anonymous
  • anonymous
prove that sqrt(5) is irrational :)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
it's not a perfect square, so it's irrational
anonymous
  • anonymous
how to prove this?
anonymous
  • anonymous
all square roots of numbers other than perfect squares yields an irrational response, because that's the definition of a perfect square

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
i know this already. but if you were given this question to work out. how would you do it?
asnaseer
  • asnaseer
One way is to prove by contradiction. i.e. start off by assuming that \(\sqrt{5}\) IS rational and show that this leads to a contradiction - thus proving that it is irrational.
anonymous
  • anonymous
if i do this assumption. i can say that \[\sqrt{5} = \frac{ a }{ b }\] then \[5 = \frac{ a ^{2} }{ b ^{2} }\] \[5b ^{2} = a ^{2}\] and then i'm stuck.
asnaseer
  • asnaseer
all steps are correct so far :)
asnaseer
  • asnaseer
you also need to state at the start that a and b are co-prime, i.e. they have no common factors so that a/b is a fraction in its simplest form. now think about what the last step means - what does it say about \(a^2\)
asnaseer
  • asnaseer
yo found:\[a^2=5b^2\]therefore \(a^2\) must be a multiple of ???
anonymous
  • anonymous
multiple of 5
asnaseer
  • asnaseer
correct. now 5 is a prime number, so if \(a^2\) is a multiple of 5, then a must also be a multiple of 5 - agreed?
anonymous
  • anonymous
yes...
asnaseer
  • asnaseer
therefore we can express a as:\[a=5m\]where m is some other integer. therefore:\[a^2=(5m)^2=25m^2\]now substitute this back into your expression that relates a^2 to b^2 and see what that leads to.
anonymous
  • anonymous
i get \[25m ^{2} = 5b ^{2}\]
asnaseer
  • asnaseer
yup - now divide both sides by 5 and see what you can say about \(b^2\)...
anonymous
  • anonymous
\[5m ^{2} = b ^{2}\]
asnaseer
  • asnaseer
... so \(b^2\) must be a multiple of ???
anonymous
  • anonymous
or 5, just like \[a ^{2}\] was
anonymous
  • anonymous
of**
asnaseer
  • asnaseer
yes - and, following the same reasoning as we did for \(a^2\), we conclude that b is also a multiple of 5.
asnaseer
  • asnaseer
so we have concluded that both a and b are multiples of 5. which means they are NOT co-prime, i.e. they have a common factor of 5. this contradicts our original assumptions, therefore \(\sqrt{5}\) must be irrational.
asnaseer
  • asnaseer
I hope that makes sense?
anonymous
  • anonymous
makes sense if we can say that a rational number is one which when expressed as a fraction has no coprime (except 1)
asnaseer
  • asnaseer
yes - that is how the proof is supposed to start off as - remember I stated it above as: "you also need to state at the start that a and b are co-prime, i.e. they have no common factors so that a/b is a fraction in its simplest form."
anonymous
  • anonymous
makes sense. thanks! goodnight!
asnaseer
  • asnaseer
yw :) and goodnight to you too.

Looking for something else?

Not the answer you are looking for? Search for more explanations.