-0.000 000 000 742 147 676 03 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 03(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 03(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 03| = 0.000 000 000 742 147 676 03


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 03.

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 03 × 2 = 0 + 0.000 000 001 484 295 352 06;
  • 2) 0.000 000 001 484 295 352 06 × 2 = 0 + 0.000 000 002 968 590 704 12;
  • 3) 0.000 000 002 968 590 704 12 × 2 = 0 + 0.000 000 005 937 181 408 24;
  • 4) 0.000 000 005 937 181 408 24 × 2 = 0 + 0.000 000 011 874 362 816 48;
  • 5) 0.000 000 011 874 362 816 48 × 2 = 0 + 0.000 000 023 748 725 632 96;
  • 6) 0.000 000 023 748 725 632 96 × 2 = 0 + 0.000 000 047 497 451 265 92;
  • 7) 0.000 000 047 497 451 265 92 × 2 = 0 + 0.000 000 094 994 902 531 84;
  • 8) 0.000 000 094 994 902 531 84 × 2 = 0 + 0.000 000 189 989 805 063 68;
  • 9) 0.000 000 189 989 805 063 68 × 2 = 0 + 0.000 000 379 979 610 127 36;
  • 10) 0.000 000 379 979 610 127 36 × 2 = 0 + 0.000 000 759 959 220 254 72;
  • 11) 0.000 000 759 959 220 254 72 × 2 = 0 + 0.000 001 519 918 440 509 44;
  • 12) 0.000 001 519 918 440 509 44 × 2 = 0 + 0.000 003 039 836 881 018 88;
  • 13) 0.000 003 039 836 881 018 88 × 2 = 0 + 0.000 006 079 673 762 037 76;
  • 14) 0.000 006 079 673 762 037 76 × 2 = 0 + 0.000 012 159 347 524 075 52;
  • 15) 0.000 012 159 347 524 075 52 × 2 = 0 + 0.000 024 318 695 048 151 04;
  • 16) 0.000 024 318 695 048 151 04 × 2 = 0 + 0.000 048 637 390 096 302 08;
  • 17) 0.000 048 637 390 096 302 08 × 2 = 0 + 0.000 097 274 780 192 604 16;
  • 18) 0.000 097 274 780 192 604 16 × 2 = 0 + 0.000 194 549 560 385 208 32;
  • 19) 0.000 194 549 560 385 208 32 × 2 = 0 + 0.000 389 099 120 770 416 64;
  • 20) 0.000 389 099 120 770 416 64 × 2 = 0 + 0.000 778 198 241 540 833 28;
  • 21) 0.000 778 198 241 540 833 28 × 2 = 0 + 0.001 556 396 483 081 666 56;
  • 22) 0.001 556 396 483 081 666 56 × 2 = 0 + 0.003 112 792 966 163 333 12;
  • 23) 0.003 112 792 966 163 333 12 × 2 = 0 + 0.006 225 585 932 326 666 24;
  • 24) 0.006 225 585 932 326 666 24 × 2 = 0 + 0.012 451 171 864 653 332 48;
  • 25) 0.012 451 171 864 653 332 48 × 2 = 0 + 0.024 902 343 729 306 664 96;
  • 26) 0.024 902 343 729 306 664 96 × 2 = 0 + 0.049 804 687 458 613 329 92;
  • 27) 0.049 804 687 458 613 329 92 × 2 = 0 + 0.099 609 374 917 226 659 84;
  • 28) 0.099 609 374 917 226 659 84 × 2 = 0 + 0.199 218 749 834 453 319 68;
  • 29) 0.199 218 749 834 453 319 68 × 2 = 0 + 0.398 437 499 668 906 639 36;
  • 30) 0.398 437 499 668 906 639 36 × 2 = 0 + 0.796 874 999 337 813 278 72;
  • 31) 0.796 874 999 337 813 278 72 × 2 = 1 + 0.593 749 998 675 626 557 44;
  • 32) 0.593 749 998 675 626 557 44 × 2 = 1 + 0.187 499 997 351 253 114 88;
  • 33) 0.187 499 997 351 253 114 88 × 2 = 0 + 0.374 999 994 702 506 229 76;
  • 34) 0.374 999 994 702 506 229 76 × 2 = 0 + 0.749 999 989 405 012 459 52;
  • 35) 0.749 999 989 405 012 459 52 × 2 = 1 + 0.499 999 978 810 024 919 04;
  • 36) 0.499 999 978 810 024 919 04 × 2 = 0 + 0.999 999 957 620 049 838 08;
  • 37) 0.999 999 957 620 049 838 08 × 2 = 1 + 0.999 999 915 240 099 676 16;
  • 38) 0.999 999 915 240 099 676 16 × 2 = 1 + 0.999 999 830 480 199 352 32;
  • 39) 0.999 999 830 480 199 352 32 × 2 = 1 + 0.999 999 660 960 398 704 64;
  • 40) 0.999 999 660 960 398 704 64 × 2 = 1 + 0.999 999 321 920 797 409 28;
  • 41) 0.999 999 321 920 797 409 28 × 2 = 1 + 0.999 998 643 841 594 818 56;
  • 42) 0.999 998 643 841 594 818 56 × 2 = 1 + 0.999 997 287 683 189 637 12;
  • 43) 0.999 997 287 683 189 637 12 × 2 = 1 + 0.999 994 575 366 379 274 24;
  • 44) 0.999 994 575 366 379 274 24 × 2 = 1 + 0.999 989 150 732 758 548 48;
  • 45) 0.999 989 150 732 758 548 48 × 2 = 1 + 0.999 978 301 465 517 096 96;
  • 46) 0.999 978 301 465 517 096 96 × 2 = 1 + 0.999 956 602 931 034 193 92;
  • 47) 0.999 956 602 931 034 193 92 × 2 = 1 + 0.999 913 205 862 068 387 84;
  • 48) 0.999 913 205 862 068 387 84 × 2 = 1 + 0.999 826 411 724 136 775 68;
  • 49) 0.999 826 411 724 136 775 68 × 2 = 1 + 0.999 652 823 448 273 551 36;
  • 50) 0.999 652 823 448 273 551 36 × 2 = 1 + 0.999 305 646 896 547 102 72;
  • 51) 0.999 305 646 896 547 102 72 × 2 = 1 + 0.998 611 293 793 094 205 44;
  • 52) 0.998 611 293 793 094 205 44 × 2 = 1 + 0.997 222 587 586 188 410 88;
  • 53) 0.997 222 587 586 188 410 88 × 2 = 1 + 0.994 445 175 172 376 821 76;
  • 54) 0.994 445 175 172 376 821 76 × 2 = 1 + 0.988 890 350 344 753 643 52;

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 03(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 03(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 03(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 03 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