Heisenberg's Uncertainty Principle I have heard various definitions of the uncertainty principle.
Yet I cannot quite comprehend how it is true. Nevertheless, something tells me, it is a consequence of the wave nature of light/electron which gives the intrinsic nature of uncertainty even if we don't measure it.
Is it true that this principle is a consequence of wave nature of particle, that the uncertainty pops up due the fact that particle acts as a wave(I find no answer which stated the exact implication of the wave characteristics which should give the uncertainty principle)?
Will it be true to assume that, if an electron acts only like a particle and not as a wave, the uncertainty principle will not be necessary(this part of the question is not asked anywhere)? 
Can you please tell me without much mathematics why this is so?
Like we understood the photoelectric effect contradicts the wave nature of light,
Could you please guide me the intuitive explanation with formal reason why we cannot absolutely know simultaneously the position and momentum of a particle?
 A: Yes, in a sense one can say that it is the wave nature of `particles' that inevitably led to the Heisenberg uncertainty principle. The original discovery of Planck, which related energy and frequency, followed by the equivalent relationship between the momentum and the propagation vector, implied that one can take the Fourier transform of the space-time distribution of a particle to get the information of its energy and momentum. While in classical physics momentum and energy are considered as information that is additional to the position information of a particle, in quantum physics we now understand that this represents the same information. One can either express the particle's wave function in position space or in momentum space. Both represent the same information.
Now, due to the properties of this relationship between the two spaces, one has the situation that if the distribution in position space is very localized, then the corresponding distribution in momentum space would be spread out, and vise versa. The products of the widths of the distributions in the two respective spaces can never be smaller than a certain quantity. This is the Heisenberg uncertainty principle.
The reason why the Fourier transform leads to this relationship is because of the different bases that are associated with the two spaces. These different bases are said to be mutually unbiased. What this means is that the magnitude of the overlap of any pair of elements of the two basis is always the same. Therefore if I have a distribution that is represented by only one of the elements of one of the bases then its representation in terms of the other basis would necessarily be spread over all the elements of that basis. This is a property that can go beyond the Fourier transform. Therefore one can find that the Heisenberg uncertainty principle also applies in situation where the different spaces are not related by a Fourier transform.
A: Contrary to some of the other answers, I think it can be quite misleading to say that the uncertainty principle relies on waves, because the uncertainty principle holds even when the state space is finite dimensional, in which case one might choose to call a state a "wave" by way of analogy, but it is certainly not a wave in the sense that the OP seems to have in mind.
Instead, the uncertainty principle is driven by the non-commutativity of the algebra of observables, quite independently of what those observables are acting on, and in particular, quite independently of whether they are acting on anything the OP would want to call a wave.
If the question is "How do I understand the uncertainty principle with a minimum of math?" then surely the answer is:  By understanding the uncertainty principle in the case of a 2-dimensional state space.  That minimizes the math, but doesn't eliminate it.  Unless you're willing to master some two-dimensional linear algebra, you're not going to understand the uncertainty principle.
If you want to eliminate math altogether, then the best answer is that an electron does not (usually) have a definite position for exactly the same reason that an electron does not have a favorite movie --- these are just not properties that electrons have.  If you want to understand what properties they do have, then it's back to the math. 
A: From the comments, you seem to want the minimum possible math. There are 4 things you have to know first:
First, what you have to know is that a basic quantum wavefunction can be imagined as exactly just a sine wave:

Second, you should know that the amplitude of the wave across an interval is related to the probability of measuring your particle's position within that interval. (This is an approximate analogy of what a probability density function does.)
Third, the wavelength of the wave is related to your particle's measured momentum. (If we want to be strict, it should be the frequency and it should also be a probability across an interval in frequency space, but it helps to imagine it with just a wavelength.)
Fourth, you can compose a more complicated quantum wavefunction just by adding together waves of different wavelengths. (This is called superposition -- see this gif: 

(Image from Wikipedia)

Now that you know these four things, we're ready to tackle the idea of Heisenberg's uncertainty principle.
Note the 4th thing we said (re: superposition). Take a look at the gif. What do you notice? When we add more and more waves of different wavelengths, a prominent central peak starts to appear.
Now remember the 2nd thing we said: amplitude is related to position. If we have a peak with a prominent amplitude, our particle's position becomes more likely to be measured within that peak. The more we make the central peak prominent, the more precisely we can predict the particle's position!
However, to make the central peak more prominent, we have to keep adding more waves of different wavelengths. Remember the 3rd thing we said? Wavelength is related to momentum. If we keep adding different wavelengths, we expect a larger range for our momentum to be measured in, which means our particle's momentum cannot be predicted as easily. The more we add waves of different wavelengths, the less precisely we can predict the particle's momentum!

