-0.000 000 000 742 147 676 619 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 619 8(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
-0.000 000 000 742 147 676 619 8(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:

|-0.000 000 000 742 147 676 619 8| = 0.000 000 000 742 147 676 619 8


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: 0.000 000 000 742 147 676 619 8.

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) 0.000 000 000 742 147 676 619 8 × 2 = 0 + 0.000 000 001 484 295 353 239 6;
  • 2) 0.000 000 001 484 295 353 239 6 × 2 = 0 + 0.000 000 002 968 590 706 479 2;
  • 3) 0.000 000 002 968 590 706 479 2 × 2 = 0 + 0.000 000 005 937 181 412 958 4;
  • 4) 0.000 000 005 937 181 412 958 4 × 2 = 0 + 0.000 000 011 874 362 825 916 8;
  • 5) 0.000 000 011 874 362 825 916 8 × 2 = 0 + 0.000 000 023 748 725 651 833 6;
  • 6) 0.000 000 023 748 725 651 833 6 × 2 = 0 + 0.000 000 047 497 451 303 667 2;
  • 7) 0.000 000 047 497 451 303 667 2 × 2 = 0 + 0.000 000 094 994 902 607 334 4;
  • 8) 0.000 000 094 994 902 607 334 4 × 2 = 0 + 0.000 000 189 989 805 214 668 8;
  • 9) 0.000 000 189 989 805 214 668 8 × 2 = 0 + 0.000 000 379 979 610 429 337 6;
  • 10) 0.000 000 379 979 610 429 337 6 × 2 = 0 + 0.000 000 759 959 220 858 675 2;
  • 11) 0.000 000 759 959 220 858 675 2 × 2 = 0 + 0.000 001 519 918 441 717 350 4;
  • 12) 0.000 001 519 918 441 717 350 4 × 2 = 0 + 0.000 003 039 836 883 434 700 8;
  • 13) 0.000 003 039 836 883 434 700 8 × 2 = 0 + 0.000 006 079 673 766 869 401 6;
  • 14) 0.000 006 079 673 766 869 401 6 × 2 = 0 + 0.000 012 159 347 533 738 803 2;
  • 15) 0.000 012 159 347 533 738 803 2 × 2 = 0 + 0.000 024 318 695 067 477 606 4;
  • 16) 0.000 024 318 695 067 477 606 4 × 2 = 0 + 0.000 048 637 390 134 955 212 8;
  • 17) 0.000 048 637 390 134 955 212 8 × 2 = 0 + 0.000 097 274 780 269 910 425 6;
  • 18) 0.000 097 274 780 269 910 425 6 × 2 = 0 + 0.000 194 549 560 539 820 851 2;
  • 19) 0.000 194 549 560 539 820 851 2 × 2 = 0 + 0.000 389 099 121 079 641 702 4;
  • 20) 0.000 389 099 121 079 641 702 4 × 2 = 0 + 0.000 778 198 242 159 283 404 8;
  • 21) 0.000 778 198 242 159 283 404 8 × 2 = 0 + 0.001 556 396 484 318 566 809 6;
  • 22) 0.001 556 396 484 318 566 809 6 × 2 = 0 + 0.003 112 792 968 637 133 619 2;
  • 23) 0.003 112 792 968 637 133 619 2 × 2 = 0 + 0.006 225 585 937 274 267 238 4;
  • 24) 0.006 225 585 937 274 267 238 4 × 2 = 0 + 0.012 451 171 874 548 534 476 8;
  • 25) 0.012 451 171 874 548 534 476 8 × 2 = 0 + 0.024 902 343 749 097 068 953 6;
  • 26) 0.024 902 343 749 097 068 953 6 × 2 = 0 + 0.049 804 687 498 194 137 907 2;
  • 27) 0.049 804 687 498 194 137 907 2 × 2 = 0 + 0.099 609 374 996 388 275 814 4;
  • 28) 0.099 609 374 996 388 275 814 4 × 2 = 0 + 0.199 218 749 992 776 551 628 8;
  • 29) 0.199 218 749 992 776 551 628 8 × 2 = 0 + 0.398 437 499 985 553 103 257 6;
  • 30) 0.398 437 499 985 553 103 257 6 × 2 = 0 + 0.796 874 999 971 106 206 515 2;
  • 31) 0.796 874 999 971 106 206 515 2 × 2 = 1 + 0.593 749 999 942 212 413 030 4;
  • 32) 0.593 749 999 942 212 413 030 4 × 2 = 1 + 0.187 499 999 884 424 826 060 8;
  • 33) 0.187 499 999 884 424 826 060 8 × 2 = 0 + 0.374 999 999 768 849 652 121 6;
  • 34) 0.374 999 999 768 849 652 121 6 × 2 = 0 + 0.749 999 999 537 699 304 243 2;
  • 35) 0.749 999 999 537 699 304 243 2 × 2 = 1 + 0.499 999 999 075 398 608 486 4;
  • 36) 0.499 999 999 075 398 608 486 4 × 2 = 0 + 0.999 999 998 150 797 216 972 8;
  • 37) 0.999 999 998 150 797 216 972 8 × 2 = 1 + 0.999 999 996 301 594 433 945 6;
  • 38) 0.999 999 996 301 594 433 945 6 × 2 = 1 + 0.999 999 992 603 188 867 891 2;
  • 39) 0.999 999 992 603 188 867 891 2 × 2 = 1 + 0.999 999 985 206 377 735 782 4;
  • 40) 0.999 999 985 206 377 735 782 4 × 2 = 1 + 0.999 999 970 412 755 471 564 8;
  • 41) 0.999 999 970 412 755 471 564 8 × 2 = 1 + 0.999 999 940 825 510 943 129 6;
  • 42) 0.999 999 940 825 510 943 129 6 × 2 = 1 + 0.999 999 881 651 021 886 259 2;
  • 43) 0.999 999 881 651 021 886 259 2 × 2 = 1 + 0.999 999 763 302 043 772 518 4;
  • 44) 0.999 999 763 302 043 772 518 4 × 2 = 1 + 0.999 999 526 604 087 545 036 8;
  • 45) 0.999 999 526 604 087 545 036 8 × 2 = 1 + 0.999 999 053 208 175 090 073 6;
  • 46) 0.999 999 053 208 175 090 073 6 × 2 = 1 + 0.999 998 106 416 350 180 147 2;
  • 47) 0.999 998 106 416 350 180 147 2 × 2 = 1 + 0.999 996 212 832 700 360 294 4;
  • 48) 0.999 996 212 832 700 360 294 4 × 2 = 1 + 0.999 992 425 665 400 720 588 8;
  • 49) 0.999 992 425 665 400 720 588 8 × 2 = 1 + 0.999 984 851 330 801 441 177 6;
  • 50) 0.999 984 851 330 801 441 177 6 × 2 = 1 + 0.999 969 702 661 602 882 355 2;
  • 51) 0.999 969 702 661 602 882 355 2 × 2 = 1 + 0.999 939 405 323 205 764 710 4;
  • 52) 0.999 939 405 323 205 764 710 4 × 2 = 1 + 0.999 878 810 646 411 529 420 8;
  • 53) 0.999 878 810 646 411 529 420 8 × 2 = 1 + 0.999 757 621 292 823 058 841 6;
  • 54) 0.999 757 621 292 823 058 841 6 × 2 = 1 + 0.999 515 242 585 646 117 683 2;

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:


0.000 000 000 742 147 676 619 8(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 619 8(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:


0.000 000 000 742 147 676 619 8(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 -0.000 000 000 742 147 676 619 8 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