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# Easy proof for rational + irrational = irrational? watch

1. Ok so I have to prove a + b = irrational, where a is a rational number and b is an irrational number.

But i get the opposite...

let,

So i get that it is rational...where am i going wrong?

+REP
2. if b is irrational and any rational number is added then its always going to be irrational...x+1/y= xy-y+1/y xy/y=rational or irrational but the addition of a irrational i.e. 1/y its going to equal irrational...dont know if that helps
3. It looks to me like h (or m+bn) in your post above is irrational.
4. your assumption that h is rational lies on m+bn being rational, i.e a rational plus an irational. So you are assuming the disproof in your disproof!
5. (Original post by Mos Def)
let,
h has to be an integer by definition of rational number, which is in fact, false.
6. Ooohh right, so basically i don't assume the h bit, and leave it as , (m + bn)/n ...which is irrational...QED?
7. Basic proof by contradiction. Suppose a is rational, and write it as p/q, for some integers p and q. Suppose b is irrational.

Now we want to prove that a + b is irrational. If it is not irrational, it is rational, and hence it can be written as: a + b = c/d, for some integers c/d; hence b = c/d - a = (cq-pd)/(dq). But then clearly b is rational, as cq-pd is an integer, as is dq. But this is a contradiction - hence a + b is irrational.
8. (Original post by GHOSH-5)
Basic proof by contradiction. Suppose a is rational, and write it as p/q, for some integers p and q. Suppose b is irrational.

Now we want to prove that a + b is irrational. If it is not irrational, it is rational, and hence it can be written as: a + b = c/d, for some integers c/d; hence b = c/d - a = (cq-pd)/(dq). But then clearly b is rational, as cq-pd is an integer, as is dq. But this is a contradiction - hence a + b is irrational.
Oh right, this is similar to what I did, cheers!
9. found this on a wiki answers page, I hope it helps -

say that 'a' is rational, and that 'b' is irrational.
assume that a + b equals a rational number, called c.
so a + b = c
subtract a from both sides.
you get b = c - a.
but c - a is a rational number subtracted from a rational number, which should equal another rational number.
However, b is an irrational number in our equation, so our assumption that a + b equals a rational number must be wrong.
10. (Original post by RVNmax)
found this on a wiki answers page, I hope it helps -

say that 'a' is rational, and that 'b' is irrational.
assume that a + b equals a rational number, called c.
so a + b = c
subtract a from both sides.
you get b = c - a.
but c - a is a rational number subtracted from a rational number, which should equal another rational number.
However, b is an irrational number in our equation, so our assumption that a + b equals a rational number must be wrong.
I like that way, it's easy
11. (Original post by CHEM1STRY)
I like that way, it's easy
Yay I guess it helped then, have some rep for quoting me.

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