-0.000 000 000 742 147 504 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 504(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
-0.000 000 000 742 147 504(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:

|-0.000 000 000 742 147 504| = 0.000 000 000 742 147 504


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: 0.000 000 000 742 147 504.

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) 0.000 000 000 742 147 504 × 2 = 0 + 0.000 000 001 484 295 008;
  • 2) 0.000 000 001 484 295 008 × 2 = 0 + 0.000 000 002 968 590 016;
  • 3) 0.000 000 002 968 590 016 × 2 = 0 + 0.000 000 005 937 180 032;
  • 4) 0.000 000 005 937 180 032 × 2 = 0 + 0.000 000 011 874 360 064;
  • 5) 0.000 000 011 874 360 064 × 2 = 0 + 0.000 000 023 748 720 128;
  • 6) 0.000 000 023 748 720 128 × 2 = 0 + 0.000 000 047 497 440 256;
  • 7) 0.000 000 047 497 440 256 × 2 = 0 + 0.000 000 094 994 880 512;
  • 8) 0.000 000 094 994 880 512 × 2 = 0 + 0.000 000 189 989 761 024;
  • 9) 0.000 000 189 989 761 024 × 2 = 0 + 0.000 000 379 979 522 048;
  • 10) 0.000 000 379 979 522 048 × 2 = 0 + 0.000 000 759 959 044 096;
  • 11) 0.000 000 759 959 044 096 × 2 = 0 + 0.000 001 519 918 088 192;
  • 12) 0.000 001 519 918 088 192 × 2 = 0 + 0.000 003 039 836 176 384;
  • 13) 0.000 003 039 836 176 384 × 2 = 0 + 0.000 006 079 672 352 768;
  • 14) 0.000 006 079 672 352 768 × 2 = 0 + 0.000 012 159 344 705 536;
  • 15) 0.000 012 159 344 705 536 × 2 = 0 + 0.000 024 318 689 411 072;
  • 16) 0.000 024 318 689 411 072 × 2 = 0 + 0.000 048 637 378 822 144;
  • 17) 0.000 048 637 378 822 144 × 2 = 0 + 0.000 097 274 757 644 288;
  • 18) 0.000 097 274 757 644 288 × 2 = 0 + 0.000 194 549 515 288 576;
  • 19) 0.000 194 549 515 288 576 × 2 = 0 + 0.000 389 099 030 577 152;
  • 20) 0.000 389 099 030 577 152 × 2 = 0 + 0.000 778 198 061 154 304;
  • 21) 0.000 778 198 061 154 304 × 2 = 0 + 0.001 556 396 122 308 608;
  • 22) 0.001 556 396 122 308 608 × 2 = 0 + 0.003 112 792 244 617 216;
  • 23) 0.003 112 792 244 617 216 × 2 = 0 + 0.006 225 584 489 234 432;
  • 24) 0.006 225 584 489 234 432 × 2 = 0 + 0.012 451 168 978 468 864;
  • 25) 0.012 451 168 978 468 864 × 2 = 0 + 0.024 902 337 956 937 728;
  • 26) 0.024 902 337 956 937 728 × 2 = 0 + 0.049 804 675 913 875 456;
  • 27) 0.049 804 675 913 875 456 × 2 = 0 + 0.099 609 351 827 750 912;
  • 28) 0.099 609 351 827 750 912 × 2 = 0 + 0.199 218 703 655 501 824;
  • 29) 0.199 218 703 655 501 824 × 2 = 0 + 0.398 437 407 311 003 648;
  • 30) 0.398 437 407 311 003 648 × 2 = 0 + 0.796 874 814 622 007 296;
  • 31) 0.796 874 814 622 007 296 × 2 = 1 + 0.593 749 629 244 014 592;
  • 32) 0.593 749 629 244 014 592 × 2 = 1 + 0.187 499 258 488 029 184;
  • 33) 0.187 499 258 488 029 184 × 2 = 0 + 0.374 998 516 976 058 368;
  • 34) 0.374 998 516 976 058 368 × 2 = 0 + 0.749 997 033 952 116 736;
  • 35) 0.749 997 033 952 116 736 × 2 = 1 + 0.499 994 067 904 233 472;
  • 36) 0.499 994 067 904 233 472 × 2 = 0 + 0.999 988 135 808 466 944;
  • 37) 0.999 988 135 808 466 944 × 2 = 1 + 0.999 976 271 616 933 888;
  • 38) 0.999 976 271 616 933 888 × 2 = 1 + 0.999 952 543 233 867 776;
  • 39) 0.999 952 543 233 867 776 × 2 = 1 + 0.999 905 086 467 735 552;
  • 40) 0.999 905 086 467 735 552 × 2 = 1 + 0.999 810 172 935 471 104;
  • 41) 0.999 810 172 935 471 104 × 2 = 1 + 0.999 620 345 870 942 208;
  • 42) 0.999 620 345 870 942 208 × 2 = 1 + 0.999 240 691 741 884 416;
  • 43) 0.999 240 691 741 884 416 × 2 = 1 + 0.998 481 383 483 768 832;
  • 44) 0.998 481 383 483 768 832 × 2 = 1 + 0.996 962 766 967 537 664;
  • 45) 0.996 962 766 967 537 664 × 2 = 1 + 0.993 925 533 935 075 328;
  • 46) 0.993 925 533 935 075 328 × 2 = 1 + 0.987 851 067 870 150 656;
  • 47) 0.987 851 067 870 150 656 × 2 = 1 + 0.975 702 135 740 301 312;
  • 48) 0.975 702 135 740 301 312 × 2 = 1 + 0.951 404 271 480 602 624;
  • 49) 0.951 404 271 480 602 624 × 2 = 1 + 0.902 808 542 961 205 248;
  • 50) 0.902 808 542 961 205 248 × 2 = 1 + 0.805 617 085 922 410 496;
  • 51) 0.805 617 085 922 410 496 × 2 = 1 + 0.611 234 171 844 820 992;
  • 52) 0.611 234 171 844 820 992 × 2 = 1 + 0.222 468 343 689 641 984;
  • 53) 0.222 468 343 689 641 984 × 2 = 0 + 0.444 936 687 379 283 968;
  • 54) 0.444 936 687 379 283 968 × 2 = 0 + 0.889 873 374 758 567 936;

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:


0.000 000 000 742 147 504(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 504(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(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:


0.000 000 000 742 147 504(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2) × 20 =

1.1001 0111 1111 1111 1111 100(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 100


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 1100 =

100 1011 1111 1111 1111 1100


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 1100


Decimal number -0.000 000 000 742 147 504 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1100

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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