Showing that irrational numbers have the same cardinality as real numbers

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So I did this but I'm not sure if it is an actual proof because it is based on a hypothesis, and also I'm not sure if I understand the Continuum Hypothesis correctly.

Because $\boldsymbol{I} \subseteq \boldsymbol{R}$, l can have either the same or less cardinality than $\boldsymbol{R}$. By the Continuum Hypothesis, l can not have a lower cardinality than $\boldsymbol{R}$, so it must have the same cardinality.

Is this correct, and/or is there a more rigorous method?

Also, can I use the notation $\operatorname{card} A < \operatorname{card} B$?

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Use the fact that for infinite cardinals $$\kappa+\lambda=\max\{\kappa,\lambda\}$$

Now $$\mathbb{R}=\mathbb{Q}\cup I$$ Thus we have $$|\mathbb{R}|=\max\{|\mathbb{Q}|,|I|\}$$ and it follows $$|\mathbb{R}|=|I|$$