-0.000 000 000 742 147 676 646 709 14 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 709 14(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
-0.000 000 000 742 147 676 646 709 14(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:

|-0.000 000 000 742 147 676 646 709 14| = 0.000 000 000 742 147 676 646 709 14


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: 0.000 000 000 742 147 676 646 709 14.

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) 0.000 000 000 742 147 676 646 709 14 × 2 = 0 + 0.000 000 001 484 295 353 293 418 28;
  • 2) 0.000 000 001 484 295 353 293 418 28 × 2 = 0 + 0.000 000 002 968 590 706 586 836 56;
  • 3) 0.000 000 002 968 590 706 586 836 56 × 2 = 0 + 0.000 000 005 937 181 413 173 673 12;
  • 4) 0.000 000 005 937 181 413 173 673 12 × 2 = 0 + 0.000 000 011 874 362 826 347 346 24;
  • 5) 0.000 000 011 874 362 826 347 346 24 × 2 = 0 + 0.000 000 023 748 725 652 694 692 48;
  • 6) 0.000 000 023 748 725 652 694 692 48 × 2 = 0 + 0.000 000 047 497 451 305 389 384 96;
  • 7) 0.000 000 047 497 451 305 389 384 96 × 2 = 0 + 0.000 000 094 994 902 610 778 769 92;
  • 8) 0.000 000 094 994 902 610 778 769 92 × 2 = 0 + 0.000 000 189 989 805 221 557 539 84;
  • 9) 0.000 000 189 989 805 221 557 539 84 × 2 = 0 + 0.000 000 379 979 610 443 115 079 68;
  • 10) 0.000 000 379 979 610 443 115 079 68 × 2 = 0 + 0.000 000 759 959 220 886 230 159 36;
  • 11) 0.000 000 759 959 220 886 230 159 36 × 2 = 0 + 0.000 001 519 918 441 772 460 318 72;
  • 12) 0.000 001 519 918 441 772 460 318 72 × 2 = 0 + 0.000 003 039 836 883 544 920 637 44;
  • 13) 0.000 003 039 836 883 544 920 637 44 × 2 = 0 + 0.000 006 079 673 767 089 841 274 88;
  • 14) 0.000 006 079 673 767 089 841 274 88 × 2 = 0 + 0.000 012 159 347 534 179 682 549 76;
  • 15) 0.000 012 159 347 534 179 682 549 76 × 2 = 0 + 0.000 024 318 695 068 359 365 099 52;
  • 16) 0.000 024 318 695 068 359 365 099 52 × 2 = 0 + 0.000 048 637 390 136 718 730 199 04;
  • 17) 0.000 048 637 390 136 718 730 199 04 × 2 = 0 + 0.000 097 274 780 273 437 460 398 08;
  • 18) 0.000 097 274 780 273 437 460 398 08 × 2 = 0 + 0.000 194 549 560 546 874 920 796 16;
  • 19) 0.000 194 549 560 546 874 920 796 16 × 2 = 0 + 0.000 389 099 121 093 749 841 592 32;
  • 20) 0.000 389 099 121 093 749 841 592 32 × 2 = 0 + 0.000 778 198 242 187 499 683 184 64;
  • 21) 0.000 778 198 242 187 499 683 184 64 × 2 = 0 + 0.001 556 396 484 374 999 366 369 28;
  • 22) 0.001 556 396 484 374 999 366 369 28 × 2 = 0 + 0.003 112 792 968 749 998 732 738 56;
  • 23) 0.003 112 792 968 749 998 732 738 56 × 2 = 0 + 0.006 225 585 937 499 997 465 477 12;
  • 24) 0.006 225 585 937 499 997 465 477 12 × 2 = 0 + 0.012 451 171 874 999 994 930 954 24;
  • 25) 0.012 451 171 874 999 994 930 954 24 × 2 = 0 + 0.024 902 343 749 999 989 861 908 48;
  • 26) 0.024 902 343 749 999 989 861 908 48 × 2 = 0 + 0.049 804 687 499 999 979 723 816 96;
  • 27) 0.049 804 687 499 999 979 723 816 96 × 2 = 0 + 0.099 609 374 999 999 959 447 633 92;
  • 28) 0.099 609 374 999 999 959 447 633 92 × 2 = 0 + 0.199 218 749 999 999 918 895 267 84;
