### Tell us what’s happening:

Not sure how to continue, I don’t thing there is enough information here to do the step?

### Your code so far

```
def square_root_bisection(square_target, tolerance=1e-7, max_iterations=100):
if square_target < 0:
raise ValueError('Square root of negative number is not defined in real numbers')
if square_target == 1:
root = 1
print(f'The square root of {square_target} is 1')
elif square_target == 0:
root = 0
print(f'The square root of {square_target} is 0')
else:
low = 0
high = max(1, square_target)
root = None
# User Editable Region
for _ in range(max_iterations):
mid = (low + high) / 2
square_mid = mid**2
if tolerance in abs
# User Editable Region
```

### Your browser information:

User Agent is: `Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/128.0.0.0 Safari/537.36`

### Challenge Information:

Learn the Bisection Method by Finding the Square Root of a Number - Step 13

Everything is here:

create an `if`

statement to check if the absolute value of the difference between `square_mid`

and `square_target`

is within the specified `tolerance`

.

create an `if`

statement

If statement needs to have a colon at the end of the line

to check if the absolute value of

The `abs()`

function returns the absolute value of a number,

`abs()`

is a function, you can look up how it works

https://www.w3schools.com/python/ref_func_abs.asp

the difference between `square_mid`

and `square_target`

Difference means you will need to use the `-`

operator to find the difference between these two variables

is within the specified `tolerance`

.

For example:

```
square_mid = 5
square_target = 10
tolerance = 2
square_mid - square_target = -5
abs(-5) = 5
(5 < tolerance) = False
```

Not within the tolerance in this case.

This is what I have now?

```
mid = (low + high) / 2
square_mid = mid**2
if num <= 0:
num = abs(square_mid - square_target) = tolerance```
```

I followed the suggestions by the help section it now reads the following and still not passing.

```
mid = (low + high) / 2
square_mid = mid**2
if abs(square_mid - square_target) < tolerance:```
```

Looking good! Remember the body of an `if`

can’t be empty, so you can use `pass`

as a placeholder for now

still not working with pass

```
mid = (low + high) / 2
square_mid = mid**2
if abs(square_mid - square_target) < tolerance:
pass
```

Keep your indentation consistent. You can’t have it all over the place like JS