And therein lies the heart of the uncertainty principle: if you try to measure position more precisely, you will consequently measure momentum less precisely, and vice versa.
So to answer your question: yes, the uncertainty principle is a necessary consequence of the 'wave-nature of particles'.
And to answer your second question (thank you for bringing it up in the comments!): yes, if the electron were a particle instead of a quantum mechanical object, the uncertainty principle wouldn't be necessary, or at least wouldn't necessarily apply. This is because the 4 basic concepts behind the uncertainty principle are uniquely wave concepts, especially the 2nd and 3rd concepts which are uniquely quantum mechanical wavefunction concepts, neither of which apply to particles.
A: Lagrange and Poisson brackets: A step back to the Classical realms.
Consider two variables $q_i, p_i$ given as a function of two parameters $u,v\,.$ The the Lagrange bracket is given by $$[~u,\, v~] ~= ~ \sum_{i~=~1}^n \left(\frac{\partial q_i}{\partial u}\frac{\partial p_i}{\partial v}- \frac{\partial q_i}{\partial v}\frac{\partial p_i}{\partial u}\right)\,.\tag{I} $$
Now, transform the variables $q_i, p_i$ to $Q_i, P_i$ such that the Lagrange bracket in the new variables remain invariant. This is known as canonical transformation.
Let the old variables be assumed to be expressed in terms of new variables in an explicit form as:
\begin{align}q_i &= f_i(Q_1,\ldots Q_n; P_1, \ldots, P_n)\\ p_i & = f^{\prime}_i(Q_1,\ldots Q_n; P_1, \ldots, P_n)\end{align}
Any pair of $Q_i, Q_k$ or $P_i, P_k$ or $Q_i, P_k$ can be replaced with $u,v$ in $\rm (I)$ considering the other variables constant.
Since, in the new coordinate system, $Q_i, P_i$ are independent of each other, from $\rm (I)$ we get:
$$[~Q_i, ~Q_k~] = 0;\qquad[~P_i, ~P_k~] = 0;\qquad[~Q_i, ~P_k~] = \delta_{ik}\,. \tag{II}$$
Now, consider $u,v$ as functions of $q_i, p_i$ as:
\begin{align}u &= u(q_1,\ldots q_n; p_1, \ldots, p_n)\\ v & = v(q_1,\ldots q_n; p_1, \ldots, p_n)\end{align}
Then the Poisson bracket is given by
$$(~u,\, v~) ~= ~ \sum_{i~=~1}^n \left(\frac{\partial u}{\partial q_i}\frac{\partial v}{\partial p_i}- \frac{\partial v}{\partial q_i}\frac{\partial u}{\partial p_i}\right)\,.\tag{III} $$
We can have $u_1, u_2, \ldots, u_{2n}$ expressed as functions of $q_i, p_i;$ alternatively $q_i, p_i$ can be expressed as functions of $u_1,u_2, \ldots, u_{2n}\,.$ We can form Poisson bracket for the first case while for the second case, we can form Lagrange bracket; thus they are related to each other. If the Lagrange bracket is invariant of an explicit transformation, then so is the Poisson bracket. The canonical transformation leaves the Poisson bracket invariant irrespective of how $u,v$ depend on $q_i, p_i\,.$
Now, express $Q_i, P_i$ in terms of old coordinates $q_i, p_i$ as:
\begin{align}Q_i &= F_i(q_1,\ldots q_n; p_1, \ldots, p_n)\\ P_i & = F^{\prime}_i(q_1,\ldots q_n; p_1, \ldots, p_n)\end{align}
Form the Poisson bracket in the new as well as the old coordinates.
By the property of invariance as explained above, 
$$(~Q_i, ~Q_k~) = 0;\qquad(~P_i, ~P_k~) = 0;\qquad(~Q_i, ~P_k~) = \delta_{ik}\,. \tag{IV}$$
Commutativity of Operators: Advent of Quantum Mechanics.
In QM, observables are represented by (Hermitian) operators. For a pair of operators $A$ and $B,$ the commutator bracket is given by $$[~A,~B~] \equiv AB -BA\tag{V}$$
It measures to what extent the operators are commutative to each other. 
Two different observables with operators $A$ and $B$ have definite values if the wave function is an eigenfunction of both $A$ and $B$. So, the question whether two quantities can be definite at the same time is really whether their operators $A$ and $B$ have common eigenfunctions.
That is

Iff two Hermitian operators commute, there is a complete set of eigenfunctions that is common to them both1. 