  • 29) 0.199 218 749 999 999 918 895 267 84 × 2 = 0 + 0.398 437 499 999 999 837 790 535 68;
  • 30) 0.398 437 499 999 999 837 790 535 68 × 2 = 0 + 0.796 874 999 999 999 675 581 071 36;
  • 31) 0.796 874 999 999 999 675 581 071 36 × 2 = 1 + 0.593 749 999 999 999 351 162 142 72;
  • 32) 0.593 749 999 999 999 351 162 142 72 × 2 = 1 + 0.187 499 999 999 998 702 324 285 44;
  • 33) 0.187 499 999 999 998 702 324 285 44 × 2 = 0 + 0.374 999 999 999 997 404 648 570 88;
  • 34) 0.374 999 999 999 997 404 648 570 88 × 2 = 0 + 0.749 999 999 999 994 809 297 141 76;
  • 35) 0.749 999 999 999 994 809 297 141 76 × 2 = 1 + 0.499 999 999 999 989 618 594 283 52;
  • 36) 0.499 999 999 999 989 618 594 283 52 × 2 = 0 + 0.999 999 999 999 979 237 188 567 04;
  • 37) 0.999 999 999 999 979 237 188 567 04 × 2 = 1 + 0.999 999 999 999 958 474 377 134 08;
  • 38) 0.999 999 999 999 958 474 377 134 08 × 2 = 1 + 0.999 999 999 999 916 948 754 268 16;
  • 39) 0.999 999 999 999 916 948 754 268 16 × 2 = 1 + 0.999 999 999 999 833 897 508 536 32;
  • 40) 0.999 999 999 999 833 897 508 536 32 × 2 = 1 + 0.999 999 999 999 667 795 017 072 64;
  • 41) 0.999 999 999 999 667 795 017 072 64 × 2 = 1 + 0.999 999 999 999 335 590 034 145 28;
  • 42) 0.999 999 999 999 335 590 034 145 28 × 2 = 1 + 0.999 999 999 998 671 180 068 290 56;
  • 43) 0.999 999 999 998 671 180 068 290 56 × 2 = 1 + 0.999 999 999 997 342 360 136 581 12;
  • 44) 0.999 999 999 997 342 360 136 581 12 × 2 = 1 + 0.999 999 999 994 684 720 273 162 24;
  • 45) 0.999 999 999 994 684 720 273 162 24 × 2 = 1 + 0.999 999 999 989 369 440 546 324 48;
  • 46) 0.999 999 999 989 369 440 546 324 48 × 2 = 1 + 0.999 999 999 978 738 881 092 648 96;
  • 47) 0.999 999 999 978 738 881 092 648 96 × 2 = 1 + 0.999 999 999 957 477 762 185 297 92;
  • 48) 0.999 999 999 957 477 762 185 297 92 × 2 = 1 + 0.999 999 999 914 955 524 370 595 84;
  • 49) 0.999 999 999 914 955 524 370 595 84 × 2 = 1 + 0.999 999 999 829 911 048 741 191 68;
  • 50) 0.999 999 999 829 911 048 741 191 68 × 2 = 1 + 0.999 999 999 659 822 097 482 383 36;
  • 51) 0.999 999 999 659 822 097 482 383 36 × 2 = 1 + 0.999 999 999 319 644 194 964 766 72;
  • 52) 0.999 999 999 319 644 194 964 766 72 × 2 = 1 + 0.999 999 998 639 288 389 929 533 44;
  • 53) 0.999 999 998 639 288 389 929 533 44 × 2 = 1 + 0.999 999 997 278 576 779 859 066 88;
  • 54) 0.999 999 997 278 576 779 859 066 88 × 2 = 1 + 0.999 999 994 557 153 559 718 133 76;

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:


0.000 000 000 742 147 676 646 709 14(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 709 14(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:


0.000 000 000 742 147 676 646 709 14(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 -0.000 000 000 742 147 676 646 709 14 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 646 708 33 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: Reputation: Bridge Between Company's Actions and Market Value. NotebookLM YouTube Video of 6.55 minutes

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