-0.000 000 000 742 147 388 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 388(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
-0.000 000 000 742 147 388(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:

|-0.000 000 000 742 147 388| = 0.000 000 000 742 147 388


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: 0.000 000 000 742 147 388.

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) 0.000 000 000 742 147 388 × 2 = 0 + 0.000 000 001 484 294 776;
  • 2) 0.000 000 001 484 294 776 × 2 = 0 + 0.000 000 002 968 589 552;
  • 3) 0.000 000 002 968 589 552 × 2 = 0 + 0.000 000 005 937 179 104;
  • 4) 0.000 000 005 937 179 104 × 2 = 0 + 0.000 000 011 874 358 208;
  • 5) 0.000 000 011 874 358 208 × 2 = 0 + 0.000 000 023 748 716 416;
  • 6) 0.000 000 023 748 716 416 × 2 = 0 + 0.000 000 047 497 432 832;
  • 7) 0.000 000 047 497 432 832 × 2 = 0 + 0.000 000 094 994 865 664;
  • 8) 0.000 000 094 994 865 664 × 2 = 0 + 0.000 000 189 989 731 328;
  • 9) 0.000 000 189 989 731 328 × 2 = 0 + 0.000 000 379 979 462 656;
  • 10) 0.000 000 379 979 462 656 × 2 = 0 + 0.000 000 759 958 925 312;
  • 11) 0.000 000 759 958 925 312 × 2 = 0 + 0.000 001 519 917 850 624;
  • 12) 0.000 001 519 917 850 624 × 2 = 0 + 0.000 003 039 835 701 248;
  • 13) 0.000 003 039 835 701 248 × 2 = 0 + 0.000 006 079 671 402 496;
  • 14) 0.000 006 079 671 402 496 × 2 = 0 + 0.000 012 159 342 804 992;
  • 15) 0.000 012 159 342 804 992 × 2 = 0 + 0.000 024 318 685 609 984;
  • 16) 0.000 024 318 685 609 984 × 2 = 0 + 0.000 048 637 371 219 968;
  • 17) 0.000 048 637 371 219 968 × 2 = 0 + 0.000 097 274 742 439 936;
  • 18) 0.000 097 274 742 439 936 × 2 = 0 + 0.000 194 549 484 879 872;
  • 19) 0.000 194 549 484 879 872 × 2 = 0 + 0.000 389 098 969 759 744;
  • 20) 0.000 389 098 969 759 744 × 2 = 0 + 0.000 778 197 939 519 488;
  • 21) 0.000 778 197 939 519 488 × 2 = 0 + 0.001 556 395 879 038 976;
  • 22) 0.001 556 395 879 038 976 × 2 = 0 + 0.003 112 791 758 077 952;
  • 23) 0.003 112 791 758 077 952 × 2 = 0 + 0.006 225 583 516 155 904;
  • 24) 0.006 225 583 516 155 904 × 2 = 0 + 0.012 451 167 032 311 808;
  • 25) 0.012 451 167 032 311 808 × 2 = 0 + 0.024 902 334 064 623 616;
  • 26) 0.024 902 334 064 623 616 × 2 = 0 + 0.049 804 668 129 247 232;
  • 27) 0.049 804 668 129 247 232 × 2 = 0 + 0.099 609 336 258 494 464;
  • 28) 0.099 609 336 258 494 464 × 2 = 0 + 0.199 218 672 516 988 928;
  • 29) 0.199 218 672 516 988 928 × 2 = 0 + 0.398 437 345 033 977 856;
  • 30) 0.398 437 345 033 977 856 × 2 = 0 + 0.796 874 690 067 955 712;
  • 31) 0.796 874 690 067 955 712 × 2 = 1 + 0.593 749 380 135 911 424;
  • 32) 0.593 749 380 135 911 424 × 2 = 1 + 0.187 498 760 271 822 848;
  • 33) 0.187 498 760 271 822 848 × 2 = 0 + 0.374 997 520 543 645 696;
  • 34) 0.374 997 520 543 645 696 × 2 = 0 + 0.749 995 041 087 291 392;
  • 35) 0.749 995 041 087 291 392 × 2 = 1 + 0.499 990 082 174 582 784;
  • 36) 0.499 990 082 174 582 784 × 2 = 0 + 0.999 980 164 349 165 568;
  • 37) 0.999 980 164 349 165 568 × 2 = 1 + 0.999 960 328 698 331 136;
  • 38) 0.999 960 328 698 331 136 × 2 = 1 + 0.999 920 657 396 662 272;
  • 39) 0.999 920 657 396 662 272 × 2 = 1 + 0.999 841 314 793 324 544;
  • 40) 0.999 841 314 793 324 544 × 2 = 1 + 0.999 682 629 586 649 088;
  • 41) 0.999 682 629 586 649 088 × 2 = 1 + 0.999 365 259 173 298 176;
  • 42) 0.999 365 259 173 298 176 × 2 = 1 + 0.998 730 518 346 596 352;
  • 43) 0.998 730 518 346 596 352 × 2 = 1 + 0.997 461 036 693 192 704;
  • 44) 0.997 461 036 693 192 704 × 2 = 1 + 0.994 922 073 386 385 408;
  • 45) 0.994 922 073 386 385 408 × 2 = 1 + 0.989 844 146 772 770 816;
  • 46) 0.989 844 146 772 770 816 × 2 = 1 + 0.979 688 293 545 541 632;
  • 47) 0.979 688 293 545 541 632 × 2 = 1 + 0.959 376 587 091 083 264;
  • 48) 0.959 376 587 091 083 264 × 2 = 1 + 0.918 753 174 182 166 528;
  • 49) 0.918 753 174 182 166 528 × 2 = 1 + 0.837 506 348 364 333 056;
  • 50) 0.837 506 348 364 333 056 × 2 = 1 + 0.675 012 696 728 666 112;
  • 51) 0.675 012 696 728 666 112 × 2 = 1 + 0.350 025 393 457 332 224;
  • 52) 0.350 025 393 457 332 224 × 2 = 0 + 0.700 050 786 914 664 448;
  • 53) 0.700 050 786 914 664 448 × 2 = 1 + 0.400 101 573 829 328 896;
  • 54) 0.400 101 573 829 328 896 × 2 = 0 + 0.800 203 147 658 657 792;

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:


0.000 000 000 742 147 388(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 10(2)

6. Positive number before normalization:

0.000 000 000 742 147 388(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 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:


0.000 000 000 742 147 388(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 10(2) × 20 =

1.1001 0111 1111 1111 1111 010(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 010


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 1010 =

100 1011 1111 1111 1111 1010


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 1010


Decimal number -0.000 000 000 742 147 388 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1010

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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