Discrete math: Prove this number is an irrational number
Prove that the number
$$\sqrt{5^0} + \sqrt{5^1} + \sqrt{5^2} $$
is an irrational number. For this problem you cannot assume that any number is irrational to begin with. You cannot use prime factorization and your solution should include a lemma demonstrating that if $a^2$ is divisible by 5 then $a$ is divisible by 5.
I'm absolutely lost in regards to how to approach this problem, and I'm not sure what the textbook means by including a lemma demonstrating that if $a^2$ is divisible by 5 then $a$ is divisible by 5. Any hints or help would be appreciated.
All these solutions were super helpful, thank you so much!!
$\endgroup$ 45 Answers
$\begingroup$The question can be written as$${\sqrt{5^0} + \sqrt{5^1} + \sqrt{5^2}},$$which is equal to$${6 + \sqrt{5}}.$$Now, the question simplifies to proving that $\sqrt{5}$ is irrational (as 6 is rational, and rational + irrational = irrational).
Now you go about doing it the normal way of proving $\sqrt{p}$ is irrational where $p$ is any prime:
Assume that $\sqrt{5}$ is rational and equal to $\frac{a}{b}$ where $a$ and $b$ are coprime and $b$ is not equal to $0$.
This gives you$${\sqrt{5} = \frac{a}{b}}$$$${\implies 5 = \frac{a^2}{b^2}}$$$${\implies 5\cdot b^2 = a^2}$$Now $a^2$ is a multiple of $5$, and since if $a^2$ is a multiple of $5$, $a$ has to be a multiple of $5$, we can write $a = 5k$:$${5\cdot b^2 = (5k)^2}$$$${\implies b^2 = 5\cdot k^2}$$Applying the same logic to $b$, we get that even $b$ is a multiple of $5$! This is a contradiction as we assumed $a$ and $b$ were co-prime!
Hence, $\sqrt{5}$ has to be irrational!
This all relies on the fact that if $a^2$ is a multiple of $5$, then $a$ has to be a multiple of $5$
We prove the contrapositive: If $a$ is not a multiple of $5$, then $a^2$ is not a multiple of $5$ either.
Let $a \equiv k$ (mod $5$), where $k$ is not $0$ ($5$ does not divide $a$).
Therefore, $a = 5n + k$, where n is an integer.
Therefore, $a^2 = 25n^2 + 10k + k^2 = 5(5n^2 + 2k) + k^2$.
We can see that this will only be a multiple of $5$ if $k^2$ is a multiple of $5$.
By putting in all possible values of $k$ $(1,2,3,4)$, you can show that it is not a multiple of $5$.
Hence, the assumption that if $a^2$ is a multiple of $5$, then $a$ is a multiple of $5,$ holds true.
Hints:
Are any of the three numbers rational anyway? Remember that rational + irrational = irrational.
You need to demonstrate that $$q^2=5\lambda \implies q=5\mu$$ for some $\lambda,\mu$
Demonstrate this by proving the opposite is true (I.e. That if $q$ is not a multiple of $5$; $q^2$ won't be either). Suppose we have $q=5\mu+D$, where $D$ is not a multiple of $5$. Evaluating $q^2$, can you see that will never be a multiple of $5$? That is an example of proof by contrapositive, and will suffice as the lemma you seek.
$\endgroup$ $\begingroup$Hint:
You'd help yourself out a lot by evaluating $\sqrt{5^0}$ and $\sqrt{5^2}$. You might as well want to prove that $\sqrt{5} \in \mathbb{R - Q}$ because you're not supposed to take that as a fact as per the question, this would include the use of the lemma demonstrating that if $5 \mid a^2 \implies 5 \mid a$. Now assume that $\sqrt{5^0}+\sqrt{5^1}+\sqrt{5^2}$ is rational and prove by contradiction that the number indeed is irrational.
$\endgroup$ $\begingroup$The sum of a rational number and an irrational number is an irrational number (otherwise the difference of two rational numbers would be irrational, but that can't be) and $\sqrt{5^0}+\sqrt{5^2} = 6$ is rational, so the problem boils down to showing that $\sqrt 5$ is irrational. To prove that, assume the contrary, i.e., $\sqrt 5 =a/b,$ where $a$ and $b$ are relatively prime integers (if they weren't relatively prime, then the fraction could be reduced). Therefore $5b^2=a^2,$ so $5$ divides $a^2=a \times a$, but since $5$ is prime, this means $5$ divides $a,$ but then $5$ divides $b$, so our assumption that $\sqrt 5=a/b$ with $a$ and $b$ relatively prime integers has led to a contradiction and $\sqrt 5$ is irrational.
$\endgroup$ $\begingroup$Assume $A=\sqrt{5^0} + \sqrt{5^1} + \sqrt{5^2}$ is rational number.
We have: $$A=6+\sqrt{5}$$$$\Leftrightarrow A-6=\sqrt{5}$$
It is easy to see $LHS$ is rational number and $RHS$ is irrational number.
Prove: $\sqrt{5}$ is irrational number.
Assume $\sqrt{5}$ is rational number, then we can let $$\sqrt 5=\frac{a}{b}(a;b\in Z;b\ne 0)$$ where $(a;b)=1 (1)$
Or $$5=\frac{a^2}{b^2}\Leftrightarrow a^2=5b^2\Leftrightarrow a\vdots5 \text{(5 is prime number)}$$
$$\Rightarrow a^2\vdots 25\Rightarrow 5b^2\vdots 25 \Rightarrow b \vdots 5\Rightarrow (a;b)\ne 1 (2)$$
$(1);(2)$ we have $\sqrt{5}$ is irrational number.
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