Prove that root 5 is irrational

If it is assumed that $\sqrt$ is irrational. Please let me know if my proof is correct, and how to use the above mentioned fact. This is from here: More generally, suppose $r =\sqrt$). If $\sqrt3+\sqrt5=x$ is rational, then $5=(x-\sqrt3)^2=x^2-2\sqrt3x+3$, and $$\sqrt3=\frac$$ But then $\sqrt3$ is also rational. However, we know that $\sqrt3$ is irrational, so $x$ is also irrational. And we know $\sqrt3$ is irrational by the usual argument: is $\sqrt3=p/q$, with coprime integers $p$ and $q$, then $p^2=3q^2$. Hence $3$ divides $p$, but then $9$ divides $p^2$ and $3$ divides $q$, so $p$ and $q$ are not coprime, contradiction. If $\sqrt 3 + \sqrt 5 = p/q; p,q \in \mathbb Z $ then... $\sqrt 5 = p/q - \sqrt 3$ $5 = (p/q)^2 - 2 (p/q)\sqrt 3 +3$ $2 (p/q)\sqrt 3 = (p/q)^2 - 2$ $\sqrt 3 = (p/2q) - (q/p) $ which is rational. Which is a contradiction as you had already proven that. ..... However if you hadn't... This is not the easiest but If $\sqrt 3 + \sqrt 5 = p/q $ $3 + 2\sqrt d$ But the LHS had an odd number of factors of 3 and 5, while the RHS has even (or 0) number of factors of 3 and 5 so they can't be equal. Hint: If $\sqrt.$$ Hence $3$ appears an odd number of times in the prime factorisation of $5,$ a contradiction. Hope this helps. P.S. I don't know what I was thinking in putting the old Hint I here. $\begingroup$ Both $5$ and $\frac$ we conclude that "3 appears an odd number of times in" the factorisation of $5.$ $\endgroup$

Statement B x 2 is a multiple of 5. Therefore x is a multiple of 5 (as 5 is a prime number). Assume Statement B to be true for now. We will prove it later. Using Statement B, we can write: x = 5a (for some integer a) Replacing x by 5a in Equation 1: 5 2a 2 = 5y 2 or 5y 2 = 5 2a 2 y 2 = 5a 2 y 2 is a multiple of 5. So y is a multiple of 5. We can write y = 5b, where b is some integer. Using x = 5a and y = 5b in Statement A: Statement C 5 ​ = y x ​ = 5 b 5 a ​ 5 is a common factor of x and y. In statement A we assumed there is no common factor of x and y. Statement C contradicts it. This means we cannot find integers x, y such that y x ​ is in the lowest form or such x, y do not exist. Therefore √5 cannot be represented as a rational number. You can use the same argument for √2. We can prove this for any prime number. ✩ Known Irrationals • Square Root of Prime ( P r im e ​ ): √2, √3, √5, √7, √11, √13, √17, √19 … • Special Numbers: Pi ( π ) , Euler’s number ( e ), Golden Ratio • log 23, log 35… x 2 is Multiple of Prime p ⇒ x is Multiple of Prime p Why does this proof work only for prime numbers? Where did we use the fact that p is a prime number? We used it in the Statement B. In the proof above, we used the fact that if x 2 is a multiple of a prime number p, then x is also a multiple of the same prime number. Let us see why this holds. Let us say x 2 is multiple of prime number p. We do a prime factorization of x: ✩ Irrational Result of Operations Following operations betwee...

I got stuck at : $a^2/b^2 = 12+2 \sqrt 35$ I understand that $12$ is rational and now I need to prove that $\sqrt$ It means that $c$ divide with $5$ and $7$? Also, how do I prove that if for example $X^2/4$ then $X/4$? Here's a different approach, assuming you know that $\sqrt5$ and/or $\sqrt7$ are irrational. If $\sqrt7+\sqrt5$ were rational, then $$$$ would be rational as well. You're on the right track. Consider the powers of $5$ that divide both sides of $c^2=35d^2$. You have an even number for the RHS but an odd number for the LHS. Indeed, if $5^m$ is the largest power of $5$ that divides $c$ and $5^n$ is the largest power of $5$ that divides $d$, then we get $2m=2n+1$, a contradiction. Assuming √5 + √7 is rational, √5 + √7 = a ∕ b Squaring both sides, 5 + 7 + 2√35 = a²/b² ⇒2√35 = a²/b² - 12 ⇒2√35 = (a²-12b²)/b² ⇒√35 = (a²-12b²)/2b² According to our assumption, (a²-12b²)/2b² should have been rational but actually it is irrational. Hence our assumption is wrong. Therefore, √5 + √7 is irrational

Let's assume that 7 √ 5 is a rational number. So, we can write this number as; 7 √ 5 = a b Here a and b are two co prime numbers and b ≠0. ⇒ √ 5 = a 7 b ----(i) R.H.S of equation (i) is rational number but we know that √ 5 is irrational number. It is not possible. That means our assumption is wrong. Therefore, 7 √ 5 is an irrational number.

Prove that 3+ √5 is an irrational number. Answer: 3+ √5 is an irrational number Let us see, how to solve. Explanation: Let us assume that 3 + √5 is a Now, 3 + √5 = a/b [Here a and b are co-prime numbers, where b ≠ 0] √5 = a/b - 3 √5 = (a - 3b)/b Here, should also be an irrational number. Hence, it is a contradiction to our assumption. Thus, 3 + √5 is an irrational number. Hence proved, 3 + √5 is an irrational number.

Games have rules and math has rules, so in that sense, math is like a game. If you follow the rules of math, you win. What do you win? You win a truth, which up till now, you have been calling “a correct answer” – but there is more. . . You already know many math rules, like keeping an equation balanced, or a negative number multiplied by another negative number results in a positive number etc. etc. . . . . There are lots of these kinds of procedural rules for playing with numbers. But there are other types of rules that will become more and more important as you get further into math and they have to do with the properties of numbers and consequences of these properties. The most amazing thing about these rules is that they, combined with other rules can lead you to a truth than no one has discovered yet, in other words, this rule based game of exploration can take you where no one has gone before – and it is that aspect which keeps us pushing the boundaries of mathematics. Many new discoveries are beautiful (fractal geometry), or useful (chaos theory), or weird (quantum mechanics) and some are even a bit disturbing (the n-body problem). Now, the path that leads to a truth in mathematics is called a proof. Guess what? You have been doing proofs all this time, right since you first started to add numbers up until now. For example, if you have a problem like 7x - 10 = 5x + 6, you can prove, using the rules of the game of math, that x can only be equal to 8 in order that 7x...

If you are not familiar with the rational-root theorem, you can still tackle the problem directly. Assume that \(\displaystyle \sqrt[3]\) If you then plug this back into the equation, you would find that 5 divides \(\displaystyle b\) as well, contradicting the assumption that \(\displaystyle a\) and \(\displaystyle b\) are coprime. Math Help Forum • Mathematics is concerned with numbers, data, quantity, structure, space, models, and change. Founded in 2005, Math Help Forum is dedicated to free math help and math discussions, and our math community welcomes students, teachers, educators, professors, mathematicians, engineers, and scientists.