If a step-up transformer increases voltage, then how current can be decreased ? is really the current in output is smaller?
A transformer essentially converts between voltage and current using a magnetic field. Because it is a conversion, then if the process is 100% efficient, then the output power and input power must be equal:
$$P_{in} = P_{out}$$
If they are not equal, then either you are losing energy in the transformer (inefficiencies), or gaining energy (perpetual motion anyone?!). The former can happen, the latter can't.
So based on this, what can we say about the voltage and current? Well, we know that:
$$P = I V$$
So: $$I_{in}V_{in} = I_{out}V_{out}$$
Lets say that you have a step up transformer with 10 turns on the primary and 50 turns on the secondary. This means you have a turns ratio of:
$$n = \frac{50}{10} = 5$$
So that means that the voltage will be increased by a factor of 5 (\$V_{out} = 5\times V_{in}\$). So what happens to the current?
$$I_{in}V_{in} = 5V_{in}\times(I_{out})$$
In order for both sides of that to stay equal (can't get energy from nothing!), then the current must be divided by 5. Basically you can say that:
$$I_{out} = \frac{1}{n}I_{in}\space\space\space\space\space\space\space\space\space V_{out} = nV_{in}$$
So what happens if you have a fixed load and change the number of turns? Lets do an example. We shall say that the input voltage is \$10\mathrm{V}\$, the transformer initially steps up by a factor of \$n=1\$, and then later by a factor of \$n=5\$. In both cases the output load is a \$2\Omega\$ resistor.
In the first case, your calculations are correct.
$$V_{out} = n\times V_{in} = 10\mathrm{V}$$ $$I_{out} = \frac{V_{out}}{R_L} = \frac{10}{2} = 5\mathrm{A}$$ $$I_{in} = n\times I_{out} = 5\mathrm{A}$$
Now lets go for \$n=5\$.
$$V_{out} = n\times V_{in} = 5\times 10 = 50\mathrm{V}$$ $$I_{out} = \frac{V_{out}}{R_L} = \frac{50}{2} = 25\mathrm{A}$$
Great, these match what you are saying. But this is where everything changes. We do the last step of the calculation:
$$I_{in} = n\times I_{out} = 5\times 25 = 125\mathrm{A}$$
Ahh, there we go. Notice that the input current goes up, quite considerably. This makes the scales balance so to speak - power in goes up in order to cope with the large power requirements of the load.
A transformer can not create power so a step-up in a sense both increases the current and reduces it.
If we have 10V ac supply and connect a 10 ohm resistor across it we will have 1 amp flowing in the resistor. If we now disconnect and replace it with a 2:1 step-up transformer connecting the same 10 ohm resistor across the secondary then the resistor will have 20 volts across it so will have 2 amps flowing in the resistor. Thus the current in the resistor has increased as you have pointed out.
However, this is not the sense in which we mean a step-up transformer reduces the current. If we consider the power in the resistor in our second case we have 2 amps and 20 volts making a total power of 40 watts. We therefore need at least 40 watts to be flowing into the primary. This means the current into the primary has to be at least 4 amps because we only have a 10 volt supply. In practice we will have slightly more than this because no transformer is 100% efficient, there are conduction losses in the windings and some power is needed to magnetise the core but the current may only be slightly higher than this as efficiencies in excess of 90% are easily achievable.
When we say a step-up transformer reduces the current we mean we have less current in the secondary than we do in the primary.