Answer :
Let's go through the solution step-by-step to understand which statement best completes the proof.
1. Define the Numbers:
Let's start by defining two numbers:
- Let [tex]\( a \)[/tex] be a rational number.
- Let [tex]\( b \)[/tex] be an irrational number.
2. Assume the Sum:
Assume that [tex]\( a + b = x \)[/tex], where [tex]\( x \)[/tex] is a rational number.
3. Subtracting to Identify [tex]\( b \)[/tex]:
By the subtraction property of equality, we get:
[tex]\[ b = x - a \][/tex]
4. Relating Rationals and Irrationals:
Now, since both [tex]\( x \)[/tex] and [tex]\( a \)[/tex] are rational numbers, their difference must also be a rational number (because the set of rational numbers is closed under subtraction). This implies that:
[tex]\[ b = x - a \quad \text{is rational} \][/tex]
5. Contradiction:
However, from our initial definition, [tex]\( b \)[/tex] is an irrational number. This leads to a contradiction because one number (b) cannot be both rational and irrational simultaneously. Therefore, the assumption that [tex]\( x \)[/tex] is rational must be incorrect.
6. Conclusion:
Since our initial assumption led to a contradiction, we conclude that the sum [tex]\( a + b \)[/tex] must not be rational. Hence, [tex]\( a + b \)[/tex] must be irrational.
In conclusion, the statement that best completes the proof is:
"it is the sum of two rational numbers."
This accurately identifies the nature of the original contradiction and provides a correct understanding of the sum of a rational and an irrational number.
1. Define the Numbers:
Let's start by defining two numbers:
- Let [tex]\( a \)[/tex] be a rational number.
- Let [tex]\( b \)[/tex] be an irrational number.
2. Assume the Sum:
Assume that [tex]\( a + b = x \)[/tex], where [tex]\( x \)[/tex] is a rational number.
3. Subtracting to Identify [tex]\( b \)[/tex]:
By the subtraction property of equality, we get:
[tex]\[ b = x - a \][/tex]
4. Relating Rationals and Irrationals:
Now, since both [tex]\( x \)[/tex] and [tex]\( a \)[/tex] are rational numbers, their difference must also be a rational number (because the set of rational numbers is closed under subtraction). This implies that:
[tex]\[ b = x - a \quad \text{is rational} \][/tex]
5. Contradiction:
However, from our initial definition, [tex]\( b \)[/tex] is an irrational number. This leads to a contradiction because one number (b) cannot be both rational and irrational simultaneously. Therefore, the assumption that [tex]\( x \)[/tex] is rational must be incorrect.
6. Conclusion:
Since our initial assumption led to a contradiction, we conclude that the sum [tex]\( a + b \)[/tex] must not be rational. Hence, [tex]\( a + b \)[/tex] must be irrational.
In conclusion, the statement that best completes the proof is:
"it is the sum of two rational numbers."
This accurately identifies the nature of the original contradiction and provides a correct understanding of the sum of a rational and an irrational number.