You are watching: Is a+b always rational

irrational-numbers

re-publishing

mention

follow

edited jan 18 "20 at 7:28

Sameer nilkhan

asked jan 18 "20 in ~ 6:16

Sameer nilkhanSameer nilkhan

48122 silver- badges88 bronze badges

$endgroup$

7

| display 2 more comments

## 3 answers 3

energetic earliest Votes

1

$egingroup$

If $a+b$ and also $a-b$ are both rational, climate so is $a$ since it is the median of those two quantities, and the average of two rational numbers is likewise rational. Thus additionally $b=(a+b)-a$ is rational.

re-publishing

mention

monitor

answered january 18 "20 in ~ 6:21

pre-kidneypre-kidney

27.8k3030 silver badges7272 bronze badges

$endgroup$

include a comment |

1

$egingroup$

Your statement in the body is not correct. For example, both $sqrt 2$ and also $sqrt 2 +1$ are irrational. Their distinction is $1$, i m sorry is rational. If both the sum and difference of two numbers space rational, the 2 numbers need to be rational, but that is not what your question says.

re-superstructure

mention

monitor

answered jan 18 "20 in ~ 6:26

Ross MillikanRoss Millikan

358k2626 yellow badges237237 silver badges424424 bronze badges

$endgroup$

1

include a comment |

1

$egingroup$

Don"t overthink it.

If $x$ irrational climate $q - x$ is irrational for every reasonable $q$.

So $x + (q-x) = q$ is the rational sum of two irrational numbers. But $q$ can be any type of rational number and doesn"t have to be $0$.

=====

This is just one of those results that when you analyze it you establish it can"t actually *mean* anything.

We think. $rational + reasonable = rational$

And $rational + irrational = irrational$.

But what the $irrational + irrational$ can that be rational.

Well, $sqrt 2 + (-sqrt 2) = 0$ and $0$ is reasonable so it deserve to be rational.

So us think... And we say... Yeah, however $sqrt 2$ and $-sqrt 2$ are pertained to each other, they room negatives of each other. Ns bet if you have two *independent* irrationals the must include to a irrational.

So who else states well if $x +y = k$ and also $k$ is rational, climate if $x$ is irrational climate $y= k - x$ is irrational and $x + y = x+(k-x) = k$. So the is 2 irrationals that include to a rational.

So we say... Yet those are *dependent* to every other; i bet if you have two *independent* irrationals the must include to a irrational.

See more: Why Should You Wear An Antistatic Wrist Strap ? Why Should You Wear An Anti Static Wrist Strap

So the other person asks. How exactly to you define if two irrationals are "dependent"?

And us say... Fine if $x$ is irrational and $y = pm x + k$ for part rational $k$ they space dependent. Therefore I case that if $x,y$ room independent 보다 $x + y$ is no rational.

And climate other person says.... For this reason in various other words, you space claiming that there is no reasonable number $k$ so that $y =pm x + k$ then $y mp x
e k$ for any rational $k$ however if over there *is* such a rational $k$ so the $y =pm x + k$ then $ymp x$ *might* be rational; is the what you are claiming?

And us say, yes, currently how have the right to I prove it?

And the other person says. Um... You just said $x mp y =k; k$ reasonable if and also only if $x = pm y +k$ for some rational $k$. Carry out you really think that is meaningful?