Anyone got a step by step approach to solve this, it aint that hard just checkin my work
Prove (There exists a real number x)(For all real numbers y)(4+6y-y^2<x)
and the 2nd one is
Prove sqrt(2) +sqrt(5) is irrational
the way i did this was by contradiction and i proved that sqrt(10) is irrational because via contradiction i said sqrt(10) i srational then for integers m, n swrt(10) = m/n and m and n are both not even integers, but as i worked it i showed that m and n were both even therefore it cantbe rational, was my approach correct?
thanks
Can you just say that for the 1st part though im sure you have to say it in more detail, i said somethinlike what u said i received 4 out of 12 pts out of it and i dunno why
What about this
Step 1: Assume y is an arbirary real number
Step 2: Then 4+6y^2-y^2 is a real number [Axiom (For all x)(x+y element of real #'s)
Step 3: Assume x is some real number... dunno how to finish it off
Show that there exists an such that for all .
Suppose otherwise, then for any there exists a such that:
,
or:
The quadratic formula can be used to find those 's that satisfy this, and these are:
This has real roots only if the discriminant or .
So if , there is no having the properties assumed, which is a contradiction, hence there exists an such that:
for all , and is one such.
RonL
Hello, ruprotein!
In #2, and are relative prime.
. . It does not involve even/odd properties.
Prove is irrational.
First, we will prove that is irrational.
Assume is rational.
Then: for relatively prime integers and
. . That is, is already reduced to lowest terms.
Square both sides: . [1]
Since is a multiple of 10, then is a multiple of 10: .
Substitute into [1]: .
Since is a multiple of 10, then is a multiple of 10: .
But this contradicts the fact that and are relatively prime.
. . Therefore, is irrational.
Assume is rational.
Then: . for relatively prime integers and
. . That is, is already reduced to lowest terms.
Square both sides: .
. . and we have: .
Hence: .
The right side is rational. .It is the product, difference and quotient of integers.
. . But we have shown that the left side is irrational.
We have reached a contradiction.
Hence, our assumption was incorrect.
. . Therefore: . is irrational.
[QUOTE=Soroban;33016]Hello, ruprotein!
In #2, and are relative prime.
. . It does not involve even/odd properties.
[size=3]
First, we will prove that is irrational.
Assume is rational.
Then: for relatively prime integers and
. . That is, is already reduced to lowest terms.
Square both sides: . [1]
Since is a multiple of 10, then is a multiple of 10: .
Substitute into [1]: . 10p)^2\;\;\Rightarrow\;\;10b^2\:=\:100p^2\;\;\Righ tarrow\;\;b^2\:=\:10p^2" alt="10b^2\:=\10p)^2\;\;\Rightarrow\;\;10b^2\:=\:100p^2\;\;\Righ tarrow\;\;b^2\:=\:10p^2" />
Since is a multiple of 10, then is a multiple of 10: .
But this contradicts the fact that and are relatively prime.
. . Therefore, is irrational.
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Here is the other proof of sqrt(10) is irrational:
Theorem. If nth root of a is not an integer , then in fact nth root of a is irrational.
sqrt(9) < sqrt(10)< sqrt(16)
3 < sqrt(10) < 4
So sqrt(10) is not an integer, hence it is irrational.