-0.000 000 000 742 147 676 141 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 141(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
-0.000 000 000 742 147 676 141(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:

|-0.000 000 000 742 147 676 141| = 0.000 000 000 742 147 676 141


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: 0.000 000 000 742 147 676 141.

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) 0.000 000 000 742 147 676 141 × 2 = 0 + 0.000 000 001 484 295 352 282;
  • 2) 0.000 000 001 484 295 352 282 × 2 = 0 + 0.000 000 002 968 590 704 564;
  • 3) 0.000 000 002 968 590 704 564 × 2 = 0 + 0.000 000 005 937 181 409 128;
  • 4) 0.000 000 005 937 181 409 128 × 2 = 0 + 0.000 000 011 874 362 818 256;
  • 5) 0.000 000 011 874 362 818 256 × 2 = 0 + 0.000 000 023 748 725 636 512;
  • 6) 0.000 000 023 748 725 636 512 × 2 = 0 + 0.000 000 047 497 451 273 024;
  • 7) 0.000 000 047 497 451 273 024 × 2 = 0 + 0.000 000 094 994 902 546 048;
  • 8) 0.000 000 094 994 902 546 048 × 2 = 0 + 0.000 000 189 989 805 092 096;
  • 9) 0.000 000 189 989 805 092 096 × 2 = 0 + 0.000 000 379 979 610 184 192;
  • 10) 0.000 000 379 979 610 184 192 × 2 = 0 + 0.000 000 759 959 220 368 384;
  • 11) 0.000 000 759 959 220 368 384 × 2 = 0 + 0.000 001 519 918 440 736 768;
  • 12) 0.000 001 519 918 440 736 768 × 2 = 0 + 0.000 003 039 836 881 473 536;
  • 13) 0.000 003 039 836 881 473 536 × 2 = 0 + 0.000 006 079 673 762 947 072;
  • 14) 0.000 006 079 673 762 947 072 × 2 = 0 + 0.000 012 159 347 525 894 144;
  • 15) 0.000 012 159 347 525 894 144 × 2 = 0 + 0.000 024 318 695 051 788 288;
  • 16) 0.000 024 318 695 051 788 288 × 2 = 0 + 0.000 048 637 390 103 576 576;
  • 17) 0.000 048 637 390 103 576 576 × 2 = 0 + 0.000 097 274 780 207 153 152;
  • 18) 0.000 097 274 780 207 153 152 × 2 = 0 + 0.000 194 549 560 414 306 304;
  • 19) 0.000 194 549 560 414 306 304 × 2 = 0 + 0.000 389 099 120 828 612 608;
  • 20) 0.000 389 099 120 828 612 608 × 2 = 0 + 0.000 778 198 241 657 225 216;
  • 21) 0.000 778 198 241 657 225 216 × 2 = 0 + 0.001 556 396 483 314 450 432;
  • 22) 0.001 556 396 483 314 450 432 × 2 = 0 + 0.003 112 792 966 628 900 864;
  • 23) 0.003 112 792 966 628 900 864 × 2 = 0 + 0.006 225 585 933 257 801 728;
  • 24) 0.006 225 585 933 257 801 728 × 2 = 0 + 0.012 451 171 866 515 603 456;
  • 25) 0.012 451 171 866 515 603 456 × 2 = 0 + 0.024 902 343 733 031 206 912;
  • 26) 0.024 902 343 733 031 206 912 × 2 = 0 + 0.049 804 687 466 062 413 824;
  • 27) 0.049 804 687 466 062 413 824 × 2 = 0 + 0.099 609 374 932 124 827 648;
  • 28) 0.099 609 374 932 124 827 648 × 2 = 0 + 0.199 218 749 864 249 655 296;
  • 29) 0.199 218 749 864 249 655 296 × 2 = 0 + 0.398 437 499 728 499 310 592;
  • 30) 0.398 437 499 728 499 310 592 × 2 = 0 + 0.796 874 999 456 998 621 184;
  • 31) 0.796 874 999 456 998 621 184 × 2 = 1 + 0.593 749 998 913 997 242 368;
  • 32) 0.593 749 998 913 997 242 368 × 2 = 1 + 0.187 499 997 827 994 484 736;
  • 33) 0.187 499 997 827 994 484 736 × 2 = 0 + 0.374 999 995 655 988 969 472;
  • 34) 0.374 999 995 655 988 969 472 × 2 = 0 + 0.749 999 991 311 977 938 944;
  • 35) 0.749 999 991 311 977 938 944 × 2 = 1 + 0.499 999 982 623 955 877 888;
  • 36) 0.499 999 982 623 955 877 888 × 2 = 0 + 0.999 999 965 247 911 755 776;
  • 37) 0.999 999 965 247 911 755 776 × 2 = 1 + 0.999 999 930 495 823 511 552;
  • 38) 0.999 999 930 495 823 511 552 × 2 = 1 + 0.999 999 860 991 647 023 104;
  • 39) 0.999 999 860 991 647 023 104 × 2 = 1 + 0.999 999 721 983 294 046 208;
  • 40) 0.999 999 721 983 294 046 208 × 2 = 1 + 0.999 999 443 966 588 092 416;
  • 41) 0.999 999 443 966 588 092 416 × 2 = 1 + 0.999 998 887 933 176 184 832;
  • 42) 0.999 998 887 933 176 184 832 × 2 = 1 + 0.999 997 775 866 352 369 664;
  • 43) 0.999 997 775 866 352 369 664 × 2 = 1 + 0.999 995 551 732 704 739 328;
  • 44) 0.999 995 551 732 704 739 328 × 2 = 1 + 0.999 991 103 465 409 478 656;
  • 45) 0.999 991 103 465 409 478 656 × 2 = 1 + 0.999 982 206 930 818 957 312;
  • 46) 0.999 982 206 930 818 957 312 × 2 = 1 + 0.999 964 413 861 637 914 624;
  • 47) 0.999 964 413 861 637 914 624 × 2 = 1 + 0.999 928 827 723 275 829 248;
  • 48) 0.999 928 827 723 275 829 248 × 2 = 1 + 0.999 857 655 446 551 658 496;
  • 49) 0.999 857 655 446 551 658 496 × 2 = 1 + 0.999 715 310 893 103 316 992;
  • 50) 0.999 715 310 893 103 316 992 × 2 = 1 + 0.999 430 621 786 206 633 984;
  • 51) 0.999 430 621 786 206 633 984 × 2 = 1 + 0.998 861 243 572 413 267 968;
  • 52) 0.998 861 243 572 413 267 968 × 2 = 1 + 0.997 722 487 144 826 535 936;
  • 53) 0.997 722 487 144 826 535 936 × 2 = 1 + 0.995 444 974 289 653 071 872;
  • 54) 0.995 444 974 289 653 071 872 × 2 = 1 + 0.990 889 948 579 306 143 744;

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:


0.000 000 000 742 147 676 141(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 141(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:


0.000 000 000 742 147 676 141(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 -0.000 000 000 742 147 676 141 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