as per me the above statement have to be true because if the sum of two irrational number is 0 , it suggests that 2 irrational numbers have to be prefer +k and also -k whereby k is an irrational number . Additionally when diffrence of 2 irrational numbers is rational it is possible only once both the irrational number are very same . Indigenous both the above cases we have the right to conclude that if amount of 2 irrational number is rational or difference of 2 irrational numbers is rational then it must be 0 . You re welcome correct me if ns am dorn . By "a" and also "b" i mean solitary or separation, personal, instance irrational numbers for instance only "pi" and also not "pi +/- something" .

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edited jan 18 "20 at 7:28
Sameer nilkhan
asked jan 18 "20 in ~ 6:16 Sameer nilkhanSameer nilkhan
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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.

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answered january 18 "20 in ~ 6:21 pre-kidneypre-kidney
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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.

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answered jan 18 "20 in ~ 6:26 Ross MillikanRoss Millikan
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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\$.

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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.

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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?