Observables with operators that do not commute cannot in general have definite values at the same time. If one has a definite value, the other is in general uncertain.
Now, the analogs of the classical equations of motion in QM can be found by substituting the commutator brackets divided by $\mathrm i\hslash$ for the Poisson bracket viz. $$(~A, B~) \rightarrow \frac1{\mathrm i\hslash}~[~A,~B~]\tag{VI}$$
Which means the classical relations imply
$$[~q_i, ~q_k~] = 0;\qquad[~p_i, ~p_k~] = 0;\qquad[~q_i, ~p_k~] = \mathrm i\hslash~\delta_{ik}\,.\tag{VII} $$
Uncertainty Principle:
Standard Deviation of operator $A$ is given by
$$\sigma_A = \sqrt{\langle A^2\rangle - \langle A\rangle ^2}$$ where $\langle \quad\rangle$ is the expectation value of the observable in the concerned state.
For an observable represented by operator $A,$  the variance of $A$ on a certain state $\Psi$  is given by
$$\sigma^2_A = \langle (A - \langle A\rangle)\Psi|(A - \langle A\rangle)\Psi\rangle\,.$$
Similarly, for observable represented by operator $B,$ we have $$\sigma^2_B = \langle (B - \langle B\rangle)\Psi|(B - \langle B\rangle)\Psi\rangle\,.$$
Therefore, \begin{align}\sigma^2_A\sigma^2_B &=\langle (A - \langle A\rangle)\Psi|(A - \langle A\rangle)\Psi\rangle\langle (B - \langle B\rangle)\Psi|(B - \langle B\rangle)\Psi\rangle\\ & \geqq \left|\langle(A - \langle A\rangle)\Psi|(B - \langle B\rangle)\Psi\rangle\right|^2~~~~~~ \textrm{Cauchy-Schwarz Inequality}\,.\tag{VIII} \end{align}
For a complex number $z,$ $$|z|^2 = ({\rm Re(z)})^2+ ({\rm Im(z) })^2 \geqq ({\rm Im(z) })^2  = \left(\frac{1}{2\mathrm i}~\left(z-z^\dagger\right)\right)^2$$ where $z^\dagger$ is the complex conjugate of $z\,.$
Take $z$ to be $\langle(A - \langle A\rangle)\Psi|(B - \langle B\rangle)\Psi\rangle;$ this means $z^\dagger =\langle(B - \langle B\rangle)\Psi|(A - \langle A\rangle)\Psi\rangle\,. $
So, we can write $\rm (VIII)$ as $$\sigma^2_A\sigma^2_B\geqq \left(\frac{1}{2\mathrm i}~\left(\langle(A - \langle A\rangle)\Psi|(B - \langle B\rangle)\Psi\rangle- \langle(B - \langle B\rangle)\Psi|(A - \langle A\rangle)\Psi\rangle\right)\right)^2\,.$$
Compute the terms; at the end, it would be found out that $$\langle(A - \langle A\rangle)\Psi|(B - \langle B\rangle)\Psi\rangle- \langle(B - \langle B\rangle)\Psi|(A - \langle A\rangle)\Psi\rangle =\langle [~A,~B~]\rangle \,.$$
And, thus it can be concluded $$\underbrace{\sigma^2_A\sigma^2_B\geqq \left(\frac{1}{2\mathrm i}~\left(\langle [~A,~B~]\rangle\right)\right)^2}_\textrm{Generalised Uncertainty Principle}\,.\tag{G.U.P}$$
Now, from $\rm(VII),$ we know $[~x, ~p_x~] = \mathrm i\hslash$ where $x$ and $p_x$ are position and momentum operators; using this and the Generalised Uncertainty Principle, we get the very familiar position-momentum Uncertainty Principle:
$$\sigma^2_x\sigma^2_{p_x} \geqq \left(\frac\hslash 2\right)^2\;.\tag{IX}$$
Uncertainty Principle is a general result; it is solely due to commutativity of operators and not due to any wave or particle nature.
Uncertainty Principle relies on the sole fact that two operators need not be commutative with each other. 
Pair of position and momentum operators is but one of the many pairs of operators which do not commute with each other and thus follows $\rm(G.U.P)\,.$
Relevance of the wave picture in the Uncertainty Principle.

If $f$ and $F$ are Fourier transforms, than the widths of the graphs of $|f(x)|^2$ and $|F(x)|^2$ cannot be both made arbitrarily small.

