Approximation for π and sqrt{2}

[mathdad]

Say whether each statement is TRUE OR FALSE. Do not use a calculator or tables; use instead the approximations sqrt{2} is about 1.4 and π is about 3.1.

1. 2 < or = (π + 1)/2

2. sqrt{7} - 2 > or = 0

For question 1, I replace π with 3.1, and then simplify, right?

How do I apply the approximation given for sqrt{2} to the sqrt{7} to answer question 2?

 

[HallsofIvy]

Say whether each statement is TRUE OR FALSE. Do not use a calculator or tables; use instead the approximations sqrt{2} is about 1.4 and π is about 3.1.

1. 2 < or = (π + 1)/2

2. sqrt{7} - 2 > or = 0

For question 1, I replace π with 3.1, and then simplify, right?

Yes.

How do I apply the approximation given for sqrt{2} to the sqrt{7} to answer question 2?

7 is larger than 4 so sqrt{7} is larger than 2.

 

[Prove It]

Say whether each statement is TRUE OR FALSE. Do not use a calculator or tables; use instead the approximations sqrt{2} is about 1.4 and π is about 3.1.

1. 2 < or = (π + 1)/2

2. sqrt{7} - 2 > or = 0

For question 1, I replace π with 3.1, and then simplify, right?

How do I apply the approximation given for sqrt{2} to the sqrt{7} to answer question 2?

For 2 all you need to remember is that since $\displaystyle \begin{align*} 4 < 7 \end{align*}$ that means $\displaystyle \begin{align*} \sqrt{4} < \sqrt{7} \end{align*}$...

 

[mathdad]

Question 1

2 < or = (π + 1)/2

Let pi be about 3.1

2 < or = (3.1 + 1)/2

2 < or = 4.1/2

2 < or = 2.05

TRUE

 

[mathdad]

Question 2

sqrt{7} - 2 > or = 0

Like you said, sqrt{4} < sqrt{7} because 4 < 7.

Then the answer is TRUE.