k = 0 $$ – PrincessEev May 28 '25 at 07:11
  • Considering the counter-example @ryang posted, we probably need to assume that $\sqrt{b}$ and $\sqrt{d}$ are irrationals? – rtybase May 28 '25 at 07:37
  • The complex number $a+bi$ is irrational if $b\ne0$. $i=\sqrt{-1}$ is irrational. – Gerry Myerson May 28 '25 at 09:08

1 Answers1

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In the case $a = c$, we get to $b = d$ almost immediately, which proves your result.

Assuming $a \neq c$,

$a+\sqrt{b} = c + \sqrt{d}$

$\sqrt{b} = c - a + \sqrt{d}$

$b = (c - a + \sqrt{d})^2$

$b = (c - a)^2 + 2(c - a)\sqrt{d} + d$

$\frac{b - (c - a)^2 - d}{2(c - a)} = \sqrt{d}$ (n.b. this is only valid because $c-a \neq 0$)

so $\sqrt{d}$ is rational. As discussed in this other question, that's only possible when $\sqrt{d}$ is an integer and $d$ is a perfect square. Which implies $\sqrt{b}$ is an integer and $b$ is a perfect square.

This holds in cases like $2+\sqrt{4} = 4+\sqrt{0}$, the counterexample @PrincessEev posted in the comments. But if we can rule out either $b$ or $d$ being a perfect square, then yes, in all other cases we have $a = c$ and $b = d$.

Toph
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