Sum of Two Irrational Numbers

Relevant wiki: Rational Numbers - Problem Solving

Take $e$ and $(3-e)$ for instance. $e+(3-e)=3$ Therefore it must not be true. The statement is false.

Relevant wiki: Rational Numbers - Problem Solving

• $\sqrt{2}$ is irrational.

• $(8 - \sqrt{2})$ is irrational.

$(8 - \sqrt{2})+ \sqrt{2} = 8$

Hence $8$ is not irrational.

Relevant wiki: Rational Numbers - Problem Solving

The sum of two positive irrational numbers sometimes Irrational, Sometimes Rational

Example:

$X=8+\sqrt{2}$

and

$Y = 9-\sqrt{2}$

so,

$X+Y$ = $8+\sqrt{2}+9-\sqrt{2}$

so,

$X+Y=8+9=\boxed{17}$ (Rational)

Again,

$X= 2+\sqrt{7}$

$Y=2+\sqrt{11}$

So,

$X+Y$ = $4+\sqrt{7}+\sqrt{11}$ (Irrational)

0.1011011101111011111... + 0.0100100010000100000...=0.1111111111111111111...= $\frac{1}{9}$

The quick way is counter example

i.e ( k) + (k-∆), where (∆=any irrational) and (k is not irrational)

A counter example is: (2- $\sqrt2$ ) + $\sqrt2$ =2

$\text{it can be proved in this way}$

$\text{irrational number +irrational number=rational}$

counter example: $(3-\sqrt3)+\sqrt3=3$

Suppose the statement is true. Then let q be rational and r be irrational, q>r. Then t=q-r must be rational, or else t+r=q will not be irrational, in contradiction with the statement. If t is rational, than r=t-q is rational, as the sum of the rational numbers t and -q is always rational (the fact that you can always choose a common denominator is a proof for it).That's another contradiction.

Therefore, the statement is false.

▄

$lg(2)$ + $lg(5)$ = $lg(2*5)$ = $lg(10)$ = $1$

Am I making a unique counter example?

Imagine having an irrational number with a decimal place like $a=0.123114958295...$

The thing to notice is how there always exist another irrational number $b$ that adds up $a$ such that it equals $\overline{m.99999999...}$ where $m$ is an integer in the range $[0;9]$ and thus this evaluates to $a+b = m+1$ , which is an integer and thus rational.

The number $b$ is basically, in any number in the decimal position in the irrational number $a$ , the corresponding decimal position's number in $b$ is exactly 9 minus that of in $a$ , and thus the two's adds up to 9. Doing this to all other decimal positions in $a$ and sum $a$ and $b$ will lend the result mentioned, $\overline{m.99999999...} = m+1$ .

Therefore, irrational + irrational can be rational.

let x be any positive irrational number (1-x) + x = 1

therefore, irrational num + irrational num not always equal to irrational number

I am so new to this but eager to learn Barney style. So when e+(3-e)= 3 you are just negating the e? Is that how I am to understand it? Sorry I'm so new!!!

$\sqrt{2}$ and $-\sqrt{2}$ are irrational, but $\sqrt{2}+(-\sqrt{2})=0=\frac{0}{1}$

Let A be an irrational number such that it is a non-repeating, non terminating decimal. Since there are only ten possible digitals for each place value of A, it is possible to find an irrational number B such that each digit of B is selected so that when the digits are added you get a fixed value e.g zero or three In this case the sum of A and B will give you a repeating decimal number which is rational!

2 positives make a negative, easy

22/7 + 20/7 =6

Take some irrational number $x$ , equal to something such as $3.1523796$ and so on. We could make a new irrational number $y$ , such that for each digit after the decimal point of $x$ , the digit of $y$ at that point is equal to the difference between the digit of $x$ at that point and $9$ . For the value of $x$ given earlier, $y$ would be $0.8476203...$ and so on. Now $x + y =ceil(x)$ , meaning that we have added two irrational numbers to produce a rational number.

Basically, it depends on the number