# How to add a tolerance to isapprox
**URL:** https://discourse.julialang.org/t/how-to-add-a-tolerance-to-isapprox/118711
**Category:** General Usage
**Tags:** question
**Created:** [August 28, 2024, 1:02pm UTC](https://discourse.julialang.org/t/how-to-add-a-tolerance-to-isapprox/118711 "2024-08-28T13:02:31Z")
**Posts on this page:** 7
**Page:** 1
### Author:  [@ufechner7](https://discourse.julialang.org/u/ufechner7)
#### Post date: [August 28, 2024, 1:02pm UTC](https://discourse.julialang.org/t/how-to-add-a-tolerance-to-isapprox/118711/1 "2024-08-28T13:02:32Z")
I have the following line of code:
```julia
@test all(forces[i,:] .β kps4.forces[i])
```
I would like to increase the relative tolerance, in other words add the parameter rtol=1e-4 . How can I do that?
---
### Author:  [@LaurentPlagne](https://discourse.julialang.org/u/LaurentPlagne)
#### Post date: [August 28, 2024, 1:05pm UTC](https://discourse.julialang.org/t/how-to-add-a-tolerance-to-isapprox/118711/2 "2024-08-28T13:05:53Z")
I think that you have to use the function form of `isapprox`
```julia
isapprox(x, y; atol::Real=0, rtol::Real=atol>0 ? 0 : βeps, nans::Bool=false[, norm::Function])
```
---
### Author:  [@ufechner7](https://discourse.julialang.org/u/ufechner7)
#### Post date: [August 28, 2024, 1:06pm UTC](https://discourse.julialang.org/t/how-to-add-a-tolerance-to-isapprox/118711/3 "2024-08-28T13:06:22Z")
The point is, I do not know how to apply this function to a pair of vectors.
---
### Author:  [@LaurentPlagne](https://discourse.julialang.org/u/LaurentPlagne)
#### Post date: [August 28, 2024, 1:08pm UTC](https://discourse.julialang.org/t/how-to-add-a-tolerance-to-isapprox/118711/4 "2024-08-28T13:08:28Z")
```julia
julia> a=[1,2,3]
3-element Vector{Int64}:
1
2
3
julia> b=[1,2,3.000001]
3-element Vector{Float64}:
1.0
2.0
3.000001
julia> all(isapprox.(a,b))
false
julia> all(isapprox.(a,b,rtol=1.e-2))
true
```
---
### Author:  [@juliohm](https://discourse.julialang.org/u/juliohm)
#### Post date: [August 28, 2024, 1:09pm UTC](https://discourse.julialang.org/t/how-to-add-a-tolerance-to-isapprox/118711/5 "2024-08-28T13:09:41Z")
Notice that the `isapprox` function has methods with vectors as well, in which case it relies on approximate norms.
---
### Author:  [@ufechner7](https://discourse.julialang.org/u/ufechner7)
#### Post date: [August 28, 2024, 1:17pm UTC](https://discourse.julialang.org/t/how-to-add-a-tolerance-to-isapprox/118711/6 "2024-08-28T13:17:44Z")
Thank you so much, that works! π
---
### Author:  [@stevengj](https://discourse.julialang.org/u/stevengj)
#### Post date: [August 28, 2024, 1:18pm UTC](https://discourse.julialang.org/t/how-to-add-a-tolerance-to-isapprox/118711/7 "2024-08-28T13:18:18Z")
> [@LaurentPlagne](#):
>
> `all(isapprox.(a,b))`
> [@juliohm](#):
>
> Notice that the `isapprox` function has methods with vectors as well, in which case it relies on approximate norms.
I would _strongly_ recommend using the vector form in general, i.e. `isapprox(a,b)` rather than `all(isapprox.(a,b))`.
Elementwise approximate comparison discards information about the scale that can be obtained from the other elements. For example, intuitively you would expect:
\begin{bmatrix} 1 \\ 0 \end{bmatrix} \approx \begin{bmatrix} 1 \\ 10^{-100} \end{bmatrix}
to return `true`, and indeed the vector form of `isapprox` does this:
```julia
julia> [1, 0] β [1, 1e-100]
true
```
because the difference `1e-100` is small (according the default `rtol=βΞ΅`) _compared to the length of the vector_ (β 1). However, if you do an elementwise `isapprox`, then it fails:
```julia
julia> all([1, 0] .β [1, 1e-100])
false
```
because `0 β 1e-100` **correctly** returns `false`: the two numbers have _no_ significant digits in common, and there is no absolute reference scale by which we can say that `1e-100` is βsmallβ (compared to what)? See also the [`isapprox` docstring](https://docs.julialang.org/en/v1/base/math/#Base.isapprox) on `x β 0` comparisons.
As I wrote [in another thread](https://discourse.julialang.org/t/approximate-equality/8952/41), people often get approximate equality wrong on their first try. Even NumPy got it wrong with their broken `isclose`.
PS. If you think this would be solved by having a default absolute tolerance, consider `[1e100, 0] β [1e100, 1]`, which should also be `true` in a relative-error sense.