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Can I prove it like this?

Proof

Case 1 Suppose $x=\sqrt2$ . Choose $y=\sqrt2-\sqrt3$ , $z=\sqrt3$

consider $y+z=(\sqrt2-\sqrt3)+\sqrt3=\sqrt2=x$

Case 2 Suppose $x\in\mathbb{R}$-{$\sqrt2$}. Choose $y=x-\sqrt2$, $z=\sqrt2$

consider $y+z=(x-\sqrt2)+\sqrt2=x$

Therefore ,For each real numbers can be expressed as sum of two irrational numbers.

Mathskid
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    Can you prove $x - \sqrt{2}$ is irrational in Case 2? For instance, what if $x = \sqrt{2} + 2$? – Dave L. Renfro Oct 22 '20 at 16:39
  • @DaveL.Renfro Thank you very much. – Mathskid Oct 22 '20 at 16:43
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    Consider different cases of whether $x$ is rational or irrational. – halrankard2 Oct 22 '20 at 16:51
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    My initial reaction was that this might require some cleverness to show when $x$ is irrational (when $x$ is rational it's easy, at least to me), so I gave it some thought just now, and I was embarrassed to find it's even easier when $x$ is irrational (if $x$ is irrational, then $x/2$ is irrational). Slightly harder, but still not very difficult, is to show that you can use different irrationals in forming the sum. – Dave L. Renfro Oct 22 '20 at 16:53
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    If $x$ is irrational you could use $x/3$ and $2x/3$ as well (or $\epsilon x$ and $\delta x$ for any nonzero rational $\epsilon$ and $\delta$ that sum to $1$). – halrankard2 Oct 22 '20 at 17:03
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    If $x$ is rational, by choosing any irrational $r$, both $,\frac12(x+r),$ and $,\frac12(x-r),$ are irrational, and we have $x=\frac12(x+r)+\frac12(x-r)$. – gpassante Oct 22 '20 at 20:21
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    Or also $x= (x+r)+(-r)$, etc. – gpassante Oct 22 '20 at 20:25

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