-0.000 000 000 742 147 676 646 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 6(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 742 147 676 646 6(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 742 147 676 646 6| = 0.000 000 000 742 147 676 646 6


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 742 147 676 646 6.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 000 000 742 147 676 646 6 × 2 = 0 + 0.000 000 001 484 295 353 293 2;
  • 2) 0.000 000 001 484 295 353 293 2 × 2 = 0 + 0.000 000 002 968 590 706 586 4;
  • 3) 0.000 000 002 968 590 706 586 4 × 2 = 0 + 0.000 000 005 937 181 413 172 8;
  • 4) 0.000 000 005 937 181 413 172 8 × 2 = 0 + 0.000 000 011 874 362 826 345 6;
  • 5) 0.000 000 011 874 362 826 345 6 × 2 = 0 + 0.000 000 023 748 725 652 691 2;
  • 6) 0.000 000 023 748 725 652 691 2 × 2 = 0 + 0.000 000 047 497 451 305 382 4;
  • 7) 0.000 000 047 497 451 305 382 4 × 2 = 0 + 0.000 000 094 994 902 610 764 8;
  • 8) 0.000 000 094 994 902 610 764 8 × 2 = 0 + 0.000 000 189 989 805 221 529 6;
  • 9) 0.000 000 189 989 805 221 529 6 × 2 = 0 + 0.000 000 379 979 610 443 059 2;
  • 10) 0.000 000 379 979 610 443 059 2 × 2 = 0 + 0.000 000 759 959 220 886 118 4;
  • 11) 0.000 000 759 959 220 886 118 4 × 2 = 0 + 0.000 001 519 918 441 772 236 8;
  • 12) 0.000 001 519 918 441 772 236 8 × 2 = 0 + 0.000 003 039 836 883 544 473 6;
  • 13) 0.000 003 039 836 883 544 473 6 × 2 = 0 + 0.000 006 079 673 767 088 947 2;
  • 14) 0.000 006 079 673 767 088 947 2 × 2 = 0 + 0.000 012 159 347 534 177 894 4;
  • 15) 0.000 012 159 347 534 177 894 4 × 2 = 0 + 0.000 024 318 695 068 355 788 8;
  • 16) 0.000 024 318 695 068 355 788 8 × 2 = 0 + 0.000 048 637 390 136 711 577 6;
  • 17) 0.000 048 637 390 136 711 577 6 × 2 = 0 + 0.000 097 274 780 273 423 155 2;
  • 18) 0.000 097 274 780 273 423 155 2 × 2 = 0 + 0.000 194 549 560 546 846 310 4;
  • 19) 0.000 194 549 560 546 846 310 4 × 2 = 0 + 0.000 389 099 121 093 692 620 8;
  • 20) 0.000 389 099 121 093 692 620 8 × 2 = 0 + 0.000 778 198 242 187 385 241 6;
  • 21) 0.000 778 198 242 187 385 241 6 × 2 = 0 + 0.001 556 396 484 374 770 483 2;
  • 22) 0.001 556 396 484 374 770 483 2 × 2 = 0 + 0.003 112 792 968 749 540 966 4;
  • 23) 0.003 112 792 968 749 540 966 4 × 2 = 0 + 0.006 225 585 937 499 081 932 8;
  • 24) 0.006 225 585 937 499 081 932 8 × 2 = 0 + 0.012 451 171 874 998 163 865 6;
  • 25) 0.012 451 171 874 998 163 865 6 × 2 = 0 + 0.024 902 343 749 996 327 731 2;
  • 26) 0.024 902 343 749 996 327 731 2 × 2 = 0 + 0.049 804 687 499 992 655 462 4;
  • 27) 0.049 804 687 499 992 655 462 4 × 2 = 0 + 0.099 609 374 999 985 310 924 8;
  • 28) 0.099 609 374 999 985 310 924 8 × 2 = 0 + 0.199 218 749 999 970 621 849 6;
  • 29) 0.199 218 749 999 970 621 849 6 × 2 = 0 + 0.398 437 499 999 941 243 699 2;
  • 30) 0.398 437 499 999 941 243 699 2 × 2 = 0 + 0.796 874 999 999 882 487 398 4;
