Irrational + Irrational = ?

Answer is false

CounterExample:

If $m = 1+\sqrt{3} \hspace{3mm} \text{and} \hspace{3mm} n = 2 - \sqrt{3}$ , then $m + n = 1 + \sqrt{3} + \left(2 - \sqrt{3}\right) = 3$

The number $3$ can be written in the form $\frac{p}{q}$ as $\frac{3}{1}$

Thus if $3$ is rational then $m+n$ is rational which contradicts the claim stated above. Therefore the claim is false.

$\sqrt{2}+(-\sqrt{2})=\boxed{0}$

The sum of any two irrational numbers of the form $a_1+b\sqrt{c}$ and $a_2-b\sqrt{c}$ where $a_1,a_2,b,c\in\mathbb Q$ is rational. We can easily verify this as: $(a_1+b\sqrt{c})+(a_2-b\sqrt{c})=(a_1+a_2)+(b\sqrt{c}-b\sqrt{c})=a_1+a_2$ And as both $a_1$ and $a_2$ are rational, therefore by the closure property of rational numbers, their sum is also rational.

$\sqrt{2}$ is irrational and $-\sqrt{2}$ is irrational too.

But $\sqrt{2}-\sqrt{2} = 0$ , which is rational.

The decimal representation of an irrational number extends to infinity without repeating. Consider the two following irrational numbers:

x = 0.101001000100001000001000000...

y = 0.010110111011110111110111111...

x + y = 0.11111111... = $\frac{1}{9}$

Therefore, the premise is false.

The counter to the claim that immediately popped into my mind was this: Take a whole number and subtract any irrational number (A) from it. The result will be irrational (B). A+B = a whole number, thus proving it is possible for two irrational numbers to add to a rational number.

Take any irrational $x$ , then $-x$ is also irrational.

Their sum is $0 \in \mathbb Q$ , a rational number.

Therefore, the claim is false.

False. The sum of two irrational numbers is NOT necessarily an irrational number. E.g. e + (1 - e) = 1

For any irrational number $k$ , $-k$ is irrational, and their sum is obviously $0$ .

(Phi^2)+(-phi)=1 as does (phi) +(phi^(-1)), both are counter examples, if you're looking for a solution that is two positive numbers you can simply add any rational number to both addends

If you add an irrational number to its negative counterpart, the result is zero.

π+(4-π)=4 or, for that matter... π+(-π)=0 These are both two irrational numbers that add up to whole numbers

Some irrational numbers can be added to equal 0, which is rational.

For example, $\pi+-(\pi)=0$

This is too easy if you don't exclude negative negative numbers. e is irrational. -e+e = 0, which is not irrational. You can do the same thing for any irrational number.

• Let x and y be irrational numbers √2 and 1 - √2 respectively.
• √2 + (1 - √2) = 1.
• x + y = 1 which is a rational number contradicting the statement. Therefore, the answer is false.

How about (Sin A)^{2} + (Cos A)^{2} = 1

For every irrationnal number $x\in \Re$ you can always do $x + (-x) = 0$ .

Answer is false.

Counter Example:

Let's take the irrational number π. $π=3.14159265358...$

Any whole number minus π is also irrational. $10-π=6.85840734641...$

$π+(10-π)=10$

Since all whole numbers are rational, the statement is false.

In general the sum of the irrational numbers $a+\sqrt{b}$ and $c-\sqrt{b}$ where $a,c \in Q$ and $b,d \in Q^+$ and b and d are square-free.

$\pi$ is irrational and $(1-\pi)$ is irrational but $\pi+(1-\pi)=1$ which is obviously rational

A rational number minus an irrational number will give another irrational number. You can prove this by contradiction. If $\frac{p}{q}$ - ${x}$ = $\frac{a}{b}$ , where ${x}$ is an irrational number, and ${p}$ , ${q}$ , ${a}$ , ${b}$ are whole numbers, you can see reach a contradiction because ${x}$ equals to ( $\frac{p}{q}$ - $\frac{a}{b}$ ), which is a rational number.

An example of sum of two irrational numbers is ${\pi}$ + (1 - ${\pi}$ ), which equals to 1. In fact, the sum of an irrational number with one minus the same irrational number always give a rational number. So the claim is false.

1/3 + 2/3 = 1

Let $x$ be any irrational number. Then $-x$ is irrational too. And $x-x=0$ , a rational number.

$\pi+(4-\pi)=4$

I thought of (4 - pi) being irrational. If you add pi, you get 4.

False. Counterexample: Pi is irrational. So is (1 - Pi). The sum of these to irrational numbers is 1, a rational number.

Square root of two - the square root of two = what?