-0.000 000 000 742 147 676 472 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 472(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
-0.000 000 000 742 147 676 472(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:

|-0.000 000 000 742 147 676 472| = 0.000 000 000 742 147 676 472


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: 0.000 000 000 742 147 676 472.

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) 0.000 000 000 742 147 676 472 × 2 = 0 + 0.000 000 001 484 295 352 944;
  • 2) 0.000 000 001 484 295 352 944 × 2 = 0 + 0.000 000 002 968 590 705 888;
  • 3) 0.000 000 002 968 590 705 888 × 2 = 0 + 0.000 000 005 937 181 411 776;
  • 4) 0.000 000 005 937 181 411 776 × 2 = 0 + 0.000 000 011 874 362 823 552;
  • 5) 0.000 000 011 874 362 823 552 × 2 = 0 + 0.000 000 023 748 725 647 104;
  • 6) 0.000 000 023 748 725 647 104 × 2 = 0 + 0.000 000 047 497 451 294 208;
  • 7) 0.000 000 047 497 451 294 208 × 2 = 0 + 0.000 000 094 994 902 588 416;
  • 8) 0.000 000 094 994 902 588 416 × 2 = 0 + 0.000 000 189 989 805 176 832;
  • 9) 0.000 000 189 989 805 176 832 × 2 = 0 + 0.000 000 379 979 610 353 664;
  • 10) 0.000 000 379 979 610 353 664 × 2 = 0 + 0.000 000 759 959 220 707 328;
  • 11) 0.000 000 759 959 220 707 328 × 2 = 0 + 0.000 001 519 918 441 414 656;
  • 12) 0.000 001 519 918 441 414 656 × 2 = 0 + 0.000 003 039 836 882 829 312;
  • 13) 0.000 003 039 836 882 829 312 × 2 = 0 + 0.000 006 079 673 765 658 624;
  • 14) 0.000 006 079 673 765 658 624 × 2 = 0 + 0.000 012 159 347 531 317 248;
  • 15) 0.000 012 159 347 531 317 248 × 2 = 0 + 0.000 024 318 695 062 634 496;
  • 16) 0.000 024 318 695 062 634 496 × 2 = 0 + 0.000 048 637 390 125 268 992;
  • 17) 0.000 048 637 390 125 268 992 × 2 = 0 + 0.000 097 274 780 250 537 984;
  • 18) 0.000 097 274 780 250 537 984 × 2 = 0 + 0.000 194 549 560 501 075 968;
  • 19) 0.000 194 549 560 501 075 968 × 2 = 0 + 0.000 389 099 121 002 151 936;
  • 20) 0.000 389 099 121 002 151 936 × 2 = 0 + 0.000 778 198 242 004 303 872;
  • 21) 0.000 778 198 242 004 303 872 × 2 = 0 + 0.001 556 396 484 008 607 744;
  • 22) 0.001 556 396 484 008 607 744 × 2 = 0 + 0.003 112 792 968 017 215 488;
  • 23) 0.003 112 792 968 017 215 488 × 2 = 0 + 0.006 225 585 936 034 430 976;
  • 24) 0.006 225 585 936 034 430 976 × 2 = 0 + 0.012 451 171 872 068 861 952;
  • 25) 0.012 451 171 872 068 861 952 × 2 = 0 + 0.024 902 343 744 137 723 904;
  • 26) 0.024 902 343 744 137 723 904 × 2 = 0 + 0.049 804 687 488 275 447 808;
  • 27) 0.049 804 687 488 275 447 808 × 2 = 0 + 0.099 609 374 976 550 895 616;
  • 28) 0.099 609 374 976 550 895 616 × 2 = 0 + 0.199 218 749 953 101 791 232;
  • 29) 0.199 218 749 953 101 791 232 × 2 = 0 + 0.398 437 499 906 203 582 464;
  • 30) 0.398 437 499 906 203 582 464 × 2 = 0 + 0.796 874 999 812 407 164 928;
  • 31) 0.796 874 999 812 407 164 928 × 2 = 1 + 0.593 749 999 624 814 329 856;
  • 32) 0.593 749 999 624 814 329 856 × 2 = 1 + 0.187 499 999 249 628 659 712;
  • 33) 0.187 499 999 249 628 659 712 × 2 = 0 + 0.374 999 998 499 257 319 424;
  • 34) 0.374 999 998 499 257 319 424 × 2 = 0 + 0.749 999 996 998 514 638 848;
  • 35) 0.749 999 996 998 514 638 848 × 2 = 1 + 0.499 999 993 997 029 277 696;
  • 36) 0.499 999 993 997 029 277 696 × 2 = 0 + 0.999 999 987 994 058 555 392;
  • 37) 0.999 999 987 994 058 555 392 × 2 = 1 + 0.999 999 975 988 117 110 784;
  • 38) 0.999 999 975 988 117 110 784 × 2 = 1 + 0.999 999 951 976 234 221 568;
  • 39) 0.999 999 951 976 234 221 568 × 2 = 1 + 0.999 999 903 952 468 443 136;
  • 40) 0.999 999 903 952 468 443 136 × 2 = 1 + 0.999 999 807 904 936 886 272;
  • 41) 0.999 999 807 904 936 886 272 × 2 = 1 + 0.999 999 615 809 873 772 544;
  • 42) 0.999 999 615 809 873 772 544 × 2 = 1 + 0.999 999 231 619 747 545 088;
  • 43) 0.999 999 231 619 747 545 088 × 2 = 1 + 0.999 998 463 239 495 090 176;
  • 44) 0.999 998 463 239 495 090 176 × 2 = 1 + 0.999 996 926 478 990 180 352;
  • 45) 0.999 996 926 478 990 180 352 × 2 = 1 + 0.999 993 852 957 980 360 704;
  • 46) 0.999 993 852 957 980 360 704 × 2 = 1 + 0.999 987 705 915 960 721 408;
  • 47) 0.999 987 705 915 960 721 408 × 2 = 1 + 0.999 975 411 831 921 442 816;
  • 48) 0.999 975 411 831 921 442 816 × 2 = 1 + 0.999 950 823 663 842 885 632;
  • 49) 0.999 950 823 663 842 885 632 × 2 = 1 + 0.999 901 647 327 685 771 264;
  • 50) 0.999 901 647 327 685 771 264 × 2 = 1 + 0.999 803 294 655 371 542 528;
  • 51) 0.999 803 294 655 371 542 528 × 2 = 1 + 0.999 606 589 310 743 085 056;
  • 52) 0.999 606 589 310 743 085 056 × 2 = 1 + 0.999 213 178 621 486 170 112;
  • 53) 0.999 213 178 621 486 170 112 × 2 = 1 + 0.998 426 357 242 972 340 224;
  • 54) 0.998 426 357 242 972 340 224 × 2 = 1 + 0.996 852 714 485 944 680 448;

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:


0.000 000 000 742 147 676 472(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 472(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:


0.000 000 000 742 147 676 472(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 -0.000 000 000 742 147 676 472 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