3.141 592 653 589 793 238 462 643 383 279 03 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.141 592 653 589 793 238 462 643 383 279 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
3.141 592 653 589 793 238 462 643 383 279 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. First, convert to binary (in base 2) the integer part: 3.
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;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

2. 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.

3(10) =

11(2)


3. Convert to binary (base 2) the fractional part: 0.141 592 653 589 793 238 462 643 383 279 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) 0.141 592 653 589 793 238 462 643 383 279 03 × 2 = 0 + 0.283 185 307 179 586 476 925 286 766 558 06;
  • 2) 0.283 185 307 179 586 476 925 286 766 558 06 × 2 = 0 + 0.566 370 614 359 172 953 850 573 533 116 12;
  • 3) 0.566 370 614 359 172 953 850 573 533 116 12 × 2 = 1 + 0.132 741 228 718 345 907 701 147 066 232 24;
  • 4) 0.132 741 228 718 345 907 701 147 066 232 24 × 2 = 0 + 0.265 482 457 436 691 815 402 294 132 464 48;
  • 5) 0.265 482 457 436 691 815 402 294 132 464 48 × 2 = 0 + 0.530 964 914 873 383 630 804 588 264 928 96;
  • 6) 0.530 964 914 873 383 630 804 588 264 928 96 × 2 = 1 + 0.061 929 829 746 767 261 609 176 529 857 92;
  • 7) 0.061 929 829 746 767 261 609 176 529 857 92 × 2 = 0 + 0.123 859 659 493 534 523 218 353 059 715 84;
  • 8) 0.123 859 659 493 534 523 218 353 059 715 84 × 2 = 0 + 0.247 719 318 987 069 046 436 706 119 431 68;
  • 9) 0.247 719 318 987 069 046 436 706 119 431 68 × 2 = 0 + 0.495 438 637 974 138 092 873 412 238 863 36;
  • 10) 0.495 438 637 974 138 092 873 412 238 863 36 × 2 = 0 + 0.990 877 275 948 276 185 746 824 477 726 72;
  • 11) 0.990 877 275 948 276 185 746 824 477 726 72 × 2 = 1 + 0.981 754 551 896 552 371 493 648 955 453 44;
  • 12) 0.981 754 551 896 552 371 493 648 955 453 44 × 2 = 1 + 0.963 509 103 793 104 742 987 297 910 906 88;
  • 13) 0.963 509 103 793 104 742 987 297 910 906 88 × 2 = 1 + 0.927 018 207 586 209 485 974 595 821 813 76;
  • 14) 0.927 018 207 586 209 485 974 595 821 813 76 × 2 = 1 + 0.854 036 415 172 418 971 949 191 643 627 52;
  • 15) 0.854 036 415 172 418 971 949 191 643 627 52 × 2 = 1 + 0.708 072 830 344 837 943 898 383 287 255 04;
  • 16) 0.708 072 830 344 837 943 898 383 287 255 04 × 2 = 1 + 0.416 145 660 689 675 887 796 766 574 510 08;
  • 17) 0.416 145 660 689 675 887 796 766 574 510 08 × 2 = 0 + 0.832 291 321 379 351 775 593 533 149 020 16;
  • 18) 0.832 291 321 379 351 775 593 533 149 020 16 × 2 = 1 + 0.664 582 642 758 703 551 187 066 298 040 32;
  • 19) 0.664 582 642 758 703 551 187 066 298 040 32 × 2 = 1 + 0.329 165 285 517 407 102 374 132 596 080 64;
  • 20) 0.329 165 285 517 407 102 374 132 596 080 64 × 2 = 0 + 0.658 330 571 034 814 204 748 265 192 161 28;
  • 21) 0.658 330 571 034 814 204 748 265 192 161 28 × 2 = 1 + 0.316 661 142 069 628 409 496 530 384 322 56;
  • 22) 0.316 661 142 069 628 409 496 530 384 322 56 × 2 = 0 + 0.633 322 284 139 256 818 993 060 768 645 12;
  • 23) 0.633 322 284 139 256 818 993 060 768 645 12 × 2 = 1 + 0.266 644 568 278 513 637 986 121 537 290 24;
  • 24) 0.266 644 568 278 513 637 986 121 537 290 24 × 2 = 0 + 0.533 289 136 557 027 275 972 243 074 580 48;

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).


4. 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:


0.141 592 653 589 793 238 462 643 383 279 03(10) =

0.0010 0100 0011 1111 0110 1010(2)

5. Positive number before normalization:

3.141 592 653 589 793 238 462 643 383 279 03(10) =

11.0010 0100 0011 1111 0110 1010(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


3.141 592 653 589 793 238 462 643 383 279 03(10) =

11.0010 0100 0011 1111 0110 1010(2) =

11.0010 0100 0011 1111 0110 1010(2) × 20 =

1.1001 0010 0001 1111 1011 0101 0(2) × 21


7. 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 0 (a positive number)

Exponent (unadjusted): 1

Mantissa (not normalized):
1.1001 0010 0001 1111 1011 0101 0


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

1 + 2(8-1) - 1 =

(1 + 127)(10) =

128(10)


9. 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;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

128(10) =

1000 0000(2)


11. 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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 100 1001 0000 1111 1101 1010 10 =

100 1001 0000 1111 1101 1010


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
1000 0000

Mantissa (23 bits) =
100 1001 0000 1111 1101 1010


Decimal number 3.141 592 653 589 793 238 462 643 383 279 03 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 1000 0000 - 100 1001 0000 1111 1101 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