# What is the cardinality of the set of all functions mapping $\{\sqrt{2},\sqrt{3}, \sqrt{5}, \sqrt{7}\}$ into the rational numbers?

Let $$T = \{\sqrt{2},\sqrt{3},\sqrt{5},\sqrt{7}\}$$ and $$\mathcal{S}$$ the set of all functions that maps $$T$$ into $$\mathbb{Q}.$$ What is the cardinality of $$\mathcal{S}$$?

So far I've been messing with the idea that we can take the subset $$T_0\subset T$$ of functions that square an element of $$T$$ and adds any other string of rational numbers to it. For example $$f(\sqrt{2})=(\sqrt{2})^2+a$$ where $$a \in \mathbb{Q}.$$ Since $$|T_0|$$ is countable and infinite $$|T_0|=\aleph_0.$$

I'm not really sure if this is going to get me anywhere.

• It is same as the cardinality of $\mathbb{Q}^{4}$, which is countable. – Seewoo Lee Nov 24 '18 at 5:15
• The particular values of your set are not important. Only that it has four elements. – badjohn Nov 24 '18 at 6:27

Could you answer the question of the cardinality of the set of all functions from $$\{ 1 \}$$ to $$\mathbb{Q}$$? I hope so, it's obviously the same as $$\mathbb{Q}$$ since any such function can be specified by a single value from $$\mathbb{Q}$$.

Now, how about the cardinality of the set of all functions from $$\{ 1, 2 \}$$ to $$\mathbb{Q}$$? Well, any such function can be specified by two values from $$\mathbb{Q}$$ so its cardinality is the same as $$\mathbb{Q} \times \mathbb{Q} = \mathbb{Q}^2$$. Do you know the cardinality of this set?

Let's jump a step to the set of functions from $$\{1, 2, 3, 4 \}$$ to $$\mathbb{Q}$$. As Seewoo says, this has cardinality $$\mathbb{Q} \times \mathbb{Q} \times \mathbb{Q} \times \mathbb{Q} = \mathbb{Q}^4$$. If you figured that $$\mathbb{Q}^2$$ is countable then you should see that this is as well.

Now your set: $$\{\sqrt{2},\sqrt{3}, \sqrt{5}, \sqrt{7}\}$$. Well there is a very simple bijection between it and my set $$\{1, 2, 3, 4 \}$$ which can be easily used to associate the maps from my set to $$\mathbb{Q}$$ with those from your set to $$\mathbb{Q}$$.

• I absolutely understand this. Thank you so much. I got so caught up in the values of the set and not how many elements are in the set. This is a new level of abstraction for me compared to what I've been exposed to so far at the undergraduate level. – Trevor Mason Nov 25 '18 at 0:36
• @TrevorMason. I expect that was the intention of whoever wrote the question. How about the set $\{ 0, 1, e, i, \pi \}$? – badjohn Nov 25 '18 at 8:42
• This should have cardinality $\mathbb{Q}^5?$ – Trevor Mason Nov 25 '18 at 21:39
• Yes, but what is that? – badjohn Nov 25 '18 at 22:12
• $\mathbb{Q}= \aleph_0?$ – Trevor Mason Nov 25 '18 at 22:13