# Why is a==b False?

Please see the following code. Why is a==b False? Thanks!

Input:

```
a=n(1/2*sqrt(5) - 1/2,digits=15)
b=n(1-a^2,digits=15)
print a
print b
a==b
```

Output:

```
0.618033988749895
0.618033988749895
False
```

Why is a==b False?

Please see the following code. Why is a==b False? Thanks!

Input:

```
a=n(1/2*sqrt(5) - 1/2,digits=15)
b=n(1-a^2,digits=15)
print a
print b
a==b
```

Output:

```
0.618033988749895
0.618033988749895
False
```

3

This comes from working with numerical approximations.

In the computer, `a`

and `b`

are represented in binary.
They are very close but not equal.

```
sage: a.sign_mantissa_exponent()
(1, 11133510565745312, -54)
sage: b.sign_mantissa_exponent()
(1, 11133510565745310, -54)
```

But the difference is very small and their decimal expansions to 15 digits coincide.

Sage lets you work with exact algebraic numbers, either by creating a number field:

```
sage: K.<a> = NumberField(x^2 + x - 1, embedding=0.6)
sage: a.numerical_approx()
0.618033988749895
sage: 1 - a^2
a
```

or by using the field of algebraic number (`QQbar`

in Sage):

```
sage: a = QQbar(1/2*sqrt(5) - 1/2)
sage: a
0.618033988749895?
sage: 1 - a^2 == a
True
sage: a.minpoly()
x^2 + x - 1
sage: a.numerical_approx()
0.618033988749895
sage: a.radical_expression()
1/2*sqrt(5) - 1/2
```

3

Note that even without defining the corresponding number field or working in `QQbar`

, SageMath correctly finds that `a == b`

if you are not working with approximations but with exact symbolic expressions¹:

```
sage: a = 1/2*sqrt(5) - 1/2
sage: b = 1 - a^2
sage: bool(a == b)
True
```

¹ Note the necessity of calling `bool(a==b)`

since `a == b`

is kept as an expression by default, and one has to explicitly asks whether this equality holds.

Asked: **
2017-12-08 17:45:51 -0500
**

Seen: **23 times**

Last updated: **Dec 09 '17**

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