-0.000 000 000 742 147 676 646 688 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 688 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
-0.000 000 000 742 147 676 646 688 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:

|-0.000 000 000 742 147 676 646 688 5| = 0.000 000 000 742 147 676 646 688 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: 0.000 000 000 742 147 676 646 688 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) 0.000 000 000 742 147 676 646 688 5 × 2 = 0 + 0.000 000 001 484 295 353 293 377;
  • 2) 0.000 000 001 484 295 353 293 377 × 2 = 0 + 0.000 000 002 968 590 706 586 754;
  • 3) 0.000 000 002 968 590 706 586 754 × 2 = 0 + 0.000 000 005 937 181 413 173 508;
  • 4) 0.000 000 005 937 181 413 173 508 × 2 = 0 + 0.000 000 011 874 362 826 347 016;
  • 5) 0.000 000 011 874 362 826 347 016 × 2 = 0 + 0.000 000 023 748 725 652 694 032;
  • 6) 0.000 000 023 748 725 652 694 032 × 2 = 0 + 0.000 000 047 497 451 305 388 064;
  • 7) 0.000 000 047 497 451 305 388 064 × 2 = 0 + 0.000 000 094 994 902 610 776 128;
  • 8) 0.000 000 094 994 902 610 776 128 × 2 = 0 + 0.000 000 189 989 805 221 552 256;
  • 9) 0.000 000 189 989 805 221 552 256 × 2 = 0 + 0.000 000 379 979 610 443 104 512;
  • 10) 0.000 000 379 979 610 443 104 512 × 2 = 0 + 0.000 000 759 959 220 886 209 024;
  • 11) 0.000 000 759 959 220 886 209 024 × 2 = 0 + 0.000 001 519 918 441 772 418 048;
  • 12) 0.000 001 519 918 441 772 418 048 × 2 = 0 + 0.000 003 039 836 883 544 836 096;
  • 13) 0.000 003 039 836 883 544 836 096 × 2 = 0 + 0.000 006 079 673 767 089 672 192;
  • 14) 0.000 006 079 673 767 089 672 192 × 2 = 0 + 0.000 012 159 347 534 179 344 384;
  • 15) 0.000 012 159 347 534 179 344 384 × 2 = 0 + 0.000 024 318 695 068 358 688 768;
  • 16) 0.000 024 318 695 068 358 688 768 × 2 = 0 + 0.000 048 637 390 136 717 377 536;
  • 17) 0.000 048 637 390 136 717 377 536 × 2 = 0 + 0.000 097 274 780 273 434 755 072;
  • 18) 0.000 097 274 780 273 434 755 072 × 2 = 0 + 0.000 194 549 560 546 869 510 144;
  • 19) 0.000 194 549 560 546 869 510 144 × 2 = 0 + 0.000 389 099 121 093 739 020 288;
  • 20) 0.000 389 099 121 093 739 020 288 × 2 = 0 + 0.000 778 198 242 187 478 040 576;
  • 21) 0.000 778 198 242 187 478 040 576 × 2 = 0 + 0.001 556 396 484 374 956 081 152;
  • 22) 0.001 556 396 484 374 956 081 152 × 2 = 0 + 0.003 112 792 968 749 912 162 304;
  • 23) 0.003 112 792 968 749 912 162 304 × 2 = 0 + 0.006 225 585 937 499 824 324 608;
  • 24) 0.006 225 585 937 499 824 324 608 × 2 = 0 + 0.012 451 171 874 999 648 649 216;
  • 25) 0.012 451 171 874 999 648 649 216 × 2 = 0 + 0.024 902 343 749 999 297 298 432;
  • 26) 0.024 902 343 749 999 297 298 432 × 2 = 0 + 0.049 804 687 499 998 594 596 864;
  • 27) 0.049 804 687 499 998 594 596 864 × 2 = 0 + 0.099 609 374 999 997 189 193 728;
  • 28) 0.099 609 374 999 997 189 193 728 × 2 = 0 + 0.199 218 749 999 994 378 387 456;
  • 29) 0.199 218 749 999 994 378 387 456 × 2 = 0 + 0.398 437 499 999 988 756 774 912;
  • 30) 0.398 437 499 999 988 756 774 912 × 2 = 0 + 0.796 874 999 999 977 513 549 824;
  • 31) 0.796 874 999 999 977 513 549 824 × 2 = 1 + 0.593 749 999 999 955 027 099 648;
  • 32) 0.593 749 999 999 955 027 099 648 × 2 = 1 + 0.187 499 999 999 910 054 199 296;
  • 33) 0.187 499 999 999 910 054 199 296 × 2 = 0 + 0.374 999 999 999 820 108 398 592;
  • 34) 0.374 999 999 999 820 108 398 592 × 2 = 0 + 0.749 999 999 999 640 216 797 184;
  • 35) 0.749 999 999 999 640 216 797 184 × 2 = 1 + 0.499 999 999 999 280 433 594 368;
  • 36) 0.499 999 999 999 280 433 594 368 × 2 = 0 + 0.999 999 999 998 560 867 188 736;
  • 37) 0.999 999 999 998 560 867 188 736 × 2 = 1 + 0.999 999 999 997 121 734 377 472;
  • 38) 0.999 999 999 997 121 734 377 472 × 2 = 1 + 0.999 999 999 994 243 468 754 944;
  • 39) 0.999 999 999 994 243 468 754 944 × 2 = 1 + 0.999 999 999 988 486 937 509 888;
  • 40) 0.999 999 999 988 486 937 509 888 × 2 = 1 + 0.999 999 999 976 973 875 019 776;
  • 41) 0.999 999 999 976 973 875 019 776 × 2 = 1 + 0.999 999 999 953 947 750 039 552;
  • 42) 0.999 999 999 953 947 750 039 552 × 2 = 1 + 0.999 999 999 907 895 500 079 104;
  • 43) 0.999 999 999 907 895 500 079 104 × 2 = 1 + 0.999 999 999 815 791 000 158 208;
  • 44) 0.999 999 999 815 791 000 158 208 × 2 = 1 + 0.999 999 999 631 582 000 316 416;
  • 45) 0.999 999 999 631 582 000 316 416 × 2 = 1 + 0.999 999 999 263 164 000 632 832;
  • 46) 0.999 999 999 263 164 000 632 832 × 2 = 1 + 0.999 999 998 526 328 001 265 664;
  • 47) 0.999 999 998 526 328 001 265 664 × 2 = 1 + 0.999 999 997 052 656 002 531 328;
  • 48) 0.999 999 997 052 656 002 531 328 × 2 = 1 + 0.999 999 994 105 312 005 062 656;
  • 49) 0.999 999 994 105 312 005 062 656 × 2 = 1 + 0.999 999 988 210 624 010 125 312;
  • 50) 0.999 999 988 210 624 010 125 312 × 2 = 1 + 0.999 999 976 421 248 020 250 624;
  • 51) 0.999 999 976 421 248 020 250 624 × 2 = 1 + 0.999 999 952 842 496 040 501 248;
  • 52) 0.999 999 952 842 496 040 501 248 × 2 = 1 + 0.999 999 905 684 992 081 002 496;
  • 53) 0.999 999 905 684 992 081 002 496 × 2 = 1 + 0.999 999 811 369 984 162 004 992;
  • 54) 0.999 999 811 369 984 162 004 992 × 2 = 1 + 0.999 999 622 739 968 324 009 984;

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:


0.000 000 000 742 147 676 646 688 5(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 688 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:


0.000 000 000 742 147 676 646 688 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 -0.000 000 000 742 147 676 646 688 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