-0.000 000 111 110 010 088 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 111 110 010 088 7(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
-0.000 000 111 110 010 088 7(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:

|-0.000 000 111 110 010 088 7| = 0.000 000 111 110 010 088 7


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: 0.000 000 111 110 010 088 7.

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) 0.000 000 111 110 010 088 7 × 2 = 0 + 0.000 000 222 220 020 177 4;
  • 2) 0.000 000 222 220 020 177 4 × 2 = 0 + 0.000 000 444 440 040 354 8;
  • 3) 0.000 000 444 440 040 354 8 × 2 = 0 + 0.000 000 888 880 080 709 6;
  • 4) 0.000 000 888 880 080 709 6 × 2 = 0 + 0.000 001 777 760 161 419 2;
  • 5) 0.000 001 777 760 161 419 2 × 2 = 0 + 0.000 003 555 520 322 838 4;
  • 6) 0.000 003 555 520 322 838 4 × 2 = 0 + 0.000 007 111 040 645 676 8;
  • 7) 0.000 007 111 040 645 676 8 × 2 = 0 + 0.000 014 222 081 291 353 6;
  • 8) 0.000 014 222 081 291 353 6 × 2 = 0 + 0.000 028 444 162 582 707 2;
  • 9) 0.000 028 444 162 582 707 2 × 2 = 0 + 0.000 056 888 325 165 414 4;
  • 10) 0.000 056 888 325 165 414 4 × 2 = 0 + 0.000 113 776 650 330 828 8;
  • 11) 0.000 113 776 650 330 828 8 × 2 = 0 + 0.000 227 553 300 661 657 6;
  • 12) 0.000 227 553 300 661 657 6 × 2 = 0 + 0.000 455 106 601 323 315 2;
  • 13) 0.000 455 106 601 323 315 2 × 2 = 0 + 0.000 910 213 202 646 630 4;
  • 14) 0.000 910 213 202 646 630 4 × 2 = 0 + 0.001 820 426 405 293 260 8;
  • 15) 0.001 820 426 405 293 260 8 × 2 = 0 + 0.003 640 852 810 586 521 6;
  • 16) 0.003 640 852 810 586 521 6 × 2 = 0 + 0.007 281 705 621 173 043 2;
  • 17) 0.007 281 705 621 173 043 2 × 2 = 0 + 0.014 563 411 242 346 086 4;
  • 18) 0.014 563 411 242 346 086 4 × 2 = 0 + 0.029 126 822 484 692 172 8;
  • 19) 0.029 126 822 484 692 172 8 × 2 = 0 + 0.058 253 644 969 384 345 6;
  • 20) 0.058 253 644 969 384 345 6 × 2 = 0 + 0.116 507 289 938 768 691 2;
  • 21) 0.116 507 289 938 768 691 2 × 2 = 0 + 0.233 014 579 877 537 382 4;
  • 22) 0.233 014 579 877 537 382 4 × 2 = 0 + 0.466 029 159 755 074 764 8;
  • 23) 0.466 029 159 755 074 764 8 × 2 = 0 + 0.932 058 319 510 149 529 6;
  • 24) 0.932 058 319 510 149 529 6 × 2 = 1 + 0.864 116 639 020 299 059 2;
  • 25) 0.864 116 639 020 299 059 2 × 2 = 1 + 0.728 233 278 040 598 118 4;
  • 26) 0.728 233 278 040 598 118 4 × 2 = 1 + 0.456 466 556 081 196 236 8;
  • 27) 0.456 466 556 081 196 236 8 × 2 = 0 + 0.912 933 112 162 392 473 6;
  • 28) 0.912 933 112 162 392 473 6 × 2 = 1 + 0.825 866 224 324 784 947 2;
  • 29) 0.825 866 224 324 784 947 2 × 2 = 1 + 0.651 732 448 649 569 894 4;
  • 30) 0.651 732 448 649 569 894 4 × 2 = 1 + 0.303 464 897 299 139 788 8;
  • 31) 0.303 464 897 299 139 788 8 × 2 = 0 + 0.606 929 794 598 279 577 6;
  • 32) 0.606 929 794 598 279 577 6 × 2 = 1 + 0.213 859 589 196 559 155 2;
  • 33) 0.213 859 589 196 559 155 2 × 2 = 0 + 0.427 719 178 393 118 310 4;
  • 34) 0.427 719 178 393 118 310 4 × 2 = 0 + 0.855 438 356 786 236 620 8;
  • 35) 0.855 438 356 786 236 620 8 × 2 = 1 + 0.710 876 713 572 473 241 6;
  • 36) 0.710 876 713 572 473 241 6 × 2 = 1 + 0.421 753 427 144 946 483 2;
  • 37) 0.421 753 427 144 946 483 2 × 2 = 0 + 0.843 506 854 289 892 966 4;
  • 38) 0.843 506 854 289 892 966 4 × 2 = 1 + 0.687 013 708 579 785 932 8;
  • 39) 0.687 013 708 579 785 932 8 × 2 = 1 + 0.374 027 417 159 571 865 6;
  • 40) 0.374 027 417 159 571 865 6 × 2 = 0 + 0.748 054 834 319 143 731 2;
  • 41) 0.748 054 834 319 143 731 2 × 2 = 1 + 0.496 109 668 638 287 462 4;
  • 42) 0.496 109 668 638 287 462 4 × 2 = 0 + 0.992 219 337 276 574 924 8;
  • 43) 0.992 219 337 276 574 924 8 × 2 = 1 + 0.984 438 674 553 149 849 6;
  • 44) 0.984 438 674 553 149 849 6 × 2 = 1 + 0.968 877 349 106 299 699 2;
  • 45) 0.968 877 349 106 299 699 2 × 2 = 1 + 0.937 754 698 212 599 398 4;
  • 46) 0.937 754 698 212 599 398 4 × 2 = 1 + 0.875 509 396 425 198 796 8;
  • 47) 0.875 509 396 425 198 796 8 × 2 = 1 + 0.751 018 792 850 397 593 6;

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:


0.000 000 111 110 010 088 7(10) =

0.0000 0000 0000 0000 0000 0001 1101 1101 0011 0110 1011 111(2)

6. Positive number before normalization:

0.000 000 111 110 010 088 7(10) =

0.0000 0000 0000 0000 0000 0001 1101 1101 0011 0110 1011 111(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 111 110 010 088 7(10) =

0.0000 0000 0000 0000 0000 0001 1101 1101 0011 0110 1011 111(2) =

0.0000 0000 0000 0000 0000 0001 1101 1101 0011 0110 1011 111(2) × 20 =

1.1101 1101 0011 0110 1011 111(2) × 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 representation:

Sign 1 (a negative number)

Exponent (unadjusted): -24

Mantissa (not normalized):
1.1101 1101 0011 0110 1011 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-24 + 2(8-1) - 1 =

(-24 + 127)(10) =

103(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;
  • 103 ÷ 2 = 51 + 1;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 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) =

103(10) =

0110 0111(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. 110 1110 1001 1011 0101 1111 =

110 1110 1001 1011 0101 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 0111

Mantissa (23 bits) =
110 1110 1001 1011 0101 1111


Decimal number -0.000 000 111 110 010 088 7 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0111 - 110 1110 1001 1011 0101 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