-0.000 000 000 742 147 676 182 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 182(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
-0.000 000 000 742 147 676 182(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:

|-0.000 000 000 742 147 676 182| = 0.000 000 000 742 147 676 182


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: 0.000 000 000 742 147 676 182.

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) 0.000 000 000 742 147 676 182 × 2 = 0 + 0.000 000 001 484 295 352 364;
  • 2) 0.000 000 001 484 295 352 364 × 2 = 0 + 0.000 000 002 968 590 704 728;
  • 3) 0.000 000 002 968 590 704 728 × 2 = 0 + 0.000 000 005 937 181 409 456;
  • 4) 0.000 000 005 937 181 409 456 × 2 = 0 + 0.000 000 011 874 362 818 912;
  • 5) 0.000 000 011 874 362 818 912 × 2 = 0 + 0.000 000 023 748 725 637 824;
  • 6) 0.000 000 023 748 725 637 824 × 2 = 0 + 0.000 000 047 497 451 275 648;
  • 7) 0.000 000 047 497 451 275 648 × 2 = 0 + 0.000 000 094 994 902 551 296;
  • 8) 0.000 000 094 994 902 551 296 × 2 = 0 + 0.000 000 189 989 805 102 592;
  • 9) 0.000 000 189 989 805 102 592 × 2 = 0 + 0.000 000 379 979 610 205 184;
  • 10) 0.000 000 379 979 610 205 184 × 2 = 0 + 0.000 000 759 959 220 410 368;
  • 11) 0.000 000 759 959 220 410 368 × 2 = 0 + 0.000 001 519 918 440 820 736;
  • 12) 0.000 001 519 918 440 820 736 × 2 = 0 + 0.000 003 039 836 881 641 472;
  • 13) 0.000 003 039 836 881 641 472 × 2 = 0 + 0.000 006 079 673 763 282 944;
  • 14) 0.000 006 079 673 763 282 944 × 2 = 0 + 0.000 012 159 347 526 565 888;
  • 15) 0.000 012 159 347 526 565 888 × 2 = 0 + 0.000 024 318 695 053 131 776;
  • 16) 0.000 024 318 695 053 131 776 × 2 = 0 + 0.000 048 637 390 106 263 552;
  • 17) 0.000 048 637 390 106 263 552 × 2 = 0 + 0.000 097 274 780 212 527 104;
  • 18) 0.000 097 274 780 212 527 104 × 2 = 0 + 0.000 194 549 560 425 054 208;
  • 19) 0.000 194 549 560 425 054 208 × 2 = 0 + 0.000 389 099 120 850 108 416;
  • 20) 0.000 389 099 120 850 108 416 × 2 = 0 + 0.000 778 198 241 700 216 832;
  • 21) 0.000 778 198 241 700 216 832 × 2 = 0 + 0.001 556 396 483 400 433 664;
  • 22) 0.001 556 396 483 400 433 664 × 2 = 0 + 0.003 112 792 966 800 867 328;
  • 23) 0.003 112 792 966 800 867 328 × 2 = 0 + 0.006 225 585 933 601 734 656;
  • 24) 0.006 225 585 933 601 734 656 × 2 = 0 + 0.012 451 171 867 203 469 312;
  • 25) 0.012 451 171 867 203 469 312 × 2 = 0 + 0.024 902 343 734 406 938 624;
  • 26) 0.024 902 343 734 406 938 624 × 2 = 0 + 0.049 804 687 468 813 877 248;
  • 27) 0.049 804 687 468 813 877 248 × 2 = 0 + 0.099 609 374 937 627 754 496;
  • 28) 0.099 609 374 937 627 754 496 × 2 = 0 + 0.199 218 749 875 255 508 992;
  • 29) 0.199 218 749 875 255 508 992 × 2 = 0 + 0.398 437 499 750 511 017 984;
  • 30) 0.398 437 499 750 511 017 984 × 2 = 0 + 0.796 874 999 501 022 035 968;
  • 31) 0.796 874 999 501 022 035 968 × 2 = 1 + 0.593 749 999 002 044 071 936;
  • 32) 0.593 749 999 002 044 071 936 × 2 = 1 + 0.187 499 998 004 088 143 872;
  • 33) 0.187 499 998 004 088 143 872 × 2 = 0 + 0.374 999 996 008 176 287 744;
  • 34) 0.374 999 996 008 176 287 744 × 2 = 0 + 0.749 999 992 016 352 575 488;
  • 35) 0.749 999 992 016 352 575 488 × 2 = 1 + 0.499 999 984 032 705 150 976;
  • 36) 0.499 999 984 032 705 150 976 × 2 = 0 + 0.999 999 968 065 410 301 952;
  • 37) 0.999 999 968 065 410 301 952 × 2 = 1 + 0.999 999 936 130 820 603 904;
  • 38) 0.999 999 936 130 820 603 904 × 2 = 1 + 0.999 999 872 261 641 207 808;
  • 39) 0.999 999 872 261 641 207 808 × 2 = 1 + 0.999 999 744 523 282 415 616;
  • 40) 0.999 999 744 523 282 415 616 × 2 = 1 + 0.999 999 489 046 564 831 232;
  • 41) 0.999 999 489 046 564 831 232 × 2 = 1 + 0.999 998 978 093 129 662 464;
  • 42) 0.999 998 978 093 129 662 464 × 2 = 1 + 0.999 997 956 186 259 324 928;
  • 43) 0.999 997 956 186 259 324 928 × 2 = 1 + 0.999 995 912 372 518 649 856;
  • 44) 0.999 995 912 372 518 649 856 × 2 = 1 + 0.999 991 824 745 037 299 712;
  • 45) 0.999 991 824 745 037 299 712 × 2 = 1 + 0.999 983 649 490 074 599 424;
  • 46) 0.999 983 649 490 074 599 424 × 2 = 1 + 0.999 967 298 980 149 198 848;
  • 47) 0.999 967 298 980 149 198 848 × 2 = 1 + 0.999 934 597 960 298 397 696;
  • 48) 0.999 934 597 960 298 397 696 × 2 = 1 + 0.999 869 195 920 596 795 392;
  • 49) 0.999 869 195 920 596 795 392 × 2 = 1 + 0.999 738 391 841 193 590 784;
  • 50) 0.999 738 391 841 193 590 784 × 2 = 1 + 0.999 476 783 682 387 181 568;
  • 51) 0.999 476 783 682 387 181 568 × 2 = 1 + 0.998 953 567 364 774 363 136;
  • 52) 0.998 953 567 364 774 363 136 × 2 = 1 + 0.997 907 134 729 548 726 272;
  • 53) 0.997 907 134 729 548 726 272 × 2 = 1 + 0.995 814 269 459 097 452 544;
  • 54) 0.995 814 269 459 097 452 544 × 2 = 1 + 0.991 628 538 918 194 905 088;

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:


0.000 000 000 742 147 676 182(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 182(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:


0.000 000 000 742 147 676 182(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 -0.000 000 000 742 147 676 182 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