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Let $P(p)$ = "there is no natural $q$ such that $(p/q)^2=2$". A simple induction argument shows that P holds for all naturals $p$ and hence that 2$\sqrt 2$ is irrational. All descent arguments are basically induction.
Let $P(p)$ = "there is no natural $q$ such that $(p/q)^2=2$". A simple induction argument shows that P holds for all naturals $p$ and hence that 2 is irrational. All descent arguments are basically induction.
Let $P(p)$ = "there is no natural $q$ such that $(p/q)^2=2$". A simple induction argument shows that P holds for all naturals $p$ and hence that $\sqrt 2$ is irrational. All descent arguments are basically induction.
Let $P(p)$ = "there is no natural $q$ such that $(p/q)^2=2$". A simple induction argument shows that P holds for all naturals $p$ and hence that 2 is irrational. All descent arguments are basically induction.
