Prove the for every rational number z and also every irrational number x, there exists a distinctive irrational number such the x + y = z.

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Assume a and also b room integers v GCD (a,b) =1, and also that c is an irrational number.There exists a number d = (a/b) + c.

Assume the d is rational. Then d - (a/b) would be rational, what is a contradiction since c is irrational. Therefore, d is irrational.

To prove uniqueness we have the right to use the reality that the addition of any kind of two real numbers has only one result, climate d is unique. Q.E.D.


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inquiry Feb 13 "15 in ~ 3:51
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BeginnerBeginner
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You don"t need to restrict the integers creating the ratio to having a greater common denominator the 1. Simply assert the they exist.

Any $z$ that is a rational number deserve to be expressed as the ratio of 2 integers. (For strictness we need the denominator integer to it is in non-zero.)

Any $x$ is one irrational number cannot be so expressed as the ratio of two integers.

For any real numbers, $x$ and $z$, over there is just a unique number $y$ where $x+y=z$.

If this $y$ to be rational it can be expressed as the ratio of some two integers. For any sum $x+y$ i beg your pardon can additionally be expressed together the proportion of two integers, it would certainly then follow the $x$ might be expressed together the sum of 2 integers. (By reason the product of any type of two integers is an integer.)

By contraposition: for any type of irrational $x$, and any reasonable $z$, the number $y$ where $x+y=z$, need to be both unique and irrational.


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answered Feb 13 "15 in ~ 4:26
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Graham KempGraham Kemp
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You have actually proven that the amount of a rational number and an irrational number is irrational, yet you have actually not proven the main result.

If $x$ is irrational and $z$ is rational, what (possibly) irrational number $y$ would certainly you wager is such that $x+y=z$? Why might this be irrational? (Hint: usage what you simply proved.)

Why should it it is in unique?


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answered Feb 13 "15 in ~ 4:00
user214321user214321
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$egingroup$ additionally note the you never ever used the fact that $gcd(a,b)=1$, so you execute not need to assume this. $endgroup$
–user214321
Feb 13 "15 at 4:01
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$egingroup$ I explained what I believe you space missing. Deserve to you tell me exactly how you think you should modify/ expand upon your proof, provided what ns wrote? $endgroup$
–user214321
Feb 13 "15 at 4:11
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$egingroup$ It would seem that this was spanned in your conversation with Yotas Trejos. As your proof is created it only proofed a peripheral result, yet when you adjust it for this paper definition it have to be fine. $endgroup$
–user214321
Feb 13 "15 at 4:31
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