-0.000 000 000 742 147 676 134 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 134(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
-0.000 000 000 742 147 676 134(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:

|-0.000 000 000 742 147 676 134| = 0.000 000 000 742 147 676 134


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: 0.000 000 000 742 147 676 134.

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) 0.000 000 000 742 147 676 134 × 2 = 0 + 0.000 000 001 484 295 352 268;
  • 2) 0.000 000 001 484 295 352 268 × 2 = 0 + 0.000 000 002 968 590 704 536;
  • 3) 0.000 000 002 968 590 704 536 × 2 = 0 + 0.000 000 005 937 181 409 072;
  • 4) 0.000 000 005 937 181 409 072 × 2 = 0 + 0.000 000 011 874 362 818 144;
  • 5) 0.000 000 011 874 362 818 144 × 2 = 0 + 0.000 000 023 748 725 636 288;
  • 6) 0.000 000 023 748 725 636 288 × 2 = 0 + 0.000 000 047 497 451 272 576;
  • 7) 0.000 000 047 497 451 272 576 × 2 = 0 + 0.000 000 094 994 902 545 152;
  • 8) 0.000 000 094 994 902 545 152 × 2 = 0 + 0.000 000 189 989 805 090 304;
  • 9) 0.000 000 189 989 805 090 304 × 2 = 0 + 0.000 000 379 979 610 180 608;
  • 10) 0.000 000 379 979 610 180 608 × 2 = 0 + 0.000 000 759 959 220 361 216;
  • 11) 0.000 000 759 959 220 361 216 × 2 = 0 + 0.000 001 519 918 440 722 432;
  • 12) 0.000 001 519 918 440 722 432 × 2 = 0 + 0.000 003 039 836 881 444 864;
  • 13) 0.000 003 039 836 881 444 864 × 2 = 0 + 0.000 006 079 673 762 889 728;
  • 14) 0.000 006 079 673 762 889 728 × 2 = 0 + 0.000 012 159 347 525 779 456;
  • 15) 0.000 012 159 347 525 779 456 × 2 = 0 + 0.000 024 318 695 051 558 912;
  • 16) 0.000 024 318 695 051 558 912 × 2 = 0 + 0.000 048 637 390 103 117 824;
  • 17) 0.000 048 637 390 103 117 824 × 2 = 0 + 0.000 097 274 780 206 235 648;
  • 18) 0.000 097 274 780 206 235 648 × 2 = 0 + 0.000 194 549 560 412 471 296;
  • 19) 0.000 194 549 560 412 471 296 × 2 = 0 + 0.000 389 099 120 824 942 592;
  • 20) 0.000 389 099 120 824 942 592 × 2 = 0 + 0.000 778 198 241 649 885 184;
  • 21) 0.000 778 198 241 649 885 184 × 2 = 0 + 0.001 556 396 483 299 770 368;
  • 22) 0.001 556 396 483 299 770 368 × 2 = 0 + 0.003 112 792 966 599 540 736;
  • 23) 0.003 112 792 966 599 540 736 × 2 = 0 + 0.006 225 585 933 199 081 472;
  • 24) 0.006 225 585 933 199 081 472 × 2 = 0 + 0.012 451 171 866 398 162 944;
  • 25) 0.012 451 171 866 398 162 944 × 2 = 0 + 0.024 902 343 732 796 325 888;
  • 26) 0.024 902 343 732 796 325 888 × 2 = 0 + 0.049 804 687 465 592 651 776;
  • 27) 0.049 804 687 465 592 651 776 × 2 = 0 + 0.099 609 374 931 185 303 552;
  • 28) 0.099 609 374 931 185 303 552 × 2 = 0 + 0.199 218 749 862 370 607 104;
  • 29) 0.199 218 749 862 370 607 104 × 2 = 0 + 0.398 437 499 724 741 214 208;
  • 30) 0.398 437 499 724 741 214 208 × 2 = 0 + 0.796 874 999 449 482 428 416;
  • 31) 0.796 874 999 449 482 428 416 × 2 = 1 + 0.593 749 998 898 964 856 832;
  • 32) 0.593 749 998 898 964 856 832 × 2 = 1 + 0.187 499 997 797 929 713 664;
  • 33) 0.187 499 997 797 929 713 664 × 2 = 0 + 0.374 999 995 595 859 427 328;
  • 34) 0.374 999 995 595 859 427 328 × 2 = 0 + 0.749 999 991 191 718 854 656;
  • 35) 0.749 999 991 191 718 854 656 × 2 = 1 + 0.499 999 982 383 437 709 312;
  • 36) 0.499 999 982 383 437 709 312 × 2 = 0 + 0.999 999 964 766 875 418 624;
  • 37) 0.999 999 964 766 875 418 624 × 2 = 1 + 0.999 999 929 533 750 837 248;
  • 38) 0.999 999 929 533 750 837 248 × 2 = 1 + 0.999 999 859 067 501 674 496;
  • 39) 0.999 999 859 067 501 674 496 × 2 = 1 + 0.999 999 718 135 003 348 992;
  • 40) 0.999 999 718 135 003 348 992 × 2 = 1 + 0.999 999 436 270 006 697 984;
  • 41) 0.999 999 436 270 006 697 984 × 2 = 1 + 0.999 998 872 540 013 395 968;
  • 42) 0.999 998 872 540 013 395 968 × 2 = 1 + 0.999 997 745 080 026 791 936;
  • 43) 0.999 997 745 080 026 791 936 × 2 = 1 + 0.999 995 490 160 053 583 872;
  • 44) 0.999 995 490 160 053 583 872 × 2 = 1 + 0.999 990 980 320 107 167 744;
  • 45) 0.999 990 980 320 107 167 744 × 2 = 1 + 0.999 981 960 640 214 335 488;
  • 46) 0.999 981 960 640 214 335 488 × 2 = 1 + 0.999 963 921 280 428 670 976;
  • 47) 0.999 963 921 280 428 670 976 × 2 = 1 + 0.999 927 842 560 857 341 952;
  • 48) 0.999 927 842 560 857 341 952 × 2 = 1 + 0.999 855 685 121 714 683 904;
  • 49) 0.999 855 685 121 714 683 904 × 2 = 1 + 0.999 711 370 243 429 367 808;
  • 50) 0.999 711 370 243 429 367 808 × 2 = 1 + 0.999 422 740 486 858 735 616;
  • 51) 0.999 422 740 486 858 735 616 × 2 = 1 + 0.998 845 480 973 717 471 232;
  • 52) 0.998 845 480 973 717 471 232 × 2 = 1 + 0.997 690 961 947 434 942 464;
  • 53) 0.997 690 961 947 434 942 464 × 2 = 1 + 0.995 381 923 894 869 884 928;
  • 54) 0.995 381 923 894 869 884 928 × 2 = 1 + 0.990 763 847 789 739 769 856;

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:


0.000 000 000 742 147 676 134(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 134(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:


0.000 000 000 742 147 676 134(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 -0.000 000 000 742 147 676 134 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