What is the truth value of this statement about numbers?

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Homework Statement


Determine the truth value of the statement.

There exists x for every (xsquared smaller than y+1)

x and y a set of non-negative real numbers

Sorry i don't know how to do the maths symbols on the computer.


Homework Equations



All the numbers i have used make the statement true. The teacher said to try and proove the statement false, rather than prooving it correct.

Another teacher said it is to do with fractions.


The Attempt at a Solution



As above
 
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Is that the question as it was given? Is the question;

Show that there exists a value of x such that for every value of y, x^2 < y+1 ?

If that is the question, the statement is actually true.
 
Ah ok.


How would i prove this?

I looked at a few of the proven workings but nothing resembles a similar question?
 
To prove it: Since y must be a non-negative real number, the smallest the RHS can be is 1. So every single time it suffices to choose a value of x^2 less than 1, which is a value of x less than 1.
 
In fact any number x < \sqrt{y + 1} would do. If you'd want to be dull you could just take x = 0 for all y.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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