  • 31) 0.796 874 999 999 882 487 398 4 × 2 = 1 + 0.593 749 999 999 764 974 796 8;
  • 32) 0.593 749 999 999 764 974 796 8 × 2 = 1 + 0.187 499 999 999 529 949 593 6;
  • 33) 0.187 499 999 999 529 949 593 6 × 2 = 0 + 0.374 999 999 999 059 899 187 2;
  • 34) 0.374 999 999 999 059 899 187 2 × 2 = 0 + 0.749 999 999 998 119 798 374 4;
  • 35) 0.749 999 999 998 119 798 374 4 × 2 = 1 + 0.499 999 999 996 239 596 748 8;
  • 36) 0.499 999 999 996 239 596 748 8 × 2 = 0 + 0.999 999 999 992 479 193 497 6;
  • 37) 0.999 999 999 992 479 193 497 6 × 2 = 1 + 0.999 999 999 984 958 386 995 2;
  • 38) 0.999 999 999 984 958 386 995 2 × 2 = 1 + 0.999 999 999 969 916 773 990 4;
  • 39) 0.999 999 999 969 916 773 990 4 × 2 = 1 + 0.999 999 999 939 833 547 980 8;
  • 40) 0.999 999 999 939 833 547 980 8 × 2 = 1 + 0.999 999 999 879 667 095 961 6;
  • 41) 0.999 999 999 879 667 095 961 6 × 2 = 1 + 0.999 999 999 759 334 191 923 2;
  • 42) 0.999 999 999 759 334 191 923 2 × 2 = 1 + 0.999 999 999 518 668 383 846 4;
  • 43) 0.999 999 999 518 668 383 846 4 × 2 = 1 + 0.999 999 999 037 336 767 692 8;
  • 44) 0.999 999 999 037 336 767 692 8 × 2 = 1 + 0.999 999 998 074 673 535 385 6;
  • 45) 0.999 999 998 074 673 535 385 6 × 2 = 1 + 0.999 999 996 149 347 070 771 2;
  • 46) 0.999 999 996 149 347 070 771 2 × 2 = 1 + 0.999 999 992 298 694 141 542 4;
  • 47) 0.999 999 992 298 694 141 542 4 × 2 = 1 + 0.999 999 984 597 388 283 084 8;
  • 48) 0.999 999 984 597 388 283 084 8 × 2 = 1 + 0.999 999 969 194 776 566 169 6;
  • 49) 0.999 999 969 194 776 566 169 6 × 2 = 1 + 0.999 999 938 389 553 132 339 2;
  • 50) 0.999 999 938 389 553 132 339 2 × 2 = 1 + 0.999 999 876 779 106 264 678 4;
  • 51) 0.999 999 876 779 106 264 678 4 × 2 = 1 + 0.999 999 753 558 212 529 356 8;
  • 52) 0.999 999 753 558 212 529 356 8 × 2 = 1 + 0.999 999 507 116 425 058 713 6;
  • 53) 0.999 999 507 116 425 058 713 6 × 2 = 1 + 0.999 999 014 232 850 117 427 2;
  • 54) 0.999 999 014 232 850 117 427 2 × 2 = 1 + 0.999 998 028 465 700 234 854 4;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 000 000 742 147 676 646 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 646 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 742 147 676 646 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(2) × 2-31


8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -31

Mantissa (not normalized):
1.1001 0111 1111 1111 1111 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(10)


10. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

96(10) =

0110 0000(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 100 1011 1111 1111 1111 1111 =

100 1011 1111 1111 1111 1111


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0110 0000

Mantissa (23 bits) =
100 1011 1111 1111 1111 1111


Decimal number -0.000 000 000 742 147 676 646 6 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111