-0.000 000 000 742 147 604 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 604(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
-0.000 000 000 742 147 604(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:

|-0.000 000 000 742 147 604| = 0.000 000 000 742 147 604


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: 0.000 000 000 742 147 604.

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) 0.000 000 000 742 147 604 × 2 = 0 + 0.000 000 001 484 295 208;
  • 2) 0.000 000 001 484 295 208 × 2 = 0 + 0.000 000 002 968 590 416;
  • 3) 0.000 000 002 968 590 416 × 2 = 0 + 0.000 000 005 937 180 832;
  • 4) 0.000 000 005 937 180 832 × 2 = 0 + 0.000 000 011 874 361 664;
  • 5) 0.000 000 011 874 361 664 × 2 = 0 + 0.000 000 023 748 723 328;
  • 6) 0.000 000 023 748 723 328 × 2 = 0 + 0.000 000 047 497 446 656;
  • 7) 0.000 000 047 497 446 656 × 2 = 0 + 0.000 000 094 994 893 312;
  • 8) 0.000 000 094 994 893 312 × 2 = 0 + 0.000 000 189 989 786 624;
  • 9) 0.000 000 189 989 786 624 × 2 = 0 + 0.000 000 379 979 573 248;
  • 10) 0.000 000 379 979 573 248 × 2 = 0 + 0.000 000 759 959 146 496;
  • 11) 0.000 000 759 959 146 496 × 2 = 0 + 0.000 001 519 918 292 992;
  • 12) 0.000 001 519 918 292 992 × 2 = 0 + 0.000 003 039 836 585 984;
  • 13) 0.000 003 039 836 585 984 × 2 = 0 + 0.000 006 079 673 171 968;
  • 14) 0.000 006 079 673 171 968 × 2 = 0 + 0.000 012 159 346 343 936;
  • 15) 0.000 012 159 346 343 936 × 2 = 0 + 0.000 024 318 692 687 872;
  • 16) 0.000 024 318 692 687 872 × 2 = 0 + 0.000 048 637 385 375 744;
  • 17) 0.000 048 637 385 375 744 × 2 = 0 + 0.000 097 274 770 751 488;
  • 18) 0.000 097 274 770 751 488 × 2 = 0 + 0.000 194 549 541 502 976;
  • 19) 0.000 194 549 541 502 976 × 2 = 0 + 0.000 389 099 083 005 952;
  • 20) 0.000 389 099 083 005 952 × 2 = 0 + 0.000 778 198 166 011 904;
  • 21) 0.000 778 198 166 011 904 × 2 = 0 + 0.001 556 396 332 023 808;
  • 22) 0.001 556 396 332 023 808 × 2 = 0 + 0.003 112 792 664 047 616;
  • 23) 0.003 112 792 664 047 616 × 2 = 0 + 0.006 225 585 328 095 232;
  • 24) 0.006 225 585 328 095 232 × 2 = 0 + 0.012 451 170 656 190 464;
  • 25) 0.012 451 170 656 190 464 × 2 = 0 + 0.024 902 341 312 380 928;
  • 26) 0.024 902 341 312 380 928 × 2 = 0 + 0.049 804 682 624 761 856;
  • 27) 0.049 804 682 624 761 856 × 2 = 0 + 0.099 609 365 249 523 712;
  • 28) 0.099 609 365 249 523 712 × 2 = 0 + 0.199 218 730 499 047 424;
  • 29) 0.199 218 730 499 047 424 × 2 = 0 + 0.398 437 460 998 094 848;
  • 30) 0.398 437 460 998 094 848 × 2 = 0 + 0.796 874 921 996 189 696;
  • 31) 0.796 874 921 996 189 696 × 2 = 1 + 0.593 749 843 992 379 392;
  • 32) 0.593 749 843 992 379 392 × 2 = 1 + 0.187 499 687 984 758 784;
  • 33) 0.187 499 687 984 758 784 × 2 = 0 + 0.374 999 375 969 517 568;
  • 34) 0.374 999 375 969 517 568 × 2 = 0 + 0.749 998 751 939 035 136;
  • 35) 0.749 998 751 939 035 136 × 2 = 1 + 0.499 997 503 878 070 272;
  • 36) 0.499 997 503 878 070 272 × 2 = 0 + 0.999 995 007 756 140 544;
  • 37) 0.999 995 007 756 140 544 × 2 = 1 + 0.999 990 015 512 281 088;
  • 38) 0.999 990 015 512 281 088 × 2 = 1 + 0.999 980 031 024 562 176;
  • 39) 0.999 980 031 024 562 176 × 2 = 1 + 0.999 960 062 049 124 352;
  • 40) 0.999 960 062 049 124 352 × 2 = 1 + 0.999 920 124 098 248 704;
  • 41) 0.999 920 124 098 248 704 × 2 = 1 + 0.999 840 248 196 497 408;
  • 42) 0.999 840 248 196 497 408 × 2 = 1 + 0.999 680 496 392 994 816;
  • 43) 0.999 680 496 392 994 816 × 2 = 1 + 0.999 360 992 785 989 632;
  • 44) 0.999 360 992 785 989 632 × 2 = 1 + 0.998 721 985 571 979 264;
  • 45) 0.998 721 985 571 979 264 × 2 = 1 + 0.997 443 971 143 958 528;
  • 46) 0.997 443 971 143 958 528 × 2 = 1 + 0.994 887 942 287 917 056;
  • 47) 0.994 887 942 287 917 056 × 2 = 1 + 0.989 775 884 575 834 112;
  • 48) 0.989 775 884 575 834 112 × 2 = 1 + 0.979 551 769 151 668 224;
  • 49) 0.979 551 769 151 668 224 × 2 = 1 + 0.959 103 538 303 336 448;
  • 50) 0.959 103 538 303 336 448 × 2 = 1 + 0.918 207 076 606 672 896;
  • 51) 0.918 207 076 606 672 896 × 2 = 1 + 0.836 414 153 213 345 792;
  • 52) 0.836 414 153 213 345 792 × 2 = 1 + 0.672 828 306 426 691 584;
  • 53) 0.672 828 306 426 691 584 × 2 = 1 + 0.345 656 612 853 383 168;
  • 54) 0.345 656 612 853 383 168 × 2 = 0 + 0.691 313 225 706 766 336;

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:


0.000 000 000 742 147 604(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.000 000 000 742 147 604(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 10(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:


0.000 000 000 742 147 604(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 10(2) × 20 =

1.1001 0111 1111 1111 1111 110(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 110


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 1110 =

100 1011 1111 1111 1111 1110


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 1110


Decimal number -0.000 000 000 742 147 604 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1110

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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