Question

for problems 25 - 31, complete each inequality using a pair of consecutive integers.

1. < √10 + 4 <
2. < √19 + 8 <
3. < √61 - 5 <
4. < √92 - 11 <
5. < ∛50 - 1 <
6. < ∛100 + 2 <
7. < ∛17 + 5 <
8. reason is there a limit to the number of decimal places you can use to approximate an irrational number? explain.

Explanation:

Step1: Estimate square - root values

Find the two perfect - squares closest to the number inside the square - root. For $\sqrt{10}$, since $9<10<16$, then $3 < \sqrt{10}<4$.

Step2: Calculate the range of $\sqrt{10}+4$

Add 4 to each part of the inequality: $3 + 4<\sqrt{10}+4<4 + 4$, so $7<\sqrt{10}+4<8$.

Step3: Estimate $\sqrt{19}$

Since $16<19<25$, then $4<\sqrt{19}<5$. Add 8 to each part: $4 + 8<\sqrt{19}+8<5 + 8$, so $12<\sqrt{19}+8<13$.

Step4: Estimate $\sqrt{61}$

Since $49<61<64$, then $7<\sqrt{61}<8$. Subtract 5 from each part: $7-5<\sqrt{61}-5<8 - 5$, so $2<\sqrt{61}-5<3$.

Step5: Estimate $\sqrt{92}$

Since $81<92<100$, then $9<\sqrt{92}<10$. Subtract 11 from each part: $9-11<\sqrt{92}-11<10 - 11$, so $-2<\sqrt{92}-11<-1$.

Step6: Estimate $\sqrt[3]{50}$

Since $27<50<64$, then $3<\sqrt[3]{50}<4$. Subtract 1 from each part: $3-1<\sqrt[3]{50}-1<4 - 1$, so $2<\sqrt[3]{50}-1<3$.

Step7: Estimate $\sqrt[3]{100}$

Since $64<100<125$, then $4<\sqrt[3]{100}<5$. Add 2 to each part: $4 + 2<\sqrt[3]{100}+2<5 + 2$, so $6<\sqrt[3]{100}+2<7$.

Step8: Estimate $\sqrt[3]{17}$

Since $8<17<27$, then $2<\sqrt[3]{17}<3$. Add 5 to each part: $2+5<\sqrt[3]{17}+5<3 + 5$, so $7<\sqrt[3]{17}+5<8$.

Step9: Answer question 32

There is no limit to the number of decimal places when approximating an irrational number. An irrational number has a non - repeating, non - terminating decimal expansion. We can keep calculating more and more decimal places as long as we have the computational power and need for more precision.

Answer:

1. $7<\sqrt{10}+4<8$
2. $12<\sqrt{19}+8<13$
3. $2<\sqrt{61}-5<3$
4. $-2<\sqrt{92}-11<-1$
5. $2<\sqrt[3]{50}-1<3$
6. $6<\sqrt[3]{100}+2<7$
7. $7<\sqrt[3]{17}+5<8$
8. There is no limit. An irrational number has a non - repeating, non - terminating decimal expansion.