-0.000 000 000 742 147 676 626 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 626 5(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
-0.000 000 000 742 147 676 626 5(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:

|-0.000 000 000 742 147 676 626 5| = 0.000 000 000 742 147 676 626 5


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: 0.000 000 000 742 147 676 626 5.

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) 0.000 000 000 742 147 676 626 5 × 2 = 0 + 0.000 000 001 484 295 353 253;
  • 2) 0.000 000 001 484 295 353 253 × 2 = 0 + 0.000 000 002 968 590 706 506;
  • 3) 0.000 000 002 968 590 706 506 × 2 = 0 + 0.000 000 005 937 181 413 012;
  • 4) 0.000 000 005 937 181 413 012 × 2 = 0 + 0.000 000 011 874 362 826 024;
  • 5) 0.000 000 011 874 362 826 024 × 2 = 0 + 0.000 000 023 748 725 652 048;
  • 6) 0.000 000 023 748 725 652 048 × 2 = 0 + 0.000 000 047 497 451 304 096;
  • 7) 0.000 000 047 497 451 304 096 × 2 = 0 + 0.000 000 094 994 902 608 192;
  • 8) 0.000 000 094 994 902 608 192 × 2 = 0 + 0.000 000 189 989 805 216 384;
  • 9) 0.000 000 189 989 805 216 384 × 2 = 0 + 0.000 000 379 979 610 432 768;
  • 10) 0.000 000 379 979 610 432 768 × 2 = 0 + 0.000 000 759 959 220 865 536;
  • 11) 0.000 000 759 959 220 865 536 × 2 = 0 + 0.000 001 519 918 441 731 072;
  • 12) 0.000 001 519 918 441 731 072 × 2 = 0 + 0.000 003 039 836 883 462 144;
  • 13) 0.000 003 039 836 883 462 144 × 2 = 0 + 0.000 006 079 673 766 924 288;
  • 14) 0.000 006 079 673 766 924 288 × 2 = 0 + 0.000 012 159 347 533 848 576;
  • 15) 0.000 012 159 347 533 848 576 × 2 = 0 + 0.000 024 318 695 067 697 152;
  • 16) 0.000 024 318 695 067 697 152 × 2 = 0 + 0.000 048 637 390 135 394 304;
  • 17) 0.000 048 637 390 135 394 304 × 2 = 0 + 0.000 097 274 780 270 788 608;
  • 18) 0.000 097 274 780 270 788 608 × 2 = 0 + 0.000 194 549 560 541 577 216;
  • 19) 0.000 194 549 560 541 577 216 × 2 = 0 + 0.000 389 099 121 083 154 432;
  • 20) 0.000 389 099 121 083 154 432 × 2 = 0 + 0.000 778 198 242 166 308 864;
  • 21) 0.000 778 198 242 166 308 864 × 2 = 0 + 0.001 556 396 484 332 617 728;
  • 22) 0.001 556 396 484 332 617 728 × 2 = 0 + 0.003 112 792 968 665 235 456;
  • 23) 0.003 112 792 968 665 235 456 × 2 = 0 + 0.006 225 585 937 330 470 912;
  • 24) 0.006 225 585 937 330 470 912 × 2 = 0 + 0.012 451 171 874 660 941 824;
  • 25) 0.012 451 171 874 660 941 824 × 2 = 0 + 0.024 902 343 749 321 883 648;
  • 26) 0.024 902 343 749 321 883 648 × 2 = 0 + 0.049 804 687 498 643 767 296;
  • 27) 0.049 804 687 498 643 767 296 × 2 = 0 + 0.099 609 374 997 287 534 592;
  • 28) 0.099 609 374 997 287 534 592 × 2 = 0 + 0.199 218 749 994 575 069 184;
  • 29) 0.199 218 749 994 575 069 184 × 2 = 0 + 0.398 437 499 989 150 138 368;
  • 30) 0.398 437 499 989 150 138 368 × 2 = 0 + 0.796 874 999 978 300 276 736;
  • 31) 0.796 874 999 978 300 276 736 × 2 = 1 + 0.593 749 999 956 600 553 472;
  • 32) 0.593 749 999 956 600 553 472 × 2 = 1 + 0.187 499 999 913 201 106 944;
  • 33) 0.187 499 999 913 201 106 944 × 2 = 0 + 0.374 999 999 826 402 213 888;
  • 34) 0.374 999 999 826 402 213 888 × 2 = 0 + 0.749 999 999 652 804 427 776;
  • 35) 0.749 999 999 652 804 427 776 × 2 = 1 + 0.499 999 999 305 608 855 552;
  • 36) 0.499 999 999 305 608 855 552 × 2 = 0 + 0.999 999 998 611 217 711 104;
  • 37) 0.999 999 998 611 217 711 104 × 2 = 1 + 0.999 999 997 222 435 422 208;
  • 38) 0.999 999 997 222 435 422 208 × 2 = 1 + 0.999 999 994 444 870 844 416;
  • 39) 0.999 999 994 444 870 844 416 × 2 = 1 + 0.999 999 988 889 741 688 832;
  • 40) 0.999 999 988 889 741 688 832 × 2 = 1 + 0.999 999 977 779 483 377 664;
  • 41) 0.999 999 977 779 483 377 664 × 2 = 1 + 0.999 999 955 558 966 755 328;
  • 42) 0.999 999 955 558 966 755 328 × 2 = 1 + 0.999 999 911 117 933 510 656;
  • 43) 0.999 999 911 117 933 510 656 × 2 = 1 + 0.999 999 822 235 867 021 312;
  • 44) 0.999 999 822 235 867 021 312 × 2 = 1 + 0.999 999 644 471 734 042 624;
  • 45) 0.999 999 644 471 734 042 624 × 2 = 1 + 0.999 999 288 943 468 085 248;
  • 46) 0.999 999 288 943 468 085 248 × 2 = 1 + 0.999 998 577 886 936 170 496;
  • 47) 0.999 998 577 886 936 170 496 × 2 = 1 + 0.999 997 155 773 872 340 992;
  • 48) 0.999 997 155 773 872 340 992 × 2 = 1 + 0.999 994 311 547 744 681 984;
  • 49) 0.999 994 311 547 744 681 984 × 2 = 1 + 0.999 988 623 095 489 363 968;
  • 50) 0.999 988 623 095 489 363 968 × 2 = 1 + 0.999 977 246 190 978 727 936;
  • 51) 0.999 977 246 190 978 727 936 × 2 = 1 + 0.999 954 492 381 957 455 872;
  • 52) 0.999 954 492 381 957 455 872 × 2 = 1 + 0.999 908 984 763 914 911 744;
  • 53) 0.999 908 984 763 914 911 744 × 2 = 1 + 0.999 817 969 527 829 823 488;
  • 54) 0.999 817 969 527 829 823 488 × 2 = 1 + 0.999 635 939 055 659 646 976;

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:


0.000 000 000 742 147 676 626 5(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 626 5(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:


0.000 000 000 742 147 676 626 5(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 -0.000 000 000 742 147 676 626 5 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