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You are watching: The sum of 2 rational numbers is rational

I know this declare is false (if ns am correct) but how to prove it"s false?

"The amount of 2 rational number is irrational."

2. I recognize this declare is true (if i am correct) yet how to prove it"s true?

"The sum of 2 irrational numbers is irrational"

I offered the example \$sqrt2+ sqrt3 = 3.14\$

But i may need to usage proof by contradiction or contaposition.  If two numbers are rational we can express their amount as\$\$fracab + fraccd\$\$which is equal to \$\$fracad + bcbd.\$\$Hence, rational.

The sum of 2 irrational numbers may be irrational. Consider \$2+sqrt2\$ and also \$3+sqrt2\$. Both room irrational, and also so is their sum \$5+2sqrt2\$. For one, that comes straight from the closure of addition on \$starrkingschool.netbbQ\$, but I don"t think that"s the price they would certainly expect.

Let \$a = dfracp_1q_1\$ and \$b = dfracp_2q_2\$ be rationals in \$starrkingschool.netbbQ\$ and \$q_1, q_2 eq 0\$:\$\$a + b = dfracp_1q_1 + dfracp_2q_2 = dfracp_1q_2 + p_2q_1q_1q_2 in starrkingschool.netbbQ\$\$

For the second one, how about \$dfracsqrt22 + dfracsqrt22 = sqrt2\$. A solitary example is adequate to prove the claim.

For bonus points, have the right to you prove that \$dfracsqrt22\$ is irrational?(Hint: Contradiction. Expect it"s rational, and use the closure of enhancement on \$starrkingschool.netbbQ\$ that was proven.) \$frac pq\$+\$frac xz\$ \$(q,z eq 0)\$(by formula of reasonable numbers).

=\$fracpz+qzqz\$,which is again in the form \$frac ab\$ so the is bound to be reasonable and additionally \$qz\$ is not equal to \$0\$.

Sum that irrational may be irrational is true but it is constantly rational if the sum consists of the irrational number and also its negative and then the amount will yield \$0\$.Sum of two irrational numbers that you expressed as a decimal is no true and only one approximation. The amount of 2 irrational numbers is no necessarily irrational. For example, \$sqrt2\$ and also \$-sqrt2\$ room two irrational numbers, but their amount is zero (\$0\$), which in turn is rational.

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