There is bandwidth theorem which states that product of length of a wave packet/ wave group $\Delta x$ with the corresponding band $\Delta k$ of wave-numbers cannot be made arbitrarily small simultaneously. Precisely, $$\Delta k \Delta x\approx \Delta \omega\Delta t \geqq 2\pi,$$ where $\Delta \omega$ is a band of angular frequencies of the wave packet and $\Delta t$ is the time taken by the packet to pass a fixed point with group velocity $v_g$  using the relation $\Delta k = \dfrac{\Delta \omega}{v_g}\,.$
Using the De-Broglie relation, we get again $$\Delta x\Delta p_x \geqq \frac{\hslash}2\,.\tag{X}$$
This is true because the position and momentum are Fourier conjugate to each other. 
Conclusion:

Is it true that this principle is a consequence of wave nature of particle, that the uncertainty pops up due the fact that particle acts as a wave?

Yes and no. 
Surely the Uncertainty Principle can be derived from the fact that quantum particles have waves associated with them.
But the Uncertainty Principle is a much more general result.
It is solely due to the commutativity of operators. 
As not all pairs of operators need to be commutative.

References:
$\bullet$ The Variational Principles of Mechanics by Cornelius Lanczos.
$\bullet$ Quantum Mechanics by Leonard I. Schiff.
$\bullet$ Introduction to Quantum Mechanics by David J. Griffiths.
$\bullet$ Waves by Frank S. Crawford Jr.
$\bullet$ Uncertainty Principle - Wikipedia.
A: 
I have heard various definitions of the uncertainty principle. Yet I cannot quite comprehend how it is true. 

It's true. As kpv said, see Wikipedia: "It has since become clear, however, that the uncertainty principle is inherent in the properties of all wave-like systems,[8] and that it arises in quantum mechanics simply due to the matter wave nature of all quantum objects". 

Nevertheless, something tells me, it is a consequence of the wave nature of light/electron which gives the intrinsic nature of uncertainty even if we don't measure it.

Correct. Remember that the photon has an E=hc/λ wave nature. So does the electron. We make electrons and positrons out of photons in pair production. The electron is a particle, but it is not a point particle. Its field is what it is. In atomic orbitals electrons "exist as standing waves". Guess what? Electrons exist as standing waves outside of atomic orbitals. That's why we can diffract electrons. Standing wave, standing field.  

Is it true that this principle is a consequence of wave nature of particle, that the uncertainty pops up due the fact that particle acts as a wave?

Yes. The point to remember is that detection involves an interaction between a wave and a wave. Neither is a point particle, they're both extended entities with no surface or edge. When the interaction occurs the wave energy tends to get localised, but it wasn't localised before that. IMHO the optical Fourier transform is useful for getting a handle on this. Two extended entities interact and what you see is some little dots:

See Steven Lehar's web page for more. And note that the input image isn't some little dot. Nor is the lens. But when they interact, that's what you see, because you're seeing a real-time Fourier transform. It isn't just some abstract mathematical thing, and neither the photon or the electron is a dot.  

Will it be true to assume that, if an electron acts only like a particle and not as a wave, the uncertainty principle will not be necessary? 

No, because the particle is the wave. There are no point-particles. The point-particle is a mathematical myth that has somehow grown legs, and has been around for so long that people take it for granted.  

Can you please tell me without much mathematics why this is so?

It's because everything is fields and waves. Because it's quantum field theory, not quantum point-particle theory. 

Like we understood the photoelectric effect contradicts the wave nature of light.

It doesn't. It demonstrates the quantum nature of light. See Leonard Susskind talking about the Higgs boson in this video. At 2 minutes 50 he rolls his whiteboard marker saying photon angular momentum is quantized. The photon is a wave with this special feature, not a point particle. 

Could you please guide me the intuitive explanation with formal reason why we cannot absolutely know simultaneously the position and momentum of a particle?

Because it doesn't have a position, because it's a spread-out extended-entity wave which doesn't have a surface or an edge. When you try to measure its momentum the only tool in the box is more of the same. Then you change it, you make it bunch up, and what you see is not what it really is. 
A: Yes, it is due to the wave nature as mentioned "that it arises in quantum mechanics simply due to the matter wave nature of all quantum objects" on https://en.wikipedia.org/wiki/Uncertainty_principle.
It makes sense that for us, the observers, uncertainty is natural at quantum levels. However, some think that even nature does not know the values with certainty. I do not think that is true. If nature did not know with certainty, then the  conservation laws would be violated - irrecoverably, not just intermittently.
